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CONTENTS TOPIC 1
ANALYTIC GEOMETRY
Outcomes
1 1
Section 1.1
Graphs
Section 1.2
Cartesian Coordinates and Graphs in the Cartesian Plane
14
Formulas we often use
34
Section 1.3
3
Topic Summary
56
Checklist
57
TOPIC 2
RELATIONS AND FUNCTIONS
Outcomes
59 59
Section 2.1
Relations and functions in R
Section 2.2
Combining Functions
×R
60 88
Topic Summary
96
Checklist
98
TOPIC 3
STRAIGHT LINE GRAPHS
Outcomes
99 99
Section 3.1
Drawing Lines
101
Section 3.2
Finding Equations of Lines
130
Section 3.3
Applications of Lines and Linear Functions
142
Topic Summary
174
Checklist
178
iv
TOPIC 4
PARABOLAS
Outcomes Section 4.1
180 180
Characteristics of Parabolas defined by y = a (x h)2 + k
−
182
Section 4.2
Sketching Parabolas
207
Section 4.3
Finding the Equation of a Parabola
224
Section 4.4
Using Parabolas and Quadratic Functions
233
Topic Summary
243
Checklist
246
TOPIC 5
HYPERBOLAS
248 248
Outcomes Section 5.1
Characteristics of Hyperbolas
249
Section 5.2
Inverse Proportion
268
Topic Summary
273
Checklist
275
TOPIC 6
COMBINATIONS OF GRAPHS
Outcomes Section 6.1
277 277
Graphs, Graphs and More Graphs
277
Topic Summary
296
Checklist
297
v
TOPIC 7
STATISTICS
Outcomes
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Section 7.1
Data Collection and Organisation
299
Section 7.2 Section 7.3
Using Graphs to Represent Data Some Statistical Measurements
311
Section 7.4
Probability
347
335
Topic Summary
361
Checklist
368
ANSWERS
370
REFERENCES
401
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ANALYTIC GEOMETRY OUTCOMES After studying this topic you should be able to do the following. SECTION 1.1: Graphs
Choose an appropriate scale (if an accurate graph is required) or mark vertical and horizontal lines in a suitable way (if an illustration of information is required) and plot given data, taking into account whether the graph consists of separate dots, or whether the dots can be joined in some way.
Interpret information conveyed by a given graph.
SECTION 1.2: Cartesian Coordinates and Graphs in the Cartesian Plane
Represent solutions of equations or inequalities in one variable by means of a one–dimensional graph (i.e. on a number line).
Find coordinates of a given point in the Cartesian plane, or, if the coordinates are known, locate the point.
Use a table of values to draw a graph.
Use an equation to set up a table of values and hence draw a graph.
If a graph represents an equation
find the x – and y –intercepts of the graph determine whether or not a given point lies on the graph, i.e. whether or not the coordinates of the point satisfy the equation of the graph
find values of x for which y > 0, y = 0 (i.e. the x –intercept) or y < 0
recognise from the quadrant in which a point lies whether x > 0 or x < 0 and whether y > 0 or y < 0.
2 SECTION 1.3: Formulas we often use
Use the Theorem of Pythagoras.
Use the distance formula.
Use the midpoint formula.
Find the standard equation of a circle with radius r and centre at ( 0, 0) .
Use the equation of a circle to find out whether a point lies on the circle or not.
Find the standard equation of a circle with radius r and centre at (h, k) = (0, 0).
Given the equation of a circle, find the centre and radius of the circle.
Given the coordinates of the end points of the diameter of a circle, find the equation of the circle.
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1.1 GRAPHS
1.1A WHAT IS A GRAPH? PICTORIAL REPRESENTATION OF INFORMATION
What do you think of when you hear the word “graph”? Most of us think of some kind of picture that represents certain information. As an example, look at the graphs shown in Figures 1.1.1(a) and (b). They illustrate reading patterns of shoppers in a shopping centre (Figure 1.1.1(a), taken from F & T Weekly, 11 September 1998), and education levels in South Africa in 1995 (Figure 1.1.1(b), taken from the South Africa Survey, 1996 - 1997). In magazines and newspapers graphs are often included because they look attractive, but the information they convey may not always be immediately clear.
Figure 1.1.1(a)
4 Education levels of people over the age of 20:1995 No education 13%
Degrees 3% Diplomas 7%
Gr 1 – Std 5 24 % Matric 19%
Std 6 – Std 9 34%
Figure 1.1.1(b)
Graphs such as these can only be drawn after a substantial amount of data has Sometimes scatterdiagrams graphs are called scatter and pie graphs are called pie charts.
been representations dataexample result inFigure scatter 1.1.1(a)). graphs, pieWe graphs (suchcollected. as FigureCertain 1.1.1(b)), or bar graphsof(for deal with these three types of graphs in the last topic of this book (Topic 7: Statistics). Graphs can help us absorb at a glance much more information than we could grasp if we were given the same informa tion in written form. Consider the following hypothetical cricket match. Suppose Paul is bowling. Michael makes 56 runs before he is caught by the wicket keeper. Figure 1.1.2 shows the different paths of the ball during the first over (six balls bowled). Obviously, the more overs there are, the more difficult it is to show the different paths on the same sketch.
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Figure 1.1.2
6 It does not matter if you are
Suppose we have the following information.
not a cricket player. You will be able to follow these ideas
Approximate path followed by the ball Towards third man
even if you do not understand cricket.
Number of times the ball travels that path 5
Towards leg leg Towardsfine square Towards mid–wicket Towardsmid–on Towardsmid–off Towardscover Towardspoint Towardsgully Towards wicket keeper
44 2 10 9 12 7 2 1
Table 1.1.1
Table 1.1.1 is called a frequency table because it shows the frequency of a particular event (i.e. the number of times the event takes place). We can represent the information in the frequency table differently, means of a graph, using dots to link corresponding items of information, asby in Figure 1.1.3.
12 Number of times ball travels a given path
10 8 6 4 2
third man
fine leg
square leg
midwicket
midon
midoff
cover
Approximate path taken by the ball
Figure 1.1.3
point
gully
wicket keeper
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In the graph we see that
SCALE
the graph is a record of only one particular cricket match and we cannot generalise the information to future matches
there is no relationship between the different dots
it would make no sense to join the dots
the distances between the marks on the horizontal line that represent the different paths of the ball are arbitrarily chosen, and it makes no difference how we space the marks (it looks better to have the spaces between the marks the same, but it makes no difference to our understanding if they are different)
the marks that represent the number of times the ball travels along the different paths are carefully chosen so that the distances between them are the same (the distance between 5 and 10 should be the same as the distance between 10 and 15).
This leads to the idea of a scale. A scale makes use of small units to denote bigger units. On a graph we use a system of marks to indicate given intervals. For example, we may consult a map where 1 centimetre represents 100 kilometres. This means that if the distance between two towns on a map is 2,3 centimetres, then the actual physical distance between the two towns is (2,3 100) kilometres, i.e. 230 kilometres.
×
In any scale drawing the lengths of successive line segments must be the same. For example, km 0
10 km
A
We define lines and line segments in Book 4.
B
20 km km 30
C
D
is an acceptable representation of distance, since the length of line segment AB is equal to the lengths of the segments BC and CD , and they all represent the same distance. However, km 0 A
10 km B
20 km
30 km C
D
is not acceptable, because line segments AB and BC both represent 10 km, but the length of AB = the length of BC.
When we draw graphs to represent certain information we need to decide what type of graph to draw. For example, should we use a pie graph, bar graph or some other type? If we draw graphs such as in Figure 1.1.3, we need to
8
decide what information should be represented along the horizontal line, and along the vertical line
decide how accurate the graph needs to be; if we need an accurate graph (possibly for the purpose of using it to predict a value), we may need to use squared paper (called graph paper)
choose a suitable scale for each set of data we need to represent.
Let us now consider a graphical representation of two different problems. In Example 1.1.1 we consider distance and time, with speed constant. In Example 1.1.2 we consider numbers of people performing a task, and the time they take to complete it, all working at the same constant rate.
1.1.1 Suppose Mishak travels by car at an average speed of 80 km/h. Determine the distance he has travelled after 1, 2, 3, 4, 5 and 6 hours. Show this information on a graph. Can we predict a pattern?
SOLUTION Time(inhours) 1 2 3 4 5 Distance (in kilometres) 80 160 240 320 Table 1.1.2
500 400 Distance (in km)
300 200 100
0
123456 Time in hours
Figure 1.1.4
6 400
480
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There are several facts to note about this graph.
For time we have chosen a scale of 1 cm represents 1 hour, and time is represented along the horizontal line.
For distance we have chosen a scale of 1 cm represents 100 km, and distance is represented along the vertical line.
There seems to be a straight line relations hip between time and distanc e, and we might predict that if additional points were to be included they would follow the same pattern.
It is logical to put a dot at zero, since it represents the fact that “at the beginning (i.e. when no time has yet passed) Mishak’s car has travelled no distance”.
We may join the dots by means of a straight line if we assume that Mishak travels at a constant speed and does not stop.
We have already learnt that distance = speed
You may want to revise direct proportion. See Book 1, Topic 3.
× time
and we know that if d = st , with s constant, then d and t are in direct proportion to each other. Hence we see a possible link betwee n straight lines and direct proportion. We will investigate this further in Topic 3.
1.1.2 Suppose Moses has a few hundred notices to fold and seal in enve lopes. If he works alone it will take him 8 hours to complete the task. He has several friends who offer to help him. If each of his friends works at the same rate as he does, the task will be completed by
him and one friend (i.e. 2 people) in
8 2
hours, i.e. in 4 hours 8 3
him and two friends (i.e. 3 people) in
him and three friends (i.e. 4 people) in
hours, i.e. in 2 23 hours 8 4
We now represent this information on a graph.
hours, i.e. in 2 hours.
10
Time (in hours) to complete the task
9 8 7 6 5 4 3 2 1
X
1
2
3
4
Number of people carrying out the task
Figure 1.1.5
There are again several facts to note about this graph.
The number of people is represented along the horizontal line, and time is
We have chosen a scale of 1,5 cm represents 1 person on the people line and 1 cm represents 2 hours on the time line.
The graph clearly illustrates that the more people who help, the less time the task will take.
It makes no sense to include a dot a zero. If no people fold notices and seal them in envelopes, the task will never be undertaken.
It makes no sense to join the dots, because a dot at the point x, for example, would indicate a time taken by 2 12 people.
The shape of the graph suggests that we could include more dots, in positions corresponding to the natural numbe rs 5, 6, 7, etc., on the horizontal line. However, we realise that in practice we cannot carry on indefinitely including more people.
represented along the vertical line.
Example 1.1.2 illustrates a situation such as “the more ... the less ...”, and we can represent this by means of the formula In Example 1.1.2 we have
c = 8.
t=
c ; c a constant, n
where t represents time (in this case measured in hours) and n represents the number of people. This formula represents indirect proportion.
11 You may revise indirect proportion (also called inverse proportion) in Topic 3 of Book 1.
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The graph of the equation t = nc that describes indirect proportion is not a straight line. The graph of this equa tion is an example of a hyperbola, and we study graphs such as these later, in Topic 5. The examples in this study unit illustrate that graphs often arise out of the need to have a “picture” to represent specific data. Much data can be quantified, and
When we “quantify” something we numerical take measurements and give information.
graphs then illustrate numerical relationships between the variables involved. At times there are only two variables involved, such as the relationship between distance and time (where spee d is constant). We may also work with three or more variables, but we do not do so in this module. In the graphs we have been discussing we have used rather vague language, such as – the line representing the number of people – dots corresponding to items of information – joining dots but in order to use graphs properly we need to be more specific. We need to step back from real–life examples and look at the basic characteristics of graphs in general. You already know (from the geometrical representations of numbers and intervals that we dealt with in Topic 1 of Book 1) that a statement about a single variable can be represented by a point , or an interval , or the union of intervals, on the real number line . In the next section we extend this idea to the representation of statements involving two variables by a subset of the Cartesian plane .
1.1 1. Month Jan Average daily 21 ◦ C temperature
Feb 24◦ C
(a) Use the given frequency table, and ske tch (not on grap h paper) a graph to represent the information.
Mar 19◦ C
Apr 16◦ C
May 12 ◦ C
June
July A ug S ep
10◦ C
7◦ C
5◦ C
Oct 8◦ C
Nov
15◦ C
Dec
19◦ C
23◦ C
(b) Does the information represent a place in the northern or southern hemisphere?
12 2. A family travels by car from Polokw ane to Pretoria. They leave Polokwane at 09:00. They stop twice, the first time to change a flat tyre, and the second time just before Pretoria for petrol and cooldrinks. The journey is illustrated by the graph in Figure 1.1.6.
300
Distance (in km)
200
100
0
1234 Time (in hours) Polokwane − Pretoria trip
Figure 1.1.6
From the graph, try to answer the following questions. (a) How far had they travelled by 10:00? (b) How long did they spend changing the tyre? (c) How far is Polokwane from Pretoria? (d) When did they reach the half–way stage between Polokwane and Pretoria? (e) How long did they take to fill up with petrol and have som ething to drink? (f) What time did they reach Pretoria?
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3. The graph in Figure 1.1.7 represents the distance a car travels in 6 seconds, from a stationary position.
90 80 70 60 50 Distance travelled (in metres)
40 30 20 10
5
1234
6
7
Time in seconds
Figure 1.1.7
Use the graph to answer the following questions. (a) What does the dot at zero rep resent? (b) How far has the car travelled in the first two seconds? (c) What distance has the car travelled between the 2 second and 4 second marks? (d) Has the car trave lled faster or slower during the phas e from 2 to 4 seconds, than during the phase from 0 to 2 seconds? (e) After how many seconds does the car begin to travel at a constant speed?
14
1.2 CARTESIAN COORDINATES AND GRAPHS IN THE CARTESIAN PLANE
1.2A THE CARTESIAN PLANE One–dimensional graphs
We discuss these in Topic 2 of Book 2.
We introduced the idea of a number line in Book 1. Every real number can be represented by a point on a number line and every point on a number line represents a real number. We also use a part or parts of a number line to represent the solution set of an equation or inequality in one variable. We can think of this as a one–dimensional graph. The graph of an equation or inequality in one variable is thus the set of points on a number line representing all real numbers which satisfy that equation or inequality. For example, when we solve an equation in one variable, such as
∈R
2x + 3 = 15, x the solution is only one number, in this case
x = 6. This solution can be represented on a number line.
0
6
↑
The only point on the real number line which represents the solution of the equation. The graph of the equation 2 x + 3 = 15 Figure 1.2.1
If we solve an inequality in one variable, such as
x+3
≤ 7, x ∈ R
the solution is an interval in R, and in this case we obtain
x
≤ 4, x ∈ R .
The solution can also be represented graphically, on a number line as in Figure 1.2.2.
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All these points represent solutions of the inequality.
4
The graph of the inequality x + 3 Figure 1.2.2
Two–dimensional graphs
7
≤
Since we often work with equations or inequalities in two variables we are interested in extending this notion of a graph to graphs in two dimensions. In order to do this we must first define the Cartesian (or rectangular) coordinate plane.
What does the word “coordinate” mean? The word is pronounced (and sometimes written) as “co–ordinate”. From the Concise Oxford Dictionary we have the idea that this is a mathematical word, relating to “a system of magnitudes used to fix position ...”. Think of a map of Pretoria. One such map locates the Unisa Pretoria campus by means of the symbols 23D4. If we do not understand this notation we will not be able to use the map. In 23D4, 23 repres ents the map (page) number and D and 4 tell us that we will find the place in the block where column D and row 4 cross each other. We could say that D and 4 are the coordinates of Unisa on this map. Since the columns are marked differently from the rows it does not matter whether we use the symbol D4 or 4D to locate the place we have identified.
Remember: we also use the word “width” for “breadth”.
What about the word “plane”? The easiest way to describe the meaning of the word is to use its characteristics: it is an unbounded flat surface . In everyday life we see examples of bounded surfaces; they have two dimensions only, namely length and breadth. Mathematically a plane has no thickness, but the objects we suggest as examples of planes obviously have some thickness. A table top, a piece of writing paper or the floor have certain common characteristics: they are bounded flat surfaces, and we can measure their length and breadth. As we have already seen, a straight line has one dimension only, and a box has three dimensions (height as well as length and breadth).
We often make use of the idea of coordinates when we give directions. Suppose you live in a city where the roads and blocks form a rectangular grid. Look at Figure 1.2.3. Suppose you live at A , and you want to explain to a friend living at B how to reach you.
16
Figure 1.2.3
The clearest instructions you could give might be the following.
Travel east until you have crossed 4 blocks. From there travel north until you have crossed 2 blocks.
If you wanted to give your friend a note for future reference, you could possibly write down something like this.
North
2
A
1 B
East 0
1
2
3
4
Figure 1.2.4
This is a very rough sketch in a two–dimensional plane that represents a given situation. In this sketch the numbers represent the ends of complete blocks, i.e. 1 indicates that we have crossed 1 whole block, 2 indicates that we have crossed 2 blocks, and so on. We also need to be sure that we specify the starting and end points. In Figure 1.2.4 we begin at B , and end at A . The important numbers are 2 and 4, and you can see that it makes a big difference to your directions whether you say “4 east, 2 north” or “2 east, 4 north”. The idea of order is also important when we consider equations or inequalities in two variables, and the graphical representation of their solutions.
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Consider the equation
y = 2x + 3, x
∈Z
and consider some possible values of the independent variable ( x) and corresponding values of the dependent variable ( y).
Possible value of x 3 2 1 0 1
− − −
Corresponding value of y 3 1 1 3 5
− −
Solution of the equation
x = 3 and x = 2 and x = 1 and and x =0 x =1 and
− − −
y= 3 y= 1 y=1 y=3 y=5
− −
Table 1.2.1
ORDERED PAIRS
In mathematical notation we make use of ordered pairs to represent the information given in the last column of Table 1.2.1. Instead of writing “ x = 2 and y = 1”, we can write this solution as follows.
−
−
( 2, 1)
−
− − ↑ ↑
possible x value (independent variable)
−
corresponding y value (dependent variable)
Order is important in this notation, because for example x = 1 and y = 2 is not a solution (when x = 1 we obtain y = 1 and not y = 2). Hence we recognise that there is a difference between ( 2, 1) and ( 1, 2).
−
CARTESIAN NATE
COORDI-
SYSTEM
We call this a rectangular system because the horizontal and vertical axes intersect each other at right angles. Note that it is also possible to set up non–rectangular coordinate systems, but we will not consider other systems in this module.
− −
− − − −
−
In the same way that we represent a single variable such as x on the real number line, we are also able to use two perpendicularly intersecting number lines to create a plane in which we can represent ordered number pairs of the form ( x, y). These perpendicular lines form a rectangular coordinate system. We call this the Cartesian coordinate system , and say that the plane that consists of points which represent all possible pairs of ordered real numbers is the Cartesian plane. The system was devised by Ren´e Descartes (1596–1650), who was a French mathematician, philosopher and scientist. In the Cartesian system we make use of specific terminology. Have a look at Figures 1.2.5, 1.2.6 and 1.2.7.
18 y
The vertical real number line is usually called the y _ axis and labelled y . Each point in the plane is identified by an ordered number pair. The point P corresponds to – 2 on the x _ axis P ( – 2,4) and 4 on the y _ axis.
4 is the y _ coordinate of the point P.
4 3 2
Lengths of intervals on the x _ axis should be the same. –5
–4
–3
–2
1 –1
– 2 is the x _ coordinate of the point P.
1
3
x
0 –1 –2
Lengths of intervals on the y _ axis should be the same.
2
The horizontal real number line is usually called the x _ axis and labelled x.
–3 –4
The zero on both lines needs to be labelled only once. The point where the x _ axis and y _ axis intersect represents the ordered number pair (0,0). It is called the srcin. We do not always label this point.
–5 –6
Figure 1.2.5
In Figure 1.2.5 we note the following.
Origin
The word srcin is used to convey the sense of a starting point, i.e. a point from which the counting process srcinates, or begins. The srcin represents ( 0, 0).
Graph
The point P can be called the graph of ( 2, 4).
Coordinates
( 2, 4) are the coordinates of P .
Coordinate axes (Note that the plural of the word axis is axes.)
−
− −2 is called the x–coordinate of P; 4 is called the y–coordinate of P. We apply the convention that we represent the independent variable (which we usually call x) on the horizontal axis, and the dependent variable (which
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we usually call y) on the vertical axis. The horizontal axis is then called the x –axis, and the vertical axis is called the y –axis. For example, in Figure 1.1.4 (in Study Unit 1.1A) we represent time on the horizontal axis, and distanc e on the vertical axis. How far you travel depends on how long you travel, i.e. we have used time as the independent variable and distance as the dependent variable. (Note that we could have done this differently, with distance as the independent variable, since how long you travel also depends on how far you go.) The xy –plane
If the axes are called the x–axis and the y–axis, then the plane is often referred to as the xy –plane.
Scale
We have the same scale on both axes. We often use the same scale on both axes, but there are times when we need to use different scales. Unless we need accurate graphs we do not need to measure the spaces between the numbers on the axes so that they are exact, but they should look exact.
Point
We identify the point P by means of the ordered pair ( 2, 4) but we also speak of the point ( 2, 4).
−
−
Quadrants
The word quadrant is asso-
The axes divide the Cartesian plane into four quadrants. Note that the points on the axes do not belong to any specific quadrant. See Figure 1.2.6.
ciated with one quarter of a circle or sphere.
y II
I
x < 0, y >0
x > 0 , y >0
We have used the Roman numerals I, II, III and IV to label the quadrants. The counting of the quadrants follows an anti–clockwise
( a , 0)
direction.
a I II
IV
x < 0 , y <0
_ All points on the y axis have an x _ coordinate of zero.
x > 0 , y <0
(0, b)
Figure 1.2.6
x
All points on the x _ axis have a y _ coordinate of zero.
20 If we are given some point P in the plane, we can determine its coordinates in the following way. Through P draw two lines, one perpendicular to the x axis and the other perpendicular to the y axis, as in Figure 1.2.7. These lines cut the When we use the word “line” we mean a straight x–axis where x = a and the y–axis where y = b . We then assign to the point P line. the ordered pair (a, b). The first number in the pair is called the x coordinate of P and the second number the y–coordinate of P. Remember that we cal l ( a, b) Finding the coordinates of
−
a given point
−
−
an ordered pair since the order of the numbers is important. We say that P is the point with coordinates (a, b) and we denote the point by P (a, b) or by P = (a, b). y 4
P ( a , b)
3 b 2 1
x –3
–2
–1
1 2
a3 4
–1 –2 –3
Figure 1.2.7
Using the coordinates to find a point
We can also reverse this process. If we are given the coordinates ( a, b) we can use them to locate the point P in the plane. We can start with a : find the number a on the x axis and draw a vertica l line through it. We then find the number b on the y axis and draw a horizon tal line through it. The point where thes e two
−
−
lines intersect is the point P . Plotting points
The process of indicating the position of a point in the plane using its coordinates is called plotting the point. In Figure 1.2.8 a few points have been plotted in the Cartesian plane. As you would expe ct, the sketch (on the next page) clearly shows that the point (1, 2) is not the same as the point ( 2, 1).
−
−
21
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4 3 (2,3) 2 ( – 2 ,1)
1 (0,0) x
–3
–2
–1
1
2
3
4
–1 –2
(1, – 2 )
(– 2 , – 2 ) –3
Figure 1.2.8
1.2.1 (a) Write down the coordinates of the points A, B, C and D given in Figure 1.2.9. y 4
B
3
2
D
1
C – 4
– 3
– 2
– 1
1
– 1 – 2 – 3 – 4
Figure 1.2.9
2
3
A
4
x
22 (b) Draw a system of axes and plot the following points.
A (3, 1); B (1, 3); C ( 1, 0); D (0, 1)
−
(c) In which quadrants do the following points lie?
A(1, 5); B ( 10, 1)
− −
(a) A = ( 2, 1) B = ( 0, 4) C = ( 4, 0) D = ( 1 12 , 2)
−
−
Note that we do not have an accurate sketch on graph paper, so we cannot be certain that the x –coordinate of D is exactly 1 12 . However, for the purpose of this activity, since it seems that the x –coordinate of D lies exactly halfway between 1 and 2, we may accept that it is 1 12 .
−
−
−
−
(b) The points are shown in the figure below. y 3
B (1,3)
2
D (0,1)
1
A (3,1)
C ( – 1 ,0) –3
–2
x
–1
1
2
3
–1 –2
Figure 1.2.10
(c) A (1, 5) lies in quadrant I. B( 10, 1) lies in quadrant III.
− −
4
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1.2B GRAPHS IN THE CARTESIAN PLANE TERMINOLOGY
Instead of the instruction “ plot the following points ...” we may also be asked to
OF GRAPHS
“textbooks. graph theHowever, points”. This use of the graph is more commo it highlights theword fact that in this sense a graphnisina American collection of points, i.e. some subse t of the Cartesian plane. We can often identify that subset by means of an equation or inequality. The figures we have given so far consist of random isolated points in the Cartesian plane. However, many of the graphs we deal with consist of separate points that follow a given pattern (for example the graph in Figure 1.1.5), or points that can be joined together to form a particular curve or line in the plane. In Topic 3 you will learn more about lines, and in Topics 4 and 5 you will learn about two specific curves, namely parabolas and hyperbolas. Table of values
In Example 1.2.1 we give a table consisting of specific numbers in the x–row, and related numbers in the y –row. When we draw a graph from the information contained in such a table, we refer to the table as a table of values . We may be given the table of values, we may given an equation we way can set up an appropriate table ofor values . Inbe either case, the table from is justwhich another of identifying the ordered pairs ( x, y) that we use when we plot points that form the graph.
1.2.1 Use the given table of values to draw a suitable graph.
x y
−3 −2 −1 0 1 2 −5 −3 −1 1 3 5
SOLUTION We plot the points ( 3, 5), ( 2, 3), ( 1, 1), (0, 1), (1, 3) and (2, 5) in the Cartesian plane. This gives us the following graph.
− − − − − −
24 y
5
4
3
2
1
x
–3
–2
–1
1
2
3
–1 –2 –3 –4 –5
Figure 1.2.11 Note
Although the points lie on the same line, we may not join them, or include additional points, because in this case the only information we have is the information that is contained in the table of values.
Consider the next example. In this case the relationship between the two quantities represented by the variables x and y has been expressed as an equation.
1.2.2 Suppose we have the equation y = x 2 + 1, x sketch a graph to represent this equation.
∈ R. Set up a table of va lues and
SOLUTION
∈
Since x R, we can choose any values of x (the independent variable) and calculate the corresponding values of y . We choose x =
−3, −2, −1, 0, 1, 2 and 3, and have the following table of values.
25
For convenience we break up the expression x 2 + 1 into its component parts. We can
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− −2 −1
x 3 x2 9 y = x2 + 1 10
4 5
0 1 2 3 1 0 1 4 9 2 1 2 5 10
thus first calculate x2 , and then x2 + 1.
From this table we plot points in the Cartesian plane as follows. y
10 9 8 7 6 5 4 3 2 1 x
–3
–2
–1
1
2
3
Figure 1.2.12
We say that the equation y = x2 + 1, x R, generates the points that give us this graph.
In this case we know that x is any real number, and hence we may also choose any other real values of x . If we choose a few more values of x and calculate the corresponding values of y , we see that all the points we find fit into the existing pattern, and we conclude that in this case we can join the points and extend the curve indefinitely to give the graph in Figure 1.2.13.
∈
y
10
8
6
4
2
x
–3
–2
–1
1
Figure 1.2.13
2
3
26 Suppose we have the equation
y = x2 + 1, x
∈ R and − 3 ≤ x ≤ 3. ∈
We may again join the points shown in Figure 1.2.12 (since x R), but we may not extend the curve any further, since we only have information relating to real values of x between and including 3 and 3. Figure 1.2.14 shows the graph that represents this situation.
−
y
10
1 x
–3
3
Figure 1.2.14
Let us investigate further the terminology of graphs. Study the comments on the opposite page in relation to Figure 1.2.15. y
( p, q)
( c, d )
a
b
x
Figure 1.2.15
27 The coordinates of a point on a graph satisfy the equation that defines the graph.
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The graph of an equation in two variables is the curve (or line) which contains the points whose coordinates ( x, y) satisfy the equation. We say that the ordered pair ( x, y) satisfies an equation if substituting the given value of x into the equation yields the corresponding value of y . For example (1, 3) satisfies y = x + 2, since if x = 1 then y = 1 + 2 = 3 (1, 3) does not satisfy y = 3x 1, since if x = 1 then y = 3(1)
−
1 = 2 = 3.
−
Thus, in Figure 1.2.15, if the graph represents an equation, then ( c, d ) satisfies the equation and ( p, q) does not satisfy the equation. Intercepts on the axes and equations of the axes
We deal with equations of lines in Topic 3.
The graph cuts the x–axis when x = b. We call this number the x–intercept of the graph.
The point of intersection of the graph with the x–axis is the point ( b, 0), since y = 0 everywhere on the x –axis. If the graph represents an equation, the coordinates (b, 0) satisfy that equation.
The equation of the x –axis is y = 0.
The graph cuts the y–axis when y = a. We call this number the y–intercept of the graph.
The point of intersection of the graph with the y–axis is the point (0 , a), since x = 0 everywhere on the y –axis. If the graph represents an equation, the coordinates (0, a) satisfy that equation.
The equation of the y–axis is x = 0.
Since (c, d ) lies in quadrant I , we have c > 0 and d > 0. Since a lies above the x axis, we have a > 0. Since b lies to the left of the y –axis, we have b < 0.
−
The graph lies above the x–axis for all values of x such that x > b. (Another way of understanding this is to interpret “above the x –axis” in the sense of “associated with positive y –values”.)
The graph lies below the x –axis for all values of x such that x < b. (We can interpret “below the x –axis” as “associated with negative y –values”.)
28
Figure 1.2.16 may be helpful. Study it together with Figure 1.2.6.
y
Positive numbers on the _ y axis are above the horizontal line
Negative numbers on the _ x axis are on the left of the vertical line x
0
Positive numbers on the _ x axis are on the right of the vertical line
Negative numbers on the _ y axis are below the horizontal line
Figure 1.2.16
The words above and below, horizontal and vertical, are used here in a two– dimensional context, although in everyday life we usually use them in a three– dimensional context. For example, the first level of a building is above ground level, and the basement is below ground level. We think of the floor as a horizontal surface (with the dimensions length and breadth) and the walls as vertical structures (whose height we can measure). In the Cartesian plane we “borrow” the words vertical, horizontal, above and below from everyday language, although they do not have exactly the same meaning.
29
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1.2.2 y
( m , n)
y = ax 2 + bx + c
r
s
x
t ( p, q )
Figure 1.2.17 This graph is an example of a parabola. We deal with graphs such as these later, in Topic 4.
The graph in Figure 1.2.17 represents the equation y = ax 2 + bx + c, where a , b and c are non–zero real numbers. Use the graph to answer the following questions. (a) What are the x –intercepts of the graph? (b) What are the points of intersection of the graph with the x –axis? (c) What is the y –intercept of the graph? (d) What is the point of inters ection of the graph with the y –axis? (e) Consider all points ( x, y) on the graph. For what value(s) of x is y
≥ 0?
(f) For what value(s) of x does the graph lie below the x –axis? (g) Explain what the following statements mean. ( p, q) satisfies the equation y = ax 2 + bx + c. (m, n) does not satisfy the equation y = ax 2 + bx + c. (h) How can you deduce from the graph that both statements in (g) are true? (i) If ( x, y) denotes a point on the graph, for what value of x will y reach its smallest value? (j) Replace ? with > or < .
p? 0 q? 0 r? 0 s? 0
m? 0 n? 0 t?0
30
(a) The x –intercepts of the graph are r and s . (b) The points of intersection of the graph with the x –axis are (r, 0) and ( s, 0). (c) The y –intercept of the graph is t . (d) The point of intersection of the graph with the y –axis is (0 , t ). (e) We know from (a) that y = 0 when x = r or when x = s . Now the points on the graph for which y > 0 correspond to the sections of the graph above the x–axis. The graph lies abo ve the x–axis when x < r , and again when x > s. Putting all this information together we have the answer Make sure you understand
y
why we use “or” and not “and”.
≥ 0 for all x ∈ R such that x ≤ r or x ≥ s. −
(f) The graph lies belo w the x axis for r < x < s (i.e. for all values of x between, but not including, r and s ). (g) ( p, q) satisfies y = ax 2 + bx + c means that when we substitute x = p into the right side of the equation, we will obtain y = q . (m, n) does not satisfy y = ax 2 + bx + c means that if we substitute x = m into the right side of the equation, we will not obtain y = n . (h) ( p, q) will satisfy the equation y = ax 2 + bx + c since the point ( p, q) lies on the graph that represents the equation. (m, n) cannot satisfy the equation since the point ( m, n) does not lie on the graph. (i) The lowest point on the graph is the poi nt ( p, q). Thus if ( x, y) denotes a point on the graph, the smallest value that y can attain is y = q. It reaches this value when x = p. (j) p > 0 q< 0 r< 0 s> 0
m< 0 n> 0 t <0
31
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1.2 The graph which represents the equation is a hyperbola
1. The graph in Figure 1.2.18 is one branch of the graph which represents the k + l, where x R, and where k, l and m are non–zero equation y = x + m real numbers.
∈
y
which has two branches. study hyperbolas in TopicWe 5.
P ( p, q ) a
b
x k +l y = _____ x +m
Figure 1.2.18
Answer the following questions related to the portion of the graph shown in Figure 1.2.18.
−
(a) What is the x intercept of the graph?
−
(b) What is the point of inters ection of the graph with the y axis? (c) If ( x, y) is a point on the graph, for what value(s) of x is y
≤ 0?
(d) If ( x, y) is a point on the graph, for what value(s) of x is 0 < y < a ? (e) How can we find out whe ther or not P = ( p, q) lies on the graph?
k > (f) What values of x satisfy the inequality x + m ber that if x + km >
−l then x +k m + l > 0.)
−l? (Hint: Remem-
32 2. Suppose the following line represents the equation
y = 2x + 1; x
∈ R, x ≥ −1.
y
1
–1 x
–1
Figure 1.2.19
(a) Does the point (3, 7) lie on the line ? (b) Does the point ( 2, 3) lie on the line?
− −
(c) If ( x, y) lies on the graph, for what values of x is
−
(d) What is the y intercept of the graph?
We will study graphs of circles in Section 1.3.
−1 < y ≤ 1?
3. The graph in Figure 1.2.20 is a circle with cen tre O = (0, 0) and radius 4 units. The circle has the equa tion x2 + y2 = 16. Each of the poi nts A, B, C , D, and E has coordinates (x, y), where x R and y R .
∈
∈
y 4
A D
4
–4
0
B
C
–4
Figure 1.2.20
E
x
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(a) Match A , B, C, D and E with the following statements. (i) 0 < x < 4 and 0 < y < 4 (ii) x > 4 and y = 0
−4 < x < 0 and 0 < y < 4 −4 < y < 0 (v) −4 < x < 0 and −4 < y < 0
(iii)
(iv) 0 < x < 4 and
(b) Let E be the point ( 5, 0). Show algebraically that E does not lie on the circle. (c) How can we show algebraically that B lies on the circle? (d) What is the distance between the srcin O and D ? (e) What is the distance between the two points of intersection of the graph with the y axis. What name do we usually give to the line that passes through the centre and joins two points on the circle?
−
−
(f) What is the distance between the origin and the negative x intercept? What do we usually call the line that joins the centre and a point on the circle?
34
1.3 FORMULAS WE OFTEN USE
1.3A THE THEOREM OF PYTHAGORAS, DISTANCE AND MIDPOINT FORMULAS
If we look at Figure 1.2.17 in the last study unit it is clear that if we know the values of p, q, r, s and t we will be able to answer questions such as the following.
How far apart are the x –intercepts?
How far below the x –axis is the y –intercept?
How far below the x –axis is the lowest point of the graph?
Let us assume p, q, r, s and t have the values shown in Figure 1.3.1 and let us try to answer these three questions.
y
1
3
1 2 14
−
2 14
−
3
(1, 3)
−
Figure 1.3.1
x 3
35
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From the graph we find the following. It should be clear that the measurement of distance cannot ever be a negative number.
−
The x–intercepts are 4 units apart (the distance from 1 on the x–axis to the srcin is 1 unit; the distance from the srcin to 3 on the x–axis is 3 units, and we thus have a total distance of 4 units). The y –intercept is 2 1 units below the srcin. 4
The lowest point on the graph is 3 units below the x –axis (the y –coordinate of the lowest point is 3, hence the distance between it and the corresponding point on the x –axis is 3 units).
−
These distances are easy to find, but suppose now that we wanted to know the length of the line from (1 , 3) to (3, 0). We need various formulas that will enable us to calculate distances such as these. The algebraic study of lengths and distances based on the coordinate representation of points is part of coordinate geometry or analytic geometry.
−
DISTANCE
BETWEEN
TWO POINTS
The distance between two points P(x1 , y1 ) and Q(x2 , y2 ) is denoted by d (P, Q). The distance between P and Q is defined as the length of the line segment joining P and Q . We denote this line segment by means of PQ . Q
Note that some books use
PQ to represent this line segment.
P
The distance between P and Q is the length of PQ Figure 1.3.2
1.3.1 Suppose farm A is 6 km west of town B, and town C is 8 km south of farm A. Farm D is 16 km west of town C , and town E is 12 km south of farm D . How will we find the distance from town B to town E if we travel via town C ?
SOLUTION Diagrams make problems like these much easier to understand. We assume north, south, west and east mean due north, south, west and east, i.e. the north– south line and the west–east line are perpendicular to each other.
36 Have a look at Figure 1.3.3.
A
6 km
B
8 km D
16 km C
12 km
E
Figure 1.3.3
Suppose we draw the dia gram to scale, so that 1 cm represents 4 km. If we measure the line segments we find that Length of EC = 5 cm Length of CB = 2 12 cm. Thus the total length is 1 1 Length of EC + Length of CB = (5 + 2 ) cm = 7 cm. 2 2 Now 7 12 cm represents (7 12 30 km.
× 4) km. Hence the total distance between E and B is
Suppose we have measurements which we cannot easily convert by means of a scale to produce lines that we can measure accurately and hence determine distance. We can then solve the problem by using the Theorem of Pythagoras. THE
THEOREM
PYTHAGORAS
OF
Pythagoras was a Greek mathematician. He was born round about 572 BC, and founded a famous school where philosophy, mathematics, music and the natural sciences were studied.
37
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Consider Figure 1.3.4. A
c b
C
a
B
A right triangle Figure 1.3.4
We note the following.
We use the symbol resent any triangle.
to rep-
◦ABC is a right triangle. (It has a right angle, i.e. an angle that measures 90 , at C . On the ske tch the rig ht angle is marked by means of a small square where the vertical and horizontal lines meet.)
By convention, we denote the length of the side opposite the angle at A by means of a; we use b for the length of the side opposite B and c for the length of the side opposite C .
The side opposite the right angle in any right triangle is called the hypotenuse. In ABC the hypotenuse is AB .
We have the following theorem.
THE THEOREM OF PYTHAGORAS
The square on the hypotenuse of a right triangle is equal in area to the sum of the areas of the squares on the other two sides. In terms of Figure 1.3.4 we thus have c2 = a 2 + b2 .
(1.3.1)
Consider Figure 1.3.5 on the next page.
Area = length
× breadth.
In (iii) we have a right triangle where a , b and c denote, respectively, the lengths of two sides and the hypotenuse. Consider the two big squares indicated by (i) and (ii) in Figure 1.3.5. Each of these squares has sides of length a + b. Hence the area of each square is the same, i.e. ( a + b)2 .
38 b
a
a
a
b
a
a
c
b c
c b
a
b c b a
b
a
c
b
a
b
(i)
(ii)
(iii) Figure 1.3.5
“Congruent” means “identical in shape and size”. There are several conditions for congruency and we discuss them in Book 4.
Draw Figure 1.3.5 onto a piece of paper, using any convenient measurements for a and b, making b larger than a. Cut up the first squa re (i), along the lines shown so that you have six pieces (four right triangles congruent to the given right triangle, one small square of side a and one big square of side b). Cut up the second square (ii), along the lines shown, so that you have five pieces (the squaretoofthe side c, i.etriangle). the square hypotenuse, four right triangles congruent given Putonallthe the squares andand triangles back together again so that you have the two big squares shown in (i) and (ii) of the sketch. In both the squares in Figure 1.3.5 we can now remove the four triangles. Since the two big squares were equal in area to begin with, it follows that what is left after removing triangles that are equal in area, must still be equal in area.
We emphasise that an illustration that a certain fact is true in one specific case is not a proof that it is true in general.
This illustrates that the two remaining squares in the diagram (i) together have the same area as the remaining square in the diagram (ii). In other words we have
a2 + b 2 = c 2 . We can state the Theorem of Pythagoras in the following way. If ABC is a right triangle with c the length of the hypotenuse and a and b the
lengths of the other two sides, then
a2 + b 2 = c 2 . In words we state this as follows.
“converse” means “reasoning in the opposite direction”.
If ABC is a right triangle then the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.
We can also formulate the converse of this theorem. If ABC is such that the lengths of the sides are a, b and c, and a2 + b2 = c 2 , then ABC is a right triangle with hypotenuse of length c .
39
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1.3.1 (a) Calculate the length of the hypotenuse in each of the following right triangles. (i)
(ii) H
12
cm
F
P
5 cm
3 cm
Q
R
4 cm
(b) Show that
G
ABC is a right triangle. 2
cm
A
B
2
2 cm
cm
C
(a)
(i) Let the length of the hypotenuse PR be q cm. Then 2
2
2
q =3 + 4
⇔ ⇔
q2 =25 q=
± 5.
But we know that length cannot be negative, hence q = 5 and thus the length of the hypotenuse is 5 cm.
40 (ii) Let the length of the hypotenuse GH be f cm. Then
f 2 =52 + 122 f 2 =169
⇔ ⇔
f=
±√169
⇔
f=
± 13.
Since f > 0 it follows that potenuse is 13 cm. We begin by considering b 2 , since it appears that the right angle is at B.
f = 13. Hence the length of the hy-
(b) In ABC, if b 2 = a2 + c2 , then ABC is a right triangle. Now b = 2 cm, c = 2 cm and a = 2 cm. Thus
√
√
√
√
a2 + c2 =( 2 cm )2 + ( 2 cm )2
=2 cm 2 + 2 cm 2 =4 cm 2 and
b2 =(2 cm )2
=4 cm 2 i.e. we have
b2 =a2 + c2 and hence
ABC is a right triangle.
Now try to do Example 1.3.1 again, in a different way.
1.3.2 Use the Theorem of Pythagoras to calculate the distance between the towns and E given in Example 1.3.1.
B
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We refer to Figure 1.3.3. By the Theorem of Pythagoras we see that in
DEC
d 2 = e2 + c 2 same letter, c, to represent
= 162 + 122 = 256 + 144
two different lengths in this activity. In the firs t case c
= 400.
Note that we have used the
represents the length of DE in
DEC, and in the second
case c represents the length of AB in
ABC.
Since C
Since d In
≥ 0 we have d = 20, i.e. the distance from E to C is 20 km.
ACB a 2 = c2 + b 2
occurs in both triangles it is
= 62 + 82 = 36 + 64 = 100.
acceptable to use c to represent the lengths of the side opposite C in each case, although we understand that c in in
DEC is not equal to ABC.
c
Since a
≥ 0 we have a = 10, i.e. the distance from B to C is 10 km. .
Thus the distance from E to B is the sum of these two distances, i.e. 30 km.
In Activities 1.3.1 and 1.3.2 we have used the Theorem of Pythagoras to calculate the length of the hypotenuse, where the lengths of the other sides have been given. We can also use this theorem to calculate the length of one side of a right triangle, if we know the length of the other side and of the hypotenuse.
1.3.2 Suppose
PQR is a right triangle, with the right angle at Q. R
15 cm
P
Calculate the length of QR .
4 cm
Q
42
SOLUTION Let the length of QR be p cm. By the Theorem of Pytha goras, p2 + 42 = 15 2 . Now
p2 + 42 =152
⇔
p2 =225
⇔
p2 =209 p=
⇔
±
− 16
√
209.
√
We discussed surds in Topic 4 of Book 1.
DISTANCE FORMULA
Thus the length of QR is 209 cm. (209 is not a perfect square, or the product of a perfect square and some other number, and we thus leave the answer as a square root, i.e. in surd form.)
Not all lengths can be calculated by considering line segments as sides of right triangles. We thus need another way of calculating distances. We have the distance formula. Consider Figure 1.3.6. y Q( x 2 , y 2 )
y2
y 2 – y1
y1
P ( x 1, y 1 )
R ( x2 , y1)
x1
x2
x x2 – x1
Figure 1.3.6
43
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We want to determine the distance between two points P and Q in the Cartesian plane. We derive a formula based on the Theorem of Pythagoras, but once we have the formula we do not need to create right triangles each time we need to calculate a length, although this might help. For convenience we use the notation (x1 , y1 ) to represent the coordinates of the point P, and (x2 , y2 ) to represent the coordinates of the point Q. We locate a third point in the plane, namely R, by completing a right triangle with PQ as the hypotenuse. Since R is as far above the x axis as P is, it has the same y coordinate as P, namely y1 . Since R is as far to the right of the y axis as Q is, it has the same x coordinate as Q , namely x 2 .
−
Note that this discussion refers to points P and Q in th e first qu adrant. We will obtain the same formula always, regardless of the quadrants in which P and Q lie.
−
−
−
−
The length of PR is x2 x1 since the distance between P and R is the distance between x 1 and x 2 on the x axis, and x 2 > x1 .
−
−
The length of QR is y 2 y1 since the distance between Q and R is the distance between y 1 and y 2 on the y axis and y 2 > y1 . Now, in the right triangle PQR , let the length of the hypotenuse be r . Then, by the Theorem of Pythagoras we have
−
r 2 = ( x2
2
−x ) 1
+ (y2
2
−y ) . 1
Hence d (P, Q) = r = (x2 x1 )2 + (y2 y1 )2 . (Since distance is non–negative we only consider the positive square root.) Hence we have the following formula.
−
−
Note that if Q lies directly above P then
DISTANCE FORMULA
x1 = x2 and y2 > y1 . Hence
y2 and thus
d (P, Q) =
−y
= y2
1>
The distance d (P, Q) between the points P (x1 , y1 ) and Q (x2 , y2 ) is given by
0 2
(y − y ) −y . 2
1
1
d(P,Q)=
(x2 – x1 )2 + (y2 – y1 )2 .
(1.3.1)
In each of the following calculate the distance between the given points. Leave answers in surd form where necessary.
− −5) (b) O(0, 0); P (−4, −3) (c) P(−1, −5); Q (−2, 3) (a) A(3, 1); B ( 4,
44
(a)
d (A, B) =
= = = =
− − − √ −
x1 )2 + (y2
(x2
3)2
( 4
2
−y ) + (−5 − 1) 1
2
( 7)2 + ( 6)2 49 + 36
−
√
85
√85 units.
Thus the distance between A and B is (b)
d (O, P) =
= = =
− − − −
x1 )2 + (y2
(x2
( 4
( 4)2
√16 + 9 √
2
−y ) 0) + (−3 − 0) + (−3) 1
2
2
2
25 = = 5 Thus the distance between O and P is 5 units. (c)
d (P, Q) =
= = = =
− − − − − (x2
x1 )2 + (y2
( 2
( 1))2
1
2
( 1)2 + (8)2
√1 + 64 √ 65
Thus the distance between P and Q is
MIDPOINT FORMULA
2
−y ) + (3 − (−5))
√65 units.
We now consider how to find the midpoint M (x, y) of the line segment that joins P(x1 , y1 ) and Q (x2 , y2 ). Consider Figure 1.3.7.
45
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y Q ( x2 , y2 )
y2
M ( x, y)
y
S x2 – x
y1
P ( x 1 , y 1)
R x – x1 x
x1
x
x2
Figure 1.3.7
One of the conditions that determines that two triang– les are congruent is that a side and two angles of the
In MQS and PMR we have the length of MQ equal to the length of PM ˆ = M PR ˆ (since PR MS , (since M is the midpoint of PQ). We also have QMS ˆ = P MR ˆ (since QS MR we have the corresponding angles are equal) and M QS
one should be equal to the corresponding side and two angles of the other. You may wish to come back to this again once you have studied Book 4.
another pair ofHence equal corresponding angles). Thus the triangles are congruent. PR and MS have the same length. Now PM R and M QS
length of PR = length of MS
⇔
x
−x
1
= x2
−x
2x =x2 + x1
⇔
x=
⇔ In a similar way we obtain y =
x1 + x2 . 2
y 1 + y2 . Thus M (x, y) = 2
x1 + x 2 , y 1 + y2 . 2 2
We have the following formula.
MIDPOINT FORMULA
The midpoint M (x, y) of the line segment from a point P (x1 , y1 ) to another point Q (x2 , y2 ) is 2 y 1 +y2 M (x, y)= x1 +x (1.3.3) 2 , 2
46
1.3.4 Show how we obtain the y –coordinate of M , i.e. y =
y1 + y 2 . 2
x + x2, we have that As given in the derivation of x = 1 MQS and PMR are 2 congruent. It follows that the length of QS is equal to the length of MR. Now
length of QS =length of MR
⇔
y2
− y =y − y
1
2 y = y 2 + y1
⇔
y=
⇔
y 1 + y2 . 2
1.3.4 Suppose P ( 1, 2) and Q (2, 5) are two points in the plane.
−
−
(a) Determine the distance d (P, Q) between P and Q . (b) Determine the midpoint M (x, y) of the line segment joining P and Q . (c) Check that M (x, y) is the midpoint by calculating the distance between M and P , and between M and Q .
SOLUTION (a)
d (P, Q) = Thus
d (P, Q) =
= = =
(x2 (2 32
2 + (y
−x ) 1
√9 + 49 √ 58.
1
2 +(
− (−1)) + (−7)
2
2
−y )
2
By (1.3.2). 2
−5 − 2)
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(b)
M (x, y) = Thus
M (x, y) =
=
x1 + x2 y1 + y 2 , 2 2 1+2 2
By equation (1.3.3) .
− − − 5
,
1 , 2
2
2
3 2
.
(c)
d (M , P) =
= = =
d (M , Q) =
= = =
−− − −
2
− −− −
2
2
1 2
1
+ 2
3 2
9 49 + 4 4
58 4 58 2
√
2
1 2
2
+
5
3 2
9 49 + 4 4
58 4 58 2
√
Thus d (M , P) = d (M , Q), and hence M is the midpoint of the line segment joining P and Q .
48
1.3B THE EQUATION OF A CIRCLE Note that we use the word “radius” in two different ways. It describes the line segment from the centre to any point on the circle; it also describes the distance from the centre to the point on the circle, i.e. the length of the line segment joining the centre and a point on the circle. Similarly the word “diameter” denotes both the line segment and its length.
We can also use the distance formula obtained in Study Unit 1.3A to find the equation a circle. know that aafixed is a set of points which).lie a fixed circlepoint distance rof(called theWe radius ) from (called the centre In at any circle the diameter is the line segment that passes through the centre, and joins any two points on the circle. Consider the circle with centre (0,0) and radius 2 units shown in Figure 1.3.8. y
( x, y) 2 x
(0,0)
Figure 1.3.8 CIRCLES WITH CENTRE AT THE ORIGIN
Let ( x, y) be any point on the circle. Then, by the distance formula, the distance from the srcin to ( x, y), i.e. the radius, is given by (x 0)2 + (y 0)2 . Thus 2 2 2 2 x + y = 2, i.e. the equation of this circle is x + y = 22 , or x 2 + y2 = 4.
−
−
In general, if the radius is given by r we have the following formula.
STANDARD FORM OF THE EQUATION OF A CIRCLE WITH CENTRE AT ( 0, 0) AND RADIUS r
The equation of a circle with centre ( 0, 0) and radius r is x2 + y2 = r2
Consider Example 1.3.5.
(1.3.4)
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1.3.5 Find the equation of the circle sketched below. The srcin is the centre. y
(– 2 ,1) x
Figure 1.3.9
SOLUTION When the centre of a circle is the srcin and its radius is r it has the equation x2 + y2 = r 2 . In Figure 1.3.9 the poin t ( 2, 1) lies on the circle, and hence the coordinates of the point satisfy the equation of the circle. Thus
−
( 2)2 + 12 =r2 .
−
We thus have
4 + 1 =r2 and hence
r 2 =5. Thus the equation of the circle is x 2 + y2 = 5.
Now try the following activity.
1.3.5 Show that the point P ( 2, 3) does not lie on the circle with centre at the srcin and radius 4 units.
−
50 The equation of the circle with centre (0, 0) and radius 4 units is
x2 + y2 = 16. If P lies on this circle then the coordinates of the point P must satisfy the equation x2 + y2 = 16. Now, if we substitute x = 2 and y = 3 into the left side of the equation, then
−
x2 + y2 = ( 2)2 + 32 = 4+9
−
= 13 = 16.
Thus the coordinates ( 2, 3) do not satisfy the equation and hence P does not lie on the circle.
−
Before we move on, try one more activity.
1.3.6 Consider the sketch below which shows the circle with centre at the srcin and radius 4 units, and the circle with cen tre at the srcin and radius 5 units. The point P lies between the two circles. y
P ( x, y)
4
5
x
Figure 1.3.10
Describe algebraically the numerical boundaries of the distance between P and the srcin.
51
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The distance from P to the srcin is
d (P, O) =
(0
2 + (0
− x)
2
− y)
=
x 2 + y2 .
P lies outside the smaller circle and inside the bigger circle. Thus the distance between P and the srcin is greater than 4 units and less than 5 units. We can thus describe the distance between P and the srcin by means of the inequalities 4 < i.e.
x 2 + y2
<
5
42 < x2 + y2 < 52 i.e. 16 < x2 + y2 < 25.
CIRCLES WHERE THE CENTRE IS NOT THE
We now consider a circle with centre ( h, k) and radius r, as shown in Figure 1.3.11.
ORIGIN
y
x
r
( h, k )
( x , y)
Figure 1.3.11
Such a circle consists of all points P (x, y) whose distance to (h, k) is r . Again, by using the distance formula, we obtain
(x - h) 2 + (y – k) 2 = r
i.e. we have the following formula.
52
THE STANDARD FORM OF THE EQUATION OF A CIRCLE WITH CENTRE AT ( h, k) AND RADIUS r
The standard form of the equation of a circle with centre ( h, k) and radius r is (x – h) 2 + (y – k) 2 = r 2 .
(1.3.5)
1.3.6 (a) Consider a circle with centre (1, 2) and radius 4 units. We substitute h = 1 and k = 2 into equation (1.3.5) and we obtain
−
−
(x
2
− 1)
+ (y
2
− (−2))
= 16.
This equation can be simplified, and we have 2
(x
− 1)
2
+ (y + 2) = 16.
(b) Consider also a circle with centre ( 3, 0) and radius 6 units. We substitute h = 3 and k = 0 into equation (1.3.5) and we obtain
−
−
(x
2
− (−3))
+ (y
2
− 0)
= 62
which we simplify and write as
(x + 3)2 + y2 = 36.
1.3.7 (a) Give the standard equation of the circl e with radius 3 units and centre (3, 2).
−
(b) What are the centre and radius of the circle with equation
x2 + (y + 2)2 = 16?
53
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(a) The standard equation is
(x We have h = 3, k = Thus we have
2
− h)
+ (y
2
− k)
= r2.
−2 and r = 3. (x
− 3) + (y − (−2)) = 3 (x − 3) + (y + 2) = 9. (b) We rewrite the equation in the form (x − h) + (y − k) 2
2
2
and hence
2
2
2
2
= r2.
We have
(x Hence the centre is (0,
2
− 0)
+ (y
2
− (−2))
= 42 .
−2) and the radius is 4 units.
In the next activity you will use the fact that if we denote the diameter of a circle by d and the radius by r , then d = 2r .
1.3.8 In Figure 1.3.12, PQ is a diameter of the circle. Find the equation of the circle. y
(– 1 ,3)
P
x Q
(4, – 1 )
Figure 1.3.12
Hint: Use the distance formula to find the diameter, and the midpoint formula to find the centre.
54
Diameter = length of line segment joining ( 1, 3) and (4,
− − −
=
( 1
Hence radius = 1 2
as
1 . 4
−1)
By (1.3.2).
√ √25 + 16 41 1√
= =
Note that we write
− (−
1))2
( 5)2 + 42
=
−
4)2 + (3
41
2
−
41 . 4 centre of circle 1+4 3 + ( 1) , 2 2 3 , 1 . 2 3 , 1 . 2
= Midpoint of diameter = Midpoint of diameter =
= Hence centre =
−
By ( 1.3.3).
The equation of the circle is
(x
− h)
where h = 32 , k = 1 and r = Hence
2
+ (y
2
= r2
− k)
By (1.3.5).
41 . 4
(x
− 32 )
4(x
− 32 )
2
+ (y
2
2
+ 4(y
− 1)
=
41 4
and we write this as 2
− 1)
= 41.
If we continue simplifying this equation we obtain 4x2
Note that when we simplify the equation it is not immediately obvious that this rep-
− 12x + 9 + 4y − 8y + 4 2
= 41
i.e. we have 4x2
resents a circle.
2
− 12x + 4y − 8y
= 28
i.e.
x2
2
− 3x + y − 2y
= 7.
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1.3 1.
ABC is a right triangle. The length of the hypotenuse is 10 units, and the length of one other side is 4 units. Calculate the length of the third side.
2. Consider the points A (1, 1), B(4, 3) and C (4, 1). Show (a) graphically (b) algebraically
ABC is a right triangle. 3. Show that ABC is not a right triangle, when that
A = ( 1, 1), B = ( 5, 3) and
C = ( 4, 2).
4. For each pai r of points P and Q determine the distance d (P, Q) between them and the midpoint M (x, y) of the line segment joining P and Q . (a) P = ( 1, 7)
Q = ( 3, 4)
(b) P = ( 1, 1)
Q = ( 2, 3)
− − (c) P = ( 1, −2)
− − Q = ( −3, 4)
5. Write down the standard equation of the circle with centre C and radius r .
√2
(a) C = ( 0, 0)
r=
(b) C = ( 1, 2)
r=3
(c) C = ( 2, 3)
r=2 3
(d) C = ( 0, 1)
r=1
−
√
6. Determine the centre C and radius r of each circle defined below. (a) x2 + y2 = 3 (b) (x
2
− 1)
+ (y + 2)2 = 36
7. Suppose AB is the line segment joining A and B, where A = (5, 4) and
B = ( 1, 2). (a) Find the midpoint C of AB .
−
(b) Find the distance d (C , A) between C and A . (c) Write down the equation of the circle that has AB as diameter. 8. Suppose AB is the diameter of a circle where A = ( 1, 1) and B = ( 2, 3). Determine the equation of the circle.
−
− −
56
The Cartesian coordinate plane R
•
R
×
Consists of points P , each represented by an ordered number pair (x, y) called the coordinates of P .
The plane (also referred to as the xy plane) is divided into four quadrants by a horizontal line (x axis or independent variable axis) and a vertical line ( y axis or dependent variable axis), such that in quadrant I, x > 0 and y > 0 in quadrant II, x < 0 and y > 0 in quadrant III, x < 0 and y < 0 in quadrant IV, x > 0 and y < 0.
−
−
−
The axes are perpendicular to each other and intersect at the O, where O = ( 0, 0).
srcin
• Plotting points (given in a table of values or generated by some equation) in R R to create graphs ×Choose an appropriate scale.
If the graph represents an equation then the coordinates of the points satisfy the equation.
If the coordinates of a point satisfy the equation of the graph then the point lies on the graph.
• Important information for any graph
the x – and y –intercepts of the graph
the equation of the y –axis is always x = 0
the equation of the x –axis is always y = 0
values of x for which y > 0 (i.e. for which the graph lies above the x–axis)
values of x for which y < 0 (i.e. for which the graph lies below the x–axis)
values of x for which y the x –axis)
≥ 0 (i.e. for which the graph lies above or on
values of x for which y the x –axis)
≤ 0 (i.e. for which the graph lies below or on
We assume the point (x, y) lies on the graph.
• The Theorem of Pythagoras
In any right triangle ABC where Cˆ = 90◦ we have
c 2 = a 2 + b2 .
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Converse If ABC is such that
then
c 2 = a 2 + b2
ABC is a right triangle with hypotenuse of length c units.
• Distance formula
The length of the line segment joining P (x1 , y1 ) and Q (x2 , y2 ) is given by
d (P, Q) =
(x2
−x ) 1
2 + (y
2
2
−y ) . 1
• Midpoint formula
The midpoint M (x, y) of the line segment joining P (x1 , y1 ) and Q (x2 , y2 ) is given by x1 + x 2 y 1 + y2 M (x, y) = , . 2 2
• Standard equation of a circle with centre at (0, 0) and radius r is x2 + y 2 = r 2 . Standard equation of a circle with centre at ( h, k) and radius r is
•
(x
2
− h)
+ (y
2
− k)
= r2.
CHECKLIST Now check that you can do the following. SECTION 1.1
1. Choose an appropriate scale (if an accurate graph is required) or mark vertical and horizontal lines in a suitable way (if an illustration of information is required) and plot given data, taking into account whether the graph consists of separate dots, or whether the dots can be joined in some way. Examples 1.1.1, 1.1.2; Exercise 1.1 (1), (2) 2. Interpret information conveyed by a given graph. Exercise 1.1 (3), (4)
SECTION 1.2
1. Represent solutions of equations or inequalities in one variable by means of a one–dimensional graph (i.e. on a number line). Book 1, Topic 1: see Figure 1.1.1 and Table 1.2.1
58 2. Find coordinates of a given point in the Cartesian plane, or, if the coordinates are known, locate the point. Activity 1.2.1 3. Use a table of values to draw a graph. Example 1.2.1 4. Use an equation to set up a table of values and hence draw a graph. Example 1.2.2 5. If a graph represents an equation
find the x – and y –intercepts of the graph
determine whether or not a given point lies on the graph, i.e. whether or not the coordinates of the point satisfy the equation of the graph
find values of x for which y > 0, y = 0 (i.e. the x –intercept) or y < 0
recognise from the quadrant in which a point lies whether x > 0 or x < 0 and whether y > 0 or y < 0. Activity 1.2.2
SECTION 1.3
1. Use the Theorem of Pythagoras. Equation (1.3.1); Activities 1.3.1, 1.3.2; Example 1.3.2 2. Use the distance formula. Equation (1.3.2); Activities 1.3.3, 1.3.8; Example 1.3.4 3. Use the midpoint formula. Equation (1.3.3); Activities 1.3.4, 1.3.8; Example 1.3.4 4. Find the standard equation of a circle with radius r and centre at ( 0, 0) . Equation (1.3.4); Example 1.3.5 5. Use the equation of a circle to find out whether a point lies on the circle or not. Activities 1.3.5, 1.3.6 6. Find the standard equation of a circle with radius r and centre at (h, k) = (0, 0). Equation (1.3.5); Example 1.3.6; Activity 1.3.7
7. Given the equation of a circle, find the centre and radius of the circle. Activity 1.3.7 8. Given the coordinates of the end points of the diameter of a circle, find the equation of the circle. Activity 1.3.8
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RELATIONS AND FUNCTIONS OUTCOMES After studying this topic you should be able to do the following.
SECTION 2.1: Relations and Functions in
R
×R
Identify the type of correspondence in any relation.
Identify the domain and range in a given relation.
Recognise when a relation is a function.
Determine function values by substituting given domain values into the formula (equation) that defines the function.
Identify the natural domain of a function.
Identify dependent and independent variables in a specific example, and use functional notation correctly to express the relationship between the variables.
Recognise when two functions are equal.
SECTION 2.2: Combining Functions
Obtain new functions from existing functions by addition, subtraction,
multiplication and division. Calculate the function value when the function is obtained from the sum, difference, product or quotient of two other functions.
Determine the natural domain of sum, difference, product or quotient functions.
60
2.1 RELATIONS AND FUNCTIONS IN R
×R
2.1A TERMINOLOGY AND NOTATION In the previous topic we looked at some specific subsets of the Cartesian plane, consisting of points which all followed a particular pattern. In some cases the points were plotted from a table of values which was either given, or set up from a specific equation. In mathematics, certain subsets of the Cartesian plane interest us more than others. In order to deal with these ideas in more detail, we need appropriate terminology and notation.
R
×R
We have already introduced unfamiliar notation in the section heading, where we refer to R R. We use this to denote the Car tesian plane. It indicates that the Cartesian plane consists of all points which represent ordered number pairs (x, y), where x is any real number, and y is any real number. We thus have the
×
following definition of the Cartesian plane. Definition 2.1.1 The Cartesian plane represents R is the set (x, y) : x R and y R .
{
RELATIONS
∈
We have already considered various subsets of special name.
× R, which ∈ } R
× R. We give these subsets a
Definition 2.1.2 A subset of R R is called a relation.
×
×
This just means that a relation in R R is any set of ordered number pairs. These ordered pairs can sometimes be described by means of a simple rule. Have a look at the next example.
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2.1.1 Write down a simple rule to specify the relations that are illustrated (a) in Figure 2.1.1 (b) in Figure 2.1.2.
y
1
x
3
–1
1 –1
Figure 2.1.1
Figure 2.1.2
SOLUTION (a) It is clear that the rule is “draw a circle with centr e at the srcin and radius 3 units”. From our knowledge of circles (discu ssed in Topic 1, Section 1.3) we can express this rule in terms of the equation x 2 + y2 = 9. (b) The relation in this case consists only of eight points, arranged in the form of a square. We could state the rule as “arran ge dots in a square around the srcin corresponding only to each of the numbers 1, 0 and 1 on the x– and y –axes”. A rule stated in this way could be ambiguous, and hence we prefer to state the rule mathematically, as follows.
−
y= y
−1
for
0
for
y= =1
for
∈ {−1, 0, 1} 1 1 ∈ {− {−1,, 0}, 1} x∈
x
x
In (a) of Example 2.1.1 we can state the rule as an equation which links the value of the dependent variable with a specific value of the independent variable.
62 If
x2 + y2 = 9 then
y2 = 9
2
−x √ y = ± 9 − x ; x ∈ R, −3 ≤ x ≤ 3.
i.e. we have the equation
2
In (b) it is more difficult to give a suitable equation. In this case it is simpler to give the relation as a set of ordered pairs:
{ (−1, −1), (−1, 0), ( −1, 1), (0, −1), (0, 1), (1, −1), (1, 0), (1, 1) }. DOMAIN AND RANGE
When we discuss the rule we use to establish a relation, we need to be very clear about the values we choose for the independent variable, i.e. the values we choose for the first coordinate of each ordered pair of the relation. We use a specific word for the set of first coordinates, and also for the set of second coordinates. These are given in the following definition.
Definition 2.1.3 The set of all first coordinates of the ordered pairs of a relation is called the domain of the
The range is sometimes re-
relation. The set of all second coordinates is called the range.
ferred to as the image set.
In Example 2.1.1 (a) we have Domain = Range =
{x ∈ R : −3 ≤ x ≤ 3} {y ∈ R : −3 ≤ y ≤ 3}.
In (b) we have Domain = Range =
{−1, 0, 1} {−1, 0, 1}.
2.1.1 Can you identify the rule that determines the relation shown in the following table?
x 1 2 3 4 5 6 y 6 9 12 15 18 21
63
The sequence of terms in the domain has a = 1 and d = 1. The sequence of terms in the range has a = 6 and d = 3.
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From your study of sequences (Topic 3 of Book 2) you will recognise that the domain and range are both arithmetic sequences, where the general term has the form a + (n 1)d . We consider two possible ways of identify ing the rule that determines the given relation.
−
First possibility
+5 1
+3+2 6
i.e.
1
6
−→
−→
+7 2
−→
3
−→
+3+2+2 9
i.e.
12
i.e.
−→
2
+9
9
+3+2+2+2
−→
3
12
From the pattern we can conclude that
y = x+3+2
× x = 3x + 3;
x
∈ N, x ≤ 6.
Second possibility
Since d = 3 in the range, we could consider the following pattern. (3 1
(3 2
(3 3
× 1)+3 −→
6
× 2)+3 −→
9
× 3)+3 −→
12
This yields the same equation as before, i.e.
y = 3x + 3;
CORRESPONDENCE
x
∈ N, x ≤ 6.
The table in Activity 2.1.1 shows a correspondence that exists between two sets of numbers. In each column in the table, a number in the x row is paired with, or associated with, another number in the y row. For example, 1 is paired with 6 to form the ordered pair (1, 6).
64 The ordered pairs in this activity are given in a table. There are many real–life cases in which variables are linked, such as the relationship between distance and time when speed is constant, discussed in Example 1.1.1 (Topic 1). In the example, Mishak was travelling at a constant speed of 80 km/h, and we make use of the formula d = 80 t to calculate the distance he had covered after he had travelled for a certain time. One–to–one correspondence
This example shows a one–to–one correspondence between two sets of data: a set of numbers representing time, and a set of numbers representing distance. Consider the data shown in Table 1.1.2 of Study Unit 1.1A. We represent this data differently, as in Figure 2.1.3.
1
80
2
160
3
240
4
320
5
400
6
480
Time (in hours) (Domain)
Distance (in kilometres) (Range)
One–to–one correspondence Figure 2.1.3
There is exactly one measurement of distance paired with each measurement of time, i.e. there is exactly one element of the range paired with each value in the domain. A correspondence of this type is called a one–to–one correspondence. Not all correspondences are of this type. For example, consider the graph of the relation sketched in Figure 2.1.4 on the next page.
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y
√3
√
(2, 5)
2 x
−3
√3 −
(2,
−√5)
Figure 2.1.4
We represent the domain and range of this relation in Figure 2.1.5. Since there are infinitely many real numbers bigger than or equal to 3, the diagram only shows selected domain and corresponding range elements.
−
5
2 –3
– 5 0
One–to–many correspondence
0
– 3 3
Domain
Range
One–to–many correspondence Figure 2.1.5
−
From the graph it is clear that all domain elements (except 3) are paired with two different elements in the range. Figure 2.1.5 thus represents a one–to–many correspondence. Many–to–one correspondence
Note that when decimals are involved we use a semi-colon (;) to separate the numbers in the ordered pairs.
We now consider a third type of correspondence. Suppose a shop has a “ten– minute special”, where selected items are sold for R5,00 each, regardless of the actual price marked on the item. We then have, for example, the following pairs (10,98; 5,00), (6,44; 5,00), (11,23; 5,00) representing the srcinal price and special price of various items. In this case we have many different domain values associated with only one range value, and we call this type of correspondence a many–to–one correspondence. Figure 2.1.6 illustrates this type of correspondence.
66 10,98 6,44
5,00
11,23
Domain
Range
Many–to–one correspondence Figure 2.1.6
See whether you have understood the different types of correspondence by trying to do the next activity.
2.1.2 Identify the correspondence represented in each of the following cases. (a) 1 2 3 4 5
2
Domain
Range
(b) Suppose x is any whole number that is a perfect square. x x – x Domain
Range
(c) a
p b
q r
c
Domain
Range
(d) The distance–time information given in Table 1.1.2 of Topic 1.
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(a) All five domain elements are paired with the same element in the range, to give the ordered pairs (1, 2), (2, 2), (3, 2), (4, 2), (5, 2). The idea “five to one” tells us that the correspondence is many–to–one. Note that each ordered pair has a different first coordinate. (b) The domain is the infinite set of all whole numbers that are perfect squares. Consider for example the first three perfect squares: 0, 1, 4. Now
√ ±√ 0 ±√ 1 ± 4
= 0 = 1 or = 2 or
−1 − 2.
We thus have 0 1
0 1 –1
4
2 –2
i.e. we have the ordered pairs ( 0, 0), (1, 1), (1, 1), (4, 2), (4, 2) and the correspondence is one–to–many. Note that there are now different ordered pairs that have the same first coordinate.
−
−
(c) The ordered pairs are (a, p), ( a, q ), (b, r ), ( c, p). One of the domain elements (i.e. a ) is paired with two different range elements, namely p and q, and the correspondence appears to be one–to– many. However, two of the domain elements ( a and c) are mapped onto one range element, namely p, which suggests a many–to–one correspondence. This example, in which a one–to–many correspondence and a many–to– one correspondence occur, illustrates a many–to–many correspondence. (d) Each time measurement is paired with a unique distance measurement and each distance measurement is paired with a unique time measurement. The correspondence is thus one–to–one. All the ordered pairs have a different first coordinate and a different second coordinate.
In mathematics we are particularly interested in relations in which the correspondence is one–to–one (see Figure 2.1.3) and many–to–one (see Figure 2.1.6). The important fact about these two types of correspondence is that each value of x in the domain is associated with a unique value of y in the range.
68 FUNCTIONS
Definition 2.1.4 A function f between two sets of real numbers A and B is a relation in which each element of A is paired with a unique element of B .
What does this definition tell us?
The function f is a set of ordered pairs.
The set A is the domain of f . For each x A there is a corresponding value of y in B (sometimes called the image of x). The set of all images is the range of f .
The elements in f can be determined by some rule, or represented by a table showing the correspondence between x and y , or represented by a set of ordered pairs.
The correspondence between the domain and the range is either one–to– one or many–to–one, but never one–to–many.
Each ordered pair in f has a different first coordinate.
If there is even just one domain value that is paired with more than one
∈
range value, the relation is not a function. Figure 2.1.7 represents some function f .
A
B
f x
Do you remember the meaning of the notation and ?
⊂
⊆
If not, see Book 1, Section 1.2.
y
A = Domain of f
C
C = Range of f (C
⊂ B)
Figure 2.1.7
In order to work with functions, we need suitable notation. We generally denote functions by lower case letters such as f , g , h , etc., although we occasionally use capital letters F , G , etc.
69 Domain and range
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Suppose we denote a function by means of f , and the ordered pairs by ( x, y).
D f denotes the domain of f .
R f denotes the range of f .
f (x) denotes the value of the function for a particular value of the independent variable x (i.e. f (x) = y ).
The equation y = f (x) defines a function in x .
The functional notation f (x) was first introduced by the Swiss mathematician Leonard Euler (1707–1783). We read
y = f (x) as “ y is equal to f of x ” or “ y is a function of x ” and we understand this to mean that y is the answer we obtain when we take the independent variable x, and apply to it the rule which determines how the ordered pairs are formed. For a function f , if we give the ordered pairs in terms of variables other than and y , for example if we have (a, b) f , then we write
x
∈
b = f (a).
We read this as “ b is equal to
f of a ”.
In this case we say that the equation b = f (a) defines a function in a , i.e. a is the independent variable and b is the dependent variable.
×
f (x) does not mean f x. The notation f (x) refers to the value in the range associated with the domain element x . If the ordered pairs ( x, y) are elements of f then we write y = f ( x) =
..... some rule ( such as an equation) which tells us what to do with x in order to obtain y.
∈
Hence f (x) is the value of y for some x D f . If x is an element of the domain of f , then f (x) is an element of the range of f .
f (x) is not the function f . The function f is the set of all ordered pairs (x, y) or ( x, f (x)) in R R, whereas f (x) is a single number in R. Remember
×
∈R ⊆ R×R f (x) = f. f (x) f
i.e.
70 For example, suppose f is the function defined by
y = f (x) = x 2 ; x
∈ N , 1 ≤ x ≤ 6.
Then
{
}
f = (1, 1), (2, 4), (3, 9), (4, 16), (5, 25), (6, 36) whereas
f (1) = 1, f (2) = 4, f (3) = 9, f (4) = 16, f (5) = 25, f (6) = 36. Functions also called mappings
Different words are sometimes used for functions. If we have a function f such that y = f (x) = x 3
−
− 3, then some authors may write f : x → x − 3.
i.e. if the function f is defined by y = x
They may refer to f as a “ mapping”, and read this statement as: “The function f maps each x D f onto y R f , where y is determined by the rule y = x 3.”
∈
−
∈
Read Definition 2.1.4 again. From the discussion immediately after Definition 2.1.4, we know that the word “each” in the definition is important. Consider the two graphs in Figure 2.1.8. y
y
x
x
Figure 2.1.8
Suppose we draw a vertical line through each graph. We then have the following figures.
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y y 4
2
2
x
x
–1
–2
Figure 2.1.9
Figure 2.1.10
In Figure 2.1.9
the definition of a function is satisfied, since the correspondence between the domain and range is one–to–one (for example 2 in the domain is paired only with 4 in the range), i.e. each element of the domain is paired with a unique element of the range.
every vertical line through the graph cuts it in only one place.
In Figure 2.1.10 The definition of a function
is not satisfied even though ments in the domain that are paired with unique elements in the range, namely the two
−
−
there are two speci fic ele-
x–intercepts.
the definition of a function is not satisfied, since the correspondence between the domain and range is one–to–many (for example 1 in the domain is paired with both 2 and 2 in the range), i.e. there are many elements in the domain that are associated with two different elements in the range
there is a vertical line (through the graph) which cuts the graph in more than one place.
We can thus apply the following useful test when a relation is represented graphically.
THE VERTICAL LINE TEST
×
Suppose a relation consists of some subset of R R, so that the elements of the relation are pairs represented by ( x, y), where x R and y R.
∈
∈
If at all x in the domain of the relation a vertical line cuts the graph of the relation only once, then the relation is a function.
If there is at least one x in the domain of the relation where a vertical line cuts the graph of the relation more than once, then the relation is not a function.
72
2.1.3 Identify which of the following relations (represented by sets of ordered pairs, a rule, or a graph) are functions? Explain why they are functions if they are, and why they are not, in the cases where they are not.
{
}
(a) f = (1, 3), (3, 1), (2, 4), (4, 2)
(b) y = g (x) = 3x 2 (c) The relation f shown in the following figure. y
−
y = f (x)
x
{ ∈ R × R: y = 1} (e) m = {(x, y) ∈ R × R: ( x − 1)
(d) k = (x, y)
2
+ (y + 2)2 = 4
}
(f) The relation g shown in the figure below. y
y = g ( x)
x –1
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(a) 1 2 3 4
1 2 3 4
The correspondence is one–to–one, hence f is a function. (b) For any real value we assign to x , we will obtain a unique value for y , and hence g is a function. (c) A vertical line through the graph at any poin t on the x axis will only cut the graph once. Hence f is a function.
−
(d)
Domain
Range
–10 1 – _2
1
0 3 100
Whatever real value we select for x , it is always mapped to the number 1. The correspondence is thus many–to–one and hence k represents a function. We discussed circles in Topic 1.
−
(e) We recognise that the equation represents a circle with centre (1, 2) and radius 2. In other words the domain of the relatio n is [ 1, 3]. A vertical line drawn through any x in the interval ( 1, 3) will cut the graph twice, and hence m does not represent a function. Note that we cannot say that because a vertical line through x = 1 or x = 3 cuts the graph only once, the relation is a function. This has to be true for all values in the domain.
−
−
−
{∈
≥− }
(f) Dg = x R : x 1 Any vertical line drawn through the graph of g for x > graph twice. Hence g is not a function.
−1 will cut the
74
2.1B SUBSTITUTION Read the Useful Hint in Study Unit 2.1A again. We have
y = f (x) = .... where the dots indicate a rule which tells us what to do with x in order to obtain y. This suggests that we may also think of a function as a process. Function considered as a process
We use the idea of a function as a machine which processes or transforms some input element to produce an output element. If f is a function then the set of input elements is the domain of f ; the set of output elements is the range of f .
Function machine
x
Processes x according to a rule for f
f (x) output element
input element Range of f
Domain of f
Figure 2.1.11
Compare this with the following example.
Figure 2.1.12
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Here tomatoes are the input elements, the machine is a liquidiser, the process is “liquidising food” and the output is tomato juice. Suppose a function is defined by a rule, for example, f (x) = x2 3x. We want to find the output when the function has operated on (or processed) a specific input element. We represent f (x) = x2 3x
−
−
in terms of “boxes” and we have
f( If the input element is the boxes. We have
2
)=
−3
.
♣, we find the output element by putting ♣ into each of
f(
♣ ↓
) =
♣ ↓
2
♣ ↓
−3
.
Then we carry out the operations indicated (i.e. squaring, multiplying by 3) and we have f( ) = 2 3 .
♣ ♣−♣
Now we dorule the same thing where the input element is a specific number. Using thewill same f (x) = x2 3x
−
and the input element 2, we find the output by putting 2 into each box.
f(
2
2
↓
↓
) =
2 2
↓
−3
.
We end up with
f (2) = 2
2
−3 2 .
Written in the usual mathematical way we have 2
f (2) = 2 = 4
=
−−63 × 2 −2.
76 Substitution
Check for yourself: do you remember what an expression is? See Book 1, Topic 1 if you have forgotten.
This process of replacing the variable x in the rule of a function with a number, or with some other variable or expression, is called substitution. Before moving on, check that you have understood the concepts of input, process and output by studying Table 2.1.1. We use the example above, where
f (x) = x 2
− 3 x.
The process or rule that defines the function can be expressed in words, as follows.
– square the number
x
• −→
– from the square, subtract the product of the number and 3
−→ •
f (x)
The function machine
Input (the number we choose)
Apply the function machine square subtract the product the number of the number and 3
–3 0 1 5
9 0 1 25
3( –9 3(0) –0 3(1) –1 25–3(5)
Output (the answer we find)
−3)
18
0 10
−2
Table 2.1.1
We note that
all possible input values form the domain of the function
the rule is expressed as y = x 2
the set of all the possible output values is the range of the function
some of the points that are generated by applying this process are ( 3, 18), (0, 0), (1, 2) and ( 5, 10).
−
− 3x
−
In Table 2.1.1 the input elements are all integers. The table shows the various operations that are carried out on these input elements to yield the output elements. Now consider the next example. In this case the input eleme nts are no longer only integers.
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2.1.2 Suppose we have the function f defined by
y = f (x) = x2 Consider the input elements
− 3x.
−1, a and x + 1. Find the corresponding function
values (i.e. output elements).
SOLUTION By substitution we know that if
f (x) = x2
− 3x f (−1) = (−1) − 3(−1)
then
2
= 1+3 = 4.
Also,
f (a) = a2
− 3a
which cannot be simplified further. Remember, if the input is a, we replace every
x with a , and then simplify if possible. To determine the output when the input is x + 1, we need to calculate f (x + 1), i.e. we replace each x with x + 1. We have
f (x + 1) = (x + 1)2
− 3(x + 1) − 3x − 3
= x2 + 2x + 1 = x2 x 2.
− −
This example shows that the output element (function value) may be a number, or an expression in the same variable as the variable that appears in the given rule, or an expression in some other variable . Now apply these ideas in the next activity.
2.1.4 (a) If f (x) =
−x
2
+ x + 4, calculate the following.
−
(i) f (1), f (2), f (3), 3 f (1), f ( 2)
− − f (x)
(ii) f ( x),
78 (iii) f (x + c), f (x) + c (iv) f (x2
− 1)
2 (b) If f (x) = 2x + 1 , determine the following. x 2
−
− − f (x − 2) f (x) − (x − 2)
(i) f ( 1), f (0), 2 f (3), f (2) (ii) f ( x) (iii) (iv)
(v) f (x2 + 1)
(a) If f (x) =
−x
2
+ x + 4, then we have the following answers.
(i)
f (1) = f (2) = f (3) =
−(1) −(2) −(3)
2
+1+4 = +2+4 = 2 +3+4 = 2
−1 + 1 + 4 = 4 −4 + 2 + 4 = 2 −9 + 3 + 4 = −2
3 f (1) = 3(4) = 12 f ( 2) = ( 2)2 + ( 2) + 4 =
−
−−
−
−4 − 2 + 4 = − 2
(ii)
−
f ( x) =
− f (x)
=
2
2
−(−x) + (−x) + 4 = −x − x + 4 −(−x + x + 4) = x − x − 4 2
2
(iii)
f (x + c) =
= = = f (x) + c =
2
−(x + c) + (x + c) + 4 −(x + 2cx + c ) + (x + c) + 4 −x − 2cx − c + x + c + 4 −x + (1 − 2c)x − c + c + 4 −x + x + 4 + c 2
2
2
2
2
2
2
(iv)
f (x2
− 1)
= = = =
2
2
4
2
2
−(x − 1) + (x − 1) + 4 −(x − 2x + 1) + x − 1 + 4 −x + 2x − 1 + x − 1 + 4 −x + 3x + 2 4
2
4
2
2
2
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2 (b) If f (x) = 2x + 1 , then we have the following. x 2
−
(i)
−
f ( 1) = f (0) =
2( 1)2 + 1 3 = = 1 2 3 2(0) + 1 1 = 0 2 2
− −− −
2 f (3) = 2
f (2) =
2 3 3
−
−1
−
1
()+ = 2
−2
2(2)2 + 1 9 = 2 2 0
2(19) = 38
which is undefined .
−
(ii)
−
f ( x) =
2( x)2 + 1 2x2 + 1 = ( x) 2 x 2
− − −
−−
(iii) 2
f (x
− 2) = 2((xx −− 22)) −+21
= =
2(x2 2x2
− 4x + 4) + 1 x−4 − 8x + 9 x−4
(iv)
f ( x)
− (x − 2)
= = = =
2x2 + 1 x 2 2x2 + 1 x 2x2 + 1
x2
− − (x − 2) − (x − 2) −2 − x + 4x − 4 x−2 + 4x − 3 x−2 2
2
(v)
f (x2 + 1) =
= =
2(x2 + 1)2 + 1 2
(x + 1) 2 2(x4 + 2x2 + 1) + 1 x2 1 2x4 + 4x2 + 3 x2 1
− −
−
80 In (b)(i) of Activity 2.1.4 you were asked to calculate f (2), but when 2 is substituted for x into the expression that represents f , we are unable to find an answer. Substituting x = 2 gives zero in the denominator of the expression, and it is thus undefined. In many cases we are told what the domain of a particular function is, for example we may have the function g defined by R
y = g(x) = 3x + 1, x . In this case we are given the information that the domain is
∈
R.
In other cases we may not be told what the domain is, but, as in (b)(i) of Activity 2.1.4, certain values of the independent variable lead to an undefined expression. For example, if the function k is defined by
y = k (x) =
1
−1
x
then, when x = 1 the denominator is zero, and k (x) will thus be undefined. How do we choose a suitabl e domain when the domain is not specifie d? For example, if k and l are functions defined respectively by 1
k(x) = x
−1
and l(x) = x + 1,
what are D k and D l ? NATURAL DOMAIN
In set notation, A
− B is the
set of all elements in A , excluding elements that are in
B. Hence
R
It is clear that x = 1 will be the only possible value of x for which k (x) is undefined; no values of x will give an undefined expression for l(x). We thus work according to the convention that the domain is the largest set for which the function is defined. This domain is called the natural domain of the function. Hence Dk = R 1 and D l = R. With reference to (b)(i) of Activity 2.1.4, x = 2 is not in the natural domain of f , i.e. D f = R 2 .
−{ }
−{ }
−{2} is the set
of all real numbers except 2.
2.1.3
Note that in this module we are only considering functions in R R.
×
(a) What is the natural domain of the function k defined below?
k(x) =
√x − 1
(b) State the natural domain of the functio n g , where
g( x ) =
x2
− 1.
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SOLUTION We know that the square root of a negative number is not a real number.
(a) The function k is defined for all x
i.e. such that
∈ R such that x − 1 ≥0 x
≥1.
{ ∈ R: x ≥ 1}.
Hence D k = x
2
∈ R such that x − 1 ≥ 0. Now − 1 ≥0
(b) The function g is defined for all x
x2
x2
⇔ ⇔ You may want to revise the solution of inequalities. See Topic 2 of Book 2.
x
Thus D g = x
≥ 1 or x ≤ −1.
{ ∈ R : x ≥ 1 or x ≤ −1}.
Try the following activity.
2.1.5 Find D f in each case.
(a) f (x) = Do not become confused because the variable given in (b) and (c) is no longer x. Remember that we can use any symbol for the variable.
≥1
2
1
(b) f (a) =
√x 1+−13a
(c) f ( p) =
p 1 5p + 1
−
82 (a) Because x2 + 1 1 we know x2 + 1 = 0, and hence f (x) exists for all x R ; i.e. we have D f = R .
≥
∈
(b) f (a) exists for all a
Dividing an inequality by
∈ R such that 1 − 3a ≥ 0
i.e. such that
a negative number changes
3a i.e. such that
the direction of the inequality sign.
− ≥ −1 a ≤ . 3 1 3
{ ∈ R : a ≤ }. f ( p) exists for all p ∈ R such that
Hence D f = a (c)
1
5p = −1
5p + 1 = 0
i.e. such that i.e. such that
− 15 .
p = 1 5
{ ∈ R : p = − }.
Hence D f = p
Before we move on to the following study unit try the next activity which deals with some concepts you learnt about in Topic 1 as well as some of the concepts discussed in this study unit.
2.1.6 The Fahrenheit and Celsius temperature scales are related by the formula
TF = 32 +
9 TC 5
where TF denotes temperature measured in degrees Fahrenheit and TC denotes temperature measured in degrees Celsius. (a) When the formula is expressed in this way, which is the independent variable and which is the dependent variable? (b) Explain why this equation defines a function. (c) Rewrite this equation in the form
TF = f (TC ) = ...... .
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(d) What is the value of the function f when TC = 0 (i.e. when the temperature is zero degrees Celsius, which is the freezing point of water)? (e) Can you describe D f ? (f) Set up a table of values by choosing six possible values of the independent variable and then using the given formula to calculate the corresponding values of the dependent variable. (g) Plot the points obtained in (f), using a suitable scale on each axis. (h) Can these points be joined and can the line or curve formed be extended in some way?
(a) The independent variable is TC . The dependent variable is TF . (b) For every possible value of TC , only one possible value of T F can be paired with it. Hence the set of ordered pairs (TC , TF ) is a relation in which each element of the domain is paired with a unique element of the range. (c) TF = f (TC ) = 32 + 95 TC (d) f (0) = 32 + 95 (0) = 32, i.e. the temperature 0◦C is equivalent to 32◦ F . (e) We know that TC will always be some real numbe r. If you have studied physics you may know that temperatures range from a theoretical minimum of approximately 273◦ C to over 200 million ◦ C (at the centres of certain stars known as massive blue stars). Mathematically we could describe D f as R, because the equation
−
9 TF = 32 + TC 5 is defined for all real numbers . However, in practice all the measurements of temperature will be rational numbers, never less than the theoretical minimum of approximately 273◦ C. (f) See the table below.
TC TF
− −10
0 1 0 2 0 3 0 40 14 32 50 68 86 104
(g) The independent variable (TC ) is represented on the horizontal axis. Values extend from 10 to 40. We mark off equal segments (the length we choose is not important) representing 10 units along the horizontal axis.
−
84 The dependent variable (TF ) is represented on the vertical axis. The values extend from 14 to 104. We mark off equal segments representing 10 units along the vertical axis.
TF 100 90 80 70 60 50 40 30 20 10 TC – 10
10
20
30
40
Figure 2.1.13 The dashed line in Figure 2.1.13 indicates that in practice many real numbers are excluded from the domain.
(h) Since it is possible to have an infinite number of variations in the measurements of TC we know that there will be infinitely many possible values of TF , and hence it makes sense to join the dots, as shown in the hashed line in Figure 2.1.13. From (e) we know that we can continue the line in both directions, up to the point that corresponds to TC = 273 on one side, and up to some unknown point on the other side.
−
EQUAL FUNCTIONS
A function may be defined by means of a particular rule or equation. Sometimes the two equations may look different but may in fact define the same function.
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For example,
y = 8 x3 + 1 and
y = ( 2x + 1)(4x2
− 2x + 1)
generate the same ordered pairs, and thus define the same function. While this seems to make sense, we need to know what we mean by “the same” function. We have the following definition.
Definition 2.1.5 Two functions f and g are equal if D f = D g and f (x) = g (x) for all x D f .
∈
2.1 1. The rule for a funct ion h states that “ h assigns to each natural number a value that is three times as big as that number”. Write down in functional notation the equation that defines h . 2. Consider the equation 2x + 3y = 2. (a) Explain why this equatio n defines a function in x. Write the rule in functional notation where
y = f (x) = ... . (b) Explain why this equation also defines a function in y . Write the rule in functional notation where
x = f (y) = ... . 3. Consider the function f defined by
y = f (x) = x 2 + 3, x
∈N . 0
(a) What is the domain of this function? (b) Use six domain values and set up a table of values for this function. (c) If you plot dots to represent the six ordered pairs you have found from the table of values, can you then complete the graph by joining the dots?
86 4. Write down D f for each of the functions f defined below.
−3x + 2 x x−1 √ f (x) = x + 2 √ f (x) = x + 2 √ f (x) = 2x 1 f (x) = √ 1− √ 2x − 1 f (x) = x + 3 + 1 √ f (x) = x − 3 + 1 √ f (x) = x
(a) f (x) = (b) f (x) = (c) (d) (e) (f) (g) (h) (i)
2
2 2
3
5. If a function f is defined by y = f (x) =
−3x + 1, find the following.
−
(a) f ( 1) (b) f (0) (c) f (x2 )
−
(d) f ( x) 2
6. If f (x) = (a) f (2) (b) (c) (d) (e)
xx 1 3, determine the following.
+−
f (−4) − f (4) f (x − 1) √ f ( x) f (a) f ( a)
−
7. The function g is defined by
g(x) =
√2x1+ 1 .
(a) Calculate (i) g( 14 ) correct to two decimal places
−
(ii) g(4). (b) Determine D g .
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8. Determine whether f = g , where the functions f and g are defined as follows. (a) f (x) =
x4 1 ; g(x) = x2 x2 + 1
(b) f (x) =
x2 1 ; g(x) = x x+1
− −
−1
−1
88
2.2 COMBINING FUNCTIONS
2.2A ADDITION, SUBTRACTION, MULTIPLICATION AND DIVISION We know that a function f in R
× R, defined by y = f (x)
consists of ordered pairs ( x, y)
has domain (input) and range (output) which are subsets of R
the elements of the domain are the values assigned to the independent variable x
the elements of the range are the values variable y , i.e. y = f (x).
f (x) obtained for the dependent
Just as machines can work together to perform more complicated tasks, so functions can be combined to form more complic ated functions. The most obvious way to produce new functions from existing ones is to use the four arithmetic operations (i.e. addition, subtraction, muliplication and division). For example, it would make sense to define the sum f + g of two functions f and g so that we have ( f + g)(x) = f (x) + g(x), See Topic 2 in Book 1.
i.e. we obtain the output element ( f + g)(x) by adding the two output elements f (x) and g(x). Since the outputs f (x) and g (x) are real numbers they obey all the rules that apply to operations on real numbers. Have a look at Figure 2.2.1.
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x
x
f ( x)
g ( x)
f ( x) + g ( x) = ( f + g) ( x )
Figure 2.2.1
Consider two functions f and g , and the four arithmetic operations, namely addition, subtraction, multiplication and division. We now define the sum , difference, product, and quotient functions.
Definition 2.2.1 The sum f + g of two functions f and g is defined by
( f + g)(x) = f (x) + g(x) for all x
∈ D f ∩ Dg .
Definition 2.2.2 The difference f
for all x
− g of two functions f and g is defined by ( f − g)(x) = f (x) − g(x)
∈ D f ∩ Dg .
90 Definition 2.2.3 The product f g (which we sometimes write as f of two functions f and g is defined by
·
·
× g or f g)
·
( f g)(x) = f (x) g(x) for all x
∈ D f ∩ Dg .
We usually write the product as f (x)g(x), without using the multiplication signs or .
· ×
Definition 2.2.4
The quotient f
÷ g (which we also write as gf ) of two functions
f and g is defined by
f g
for all x
(x) =
f (x) g(x)
∈ D f ∩ Dg such that g(x) = 0.
In these definitions note the following.
f + g, f
− g, f · g and gf are also functions.
We must think carefully about the domain of the new function. For example Definition 2.2.1 makes sense only if x is a number for which both f and g are defined, i.e. the domain of f + g is the set of all elements which are common to the domain of f and to the domain of g . Thus the domain of f + g is the intersection of the domains of f and g . We write
D f +g = D f
∩ Dg .
·
·
f + g = g + f and f g = g f but
f
− g = g − f and gf = gf
This is consistent with what we already know: addition and multiplication in R are commutative operations, but subtraction and division are not.
Commutativity is discussed in Topic 2 of Book 1.
In the case of the quotient of two functions we have to apply the additional condition that the denominator may not be zero.
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Table 2.2.1 summarises the definitions of the sum, difference, product and quotient functions generated by f and g , as well as the respective domains. It is important to memorise and understand this Table 2.2.1
Name Sum of f and g Difference f g between f and g Product of f and g
−
Quotient of f and g Note that we sometimes write
f g
Definition/rule Domain ( f + g)(x) = f (x) + g(x) D f +g = D f ( f g)(x) = f (x) g(x) D f −g = D f
−
−
( f g)(x) = f (x)g(x) f f (x) (x) = g( x ) g
·
as f /g.
Df g = Df
∩ Dg ∩ Dg Dg
∩ D f / g = D f ∩ Dg −{x ∈ Dg : g(x) = 0} ·
Table 2.2.1
The following definitions arise from the definitions summarised in Table 2.2.1.
The reciprocal of g , denoted by 1g , is defined as
1 g
(x) =
1 . g(x)
Note that this arises from the definition of the quotient function, where we have f equal to the constant function defined by f (x) = 1. We thus have
D 1/ g = R This applies to all powers of functions.
∩ Dg −{x ∈ Dg : g(x) = 0} = Dg −{x ∈ Dg : g(x) = 0}.
The square of f , denoted by f 2 , is defined as
f 2 (x) = f (x) f (x).
·
This arises from the definition of the product function. It is clear that Df2 = Df .
2.2.1 Consider the functions f and g defined by f (x) = 2 x Calculate ( f + g)( 6).
−
1 and g(x) = x2 + 3.
−
SOLUTION
−
− − − − −
( f + g)( 6) = f ( 6) + g( 6) = 2( 6) 1 + ( 6)2 + 3 = 13 + 39 = 26
−
92 Now try the next activity.
2.2.1 Suppose f and g are functions defined by
f (x) =
(a) Write down D f and D g .
1
x
and g(x) =
3
√x .
−
(b) In each of the following cases determine the equation of the function and write down the corresponding domain. (i) f + g
(ii) f
−g
·
(iii) f g
{ ∈ R : x = 3} { ∈ R : x ≥ 0}
(a) D f = x Dg = x
(b)
(iv)
f g
Division by zero is undefined. The square root of a negative number is undefined.
(i)
√ 1 −3 + x 3} D f ∩ Dg = { x ∈ R : x ≥ 0 and x =
( f + g)(x) = f (x) + g(x) = D f +g =
x
(ii)
(f
− g)(x)
− g(x) = x −1 3 − √x D f ∩ Dg = { x ∈ R : x ≥ 0 and x = 3}
= f (x)
D f −g = (iii)
·
√x = √x × x 3 x 3 f g D ∩ D = { x− ∈ R : x ≥ 0 and− x = 3}
( f g)(x) = f (x)g(x) = D f ·g =
1
(iv)
f g
(x) =
√−
f (x) 1/(x 3) = = g(x) x
∩ Dg −{x ∈ Dg : g(x) = 0} {x ∈ R : x > 0 and x = 3}
D f /g = D f
=
√x(x1− 3)
Note that x = 0 is excluded since g (0) = 0.
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Now that you have studied the solution for the activity, it may be a good idea to consolidate your understanding by trying the next activity. Note that the domain in each answer in Activity 2.2.1 is a set, so it must be expressed in suitable notation. For example, in (b)(ii) it is incorrect to write
D f −g = x D f −g is a set in
R, whereas
≥0
and x = 3.
x is a number in
R.
2.2.2 Suppose f , g and h are functions defined by
f (x) =
√x − 2,
g(x) =
− − −
1 x 2 and h(x) = 2 . x+1 x x 2
(a) Write down D f , D g and D h . (b) In each of the following cases determine the equation of the function and the corresponding domain. (i) 1 / f (ii) h /g (iii) f /g (iv) g h
−
{ ∈ R : x ≥ 2} Dg = { x ∈ R : x = −1}
(a) D f = x
The square root of a negative number is undefined. Division by zero is undefined.
{ ∈ R : x = −1 and x = 2}
Dh = x
(b)
h(x) =
(x
x−2 − 2)(x + 1) ,
and division by zero is undefined.
(i)
(1/ f )(x) = D 1/ f
=
(x) =
h( x ) g( x )
1 1 = f (x) x 2 x R : x>2
√−
{∈
}
(ii)
h g
=
(x
x−2 − 2)(x + 1)
1 x+1 1 x+1 = x+1 1 = 1
×
Dh / g = D h
∩ Dg −{x ∈ Dg : g(x) = 0}
94 Consider the two diagrams below. –1
2
Dh –1
Dg
Thus Dh Dg = x R : x = 1 and x = 2 . Now we co nsider x Dg : g (x) = 0 . Since g (x) = 1 we see that g (x) = 0 for all x+1 x R 1 , i.e. for all x Dg , and hence x Dg : g(x) = 0 = 0/ . Thus D h/g = R 1, 2 .
{} ∈ − {∈ ∩ ∈ −{− } ∈ − {− }
} {∈
}
Note
When we determine ( h/g) and find that (h/g)(x) = 1, we may be tempted to say that D h/g = R , but we have to keep in mind the way in which D h/g is defined. (iii)
f g
f (x) (x) = g(x)
=
√x − 2 1 x+1 x 1
√x
2
− D f ∩ Dg −{x ∈ Dg : g(x) = 0} {x ∈ R : x ≥ 2}∩{x ∈ R : x = −1}− 0/ { x ∈ R : x ≥ 2} = ( + )
D f /g =
= = (iv)
(g
− h)(x)
− h(x) − (x −x2−)(x2+ 1) x+1
= g(x) =
1
1 x+1 = 0
=
∩ Dh {x ∈ R : x = −1}∩{x ∈ R : x = −1 and x = 2} {x ∈ R : x = −1 and x = 2} R − {−1, 2}
Dg − h = Dg
= = =
− x +1 1
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2.2
√−
1. If f (x) = x 2 1 and g (x) = x 2, find each of the following (leave your answers in surd form where necessary).
−
(a) ( f + g)(2)
− g)(3) (c) ( f · g)(4)
(b) ( f
(d) ( f /g)(6)
2. If p(t ) = 1t and q (t ) = 3t + 1, find each of the following. (a) ( p + q)(1)
− q)(−3) · (d) ( p/q)(−4) (b) ( p
(c) ( p q)(2)
In Questions 3–6, for the given functions f and g, find an equation for each of the following.
(a) ( f + g)(x) (b) ( f
− g)(x) ·
(c) ( f g)(x) (d) ( f /g)(x) (e) ( f 2 2
2
− g )(x)
(f) ( f + g2 )(x) (g) D f +g (h) D f /g 3. f (x) = x 2 + 3x 4.
√ f (x) = x
g(x) = x + 5 g(x) = 3x
2
√− g(x) = 3x − 2
5. f (x) = x 6. f (x) = x 2
− 10
−1
g( x ) =
x2
1
−1
96
• The Cartesian plane is represented by R × R. R × R = {(x, y) : x ∈ R and y ∈ R} • A subset of R × R is called a relation. • Consider any relation r = {(x, y) : x ∈ A and y ∈ B}. A is the domain of the relation. The range of the relation is the set
{y : (x, y) ∈ r and x ∈ A}.
• Types of correspondence Type one–to–one
Example
1
4
3
2
5
0
one–to–many 2 4 –2
many–to–one 10
5
5 0
0
many–to–many 2 4
1 2
6 8
3 4
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• A function f is a set of ordered pairs in R × R in which different ordered pairs have different x −coordinates, i.e. each element of D f is paired with a unique element of Rf .
The equation y = f (x) = ... (i.e. some rule which tells us what to do with x in order to obtain y) defines a function in x. We call f a function in x .
Substitution
• Example If y = f ( x) = we may substitute x =
√x − 3 x
−1 and we obtain −1 − 3(−1) f (−1) = √ −1 + 3 = √ =
2.
• Natural domains of functions
When no domain is specified we work according to the convention that we use the biggest set of real numbers for which the equation that defines the function makes sense.
Equality of functions
• The functions f and g are equal if D f = Dg and
f (x) = g(x) for all x
∈ D f (or x ∈ Dg).
• The sum of functions ( f + g)(x) = f (x) + g(x) for all x
∈ D f ∩ Dg .
• The difference of functions ( f − g)(x) = f (x) − g(x) for all x ∈ D f ∩ Dg . • The product of functions ( f g)(x) = ( f
·
g)(x) = ( f g)(x) = f (x)g(x) for all x
×
• The quotient of functions f f ( x) ( f ÷ g)(x) = (x) = g g(x) for all x ∈ D f ∩ Dg such that g (x) = 0.
Df
∈ ∩
Dg .
98
CHECKLIST Now check that you can do the following.
SECTION 2.1 1. Identify the type of correspondence in any relation. Activity 2.1.2 2. Identify the domain and range in a given relation. Definition 2.1.3 3. Recognise when a relation is a function. Definition 2.1.4, Activity 2.1.3 4. Determine function values by substituting given domain values into the formula (equation) that defines the function. Example 2.1.2, Activity 2.1.4 5. Identify the natural domain of a function. Example 2.1.3, Activity 2.1.5 6. Identify dependent and independent variables in a specific example, and use functional notation correctly to express the relationship between the variables. Activity 2.1.6 7. Recognise when two functions are equal. Definition 2.1.5
SECTION 2.2 1. Obtain new functions from existing functions by addition, subtraction, multiplication and division. Definitions 2.2.1, 2.2.2, 2.2.3 and 2.2.4; Activities 2.2.1, 2.2.2 2. Calculate the function value when the function is obtain ed from the sum, difference, product or quotient of two other functions. Example 2.2.1 3. Determine the natural domain of sum, difference, product or quotient functions. Activity 2.2.2
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STRAIGHT LINES OUTCOMES After studying this topic you should be able to do the following.
SECTION 3.1: Drawing Lines
Draw a line using a table of values.
Find the slope of a line using the coordinates of two points on the line.
Draw a line using two points (we usually use the x – and y –intercepts).
Draw a line using one point on the line and the slope of the line.
Draw a line which is parallel or perpendicular to a given line.
Draw vertical and horizontal lines.
Draw graphs of linear functions with restricted domains.
SECTION 3.2: Finding Equations of Lines
Find the slope–intercept form of the equation of a line.
Find the point–slope form of the equation of a line.
Find the two–point form of the equation of a line.
Find the general equation of a line.
100 SECTION 3.3: Application of Lines and Linear Functions
Find the point of intersection of two lines.
Find the vertical distance between corresponding points on two lines, or between a point on a line and the corresponding point on the horizontal axis.
Apply your knowledge of linear functions and straight lines to simple problems involving linear cost and income functions, break–even points, profit and loss.
Sketch the graphs of linear inequalities.
Sketch the graph of the solution set of a system of linear inequalities.
Recognise equations which show direct or joint proportion and apply these concepts to various real–life situations.
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3.1 DRAWING LINES
3.1A GRAPHICAL REPRESENTATION OF A LINEAR EQUATION USING A TABLE OF VALUES From Definition 2.5.1 in Section 2.5 of Book 2 it follows that equation (3.1.1) is linear since it can be written as 2 x y 1 = 0.
We consider the following linear equation in two variables
y = 2x
− 1; x ∈ R.
(3.1.1)
− −
How can we represent this equation graphically? In order to answer this question we
For convenience we use integer values of x from 3.
−3 to
set up a table of values
plot the corresponding points on a system of axes
join the points in a meaningful way.
x y = 2x
−3 − 2 −1 0 1 2 3 − 1 −7 − 5 −3 −1 1 3 5 Table 3.1.1
102
y 5 4 3 2 1
–4
–3
–2
0
–1
x 1
2
3
4
–1 –2 –3 –4 –5 –6 –7
Figure 3.1.1
− −
− − − −
−
In Figure 3.1.1 we plot the points ( 3, 7), ( 2, 5), ( 1, 3), (0, 1), (1 , 1), (2, 3) and (3, 5) which we obtain from Table 3.1.1. It is obvious from Figure 3.1.1 that we can join all the points by means of one straight line that passes through all of them (see Figure 3.1.2). If you are not sure that this is so, you can choose any non–integer value of x between 3 and 3, calculate the corresponding value of y and plot the point you find. Repeat this process until you are convinced that the points whose coordinates satisfy the equation all lie on the straight line.
−
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y 5
y = 2x – 1
4 3 2 1
–5
–4
–3
–2
–1
0
x 12345
–1 –2 –3 –4
–5 –6 –7
Figure 3.1.2
In Figure 3.1.2 we note the following.
We have already discussed constants. See the discussion regarding polynomials in one variable after Activity 1.1.1 in Topic 1 of Book 2.
− 1,
The graph is a straight line and hence the equation y = 2x i.e. 2x y 1 = 0, is called a linear equation.
The y –intercept is
As the value of x increases so the value of y increases, i.e. the line slants upwards from left to right .
Since x R the dots can be joined and the line can be extended indefinitely in both directions.
− − ∈
−1, and this is the constant in the equation y = 2x − 1.
104 LINEAR FUNCTIONS
−
We also note that the equation y = 2x 1 expresses a relationship between x and y in which y is given in terms of x . From the graph in Figure 3.1.2 it is obvious that for each x –value there is a unique y –value and thus the equation defines a function in x . Hence we can write
y = f ( x) = 2 x The definition of a linear function is given in the next study unit. It is incorrect to write
l = y = 2x
− 1, since l is not
−1
and f is called a linear function . Figure 3.1.2 represents the graph of f . We sometimes refer to a graph such as this as the line l defined by y = 2 x 1, and write l : y = 2x 1,
−
−
where the colon means “defined by” or “represented by”.
equal to the equation.
Now try the following activity.
3.1.1 Consider the equation y =
−2x + 3.
(a) Set up a table of values, using integer values of x from the line l , where
l : y=
−3 to 3, and draw
−2x + 3.
(b) From the graph give the y –intercept of the line. (c) Does the line slant upwards or downwards?
(a) Table of values
x y=
−2 x + 3
−3 − 2 − 1 0 1 9
7
5 3 1
2
3
−1 −3
The line l is sketched in Figure 3.1.3. on the next page.
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y 9 8
y = – 2x + 3
7 6 5 4 3 2 1
x –4
–3
–2
–1
1234 –1 –2 –3
Figure 3.1.3
(b) The y –intercept is 3 (the same as the constant in y =
−2x + 3).
(c) The line slopes downwards from left to right, i.e. as the value of x increases so the value of y decreases.
In the next few study units we shall consider various techniques that will help us to draw lines without having to set up tables of values.
106
3.1B LINEAR FUNCTIONS AND LINES Equation (3.1.1) and the equation in Activity 3.1.1 are two examples of linear equations of the form y = mx + c, where m and c are constants. Both equations represent functions. In general we have the following definition.
Definition 3.1.1 Suppose m and c are fixed real numbers (i.e. constants). A function f defined by
y = f (x) = mx + c is called a linear function.
There are two special linear functions which have specific names.
If m = 0 then f (x) = c , and f is called a constant function . Note that f maps every real number onto the real number c .
If m = 1 and c = 0 then f (x) = x, and f is called the identity function. This function is called the identity function because it maps each real number onto itself. It is sometimes denoted by i , i.e.
i(x) = x for all x
See Exercise 3.1, question 13.
Note that when we speak about a “line” we mean a “straight line”.
∈ R.
Once you have worked through this section you will be able to draw the graphs of the functions defined by f (x) = c and f (x) = x . We shall leave this as an exercise. In Definition 3.1.1, the term “linear” is used because the graphs of all linear functions are straight lines. However, not all straight lines are the graphs of linear functions. Think, for example, of a line that is parallel to the y–axis. It has only one x –value in its domain, but this x –value is associated with infinitely many y –values. It therefore does not represent a function. In the next study unit we consider lines defined by the equation y = mx + c and consider how we interpret c and m .
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MAT0511/003
3.1C THE y–INTERCEPT AND SLOPE OF A LINE y–INTERCEPT The y–intercept of a graph is the y–coordinate of the point where the graph cuts the y–axis.
From Figures 3.1.2 and 3.1.3 in Study Unit 3.1A we see that the y –intercepts of the lines defined by y = 2x 1 and y = 2x + 3 are respectively 1 and 3. We can check this algebraically by substituting x = 0 into each of the equations since
−
−
−
xthen for yevery onobtain the y–axis. By substituting x = 0 into y = 2 x = 0into = 2xpoint + 3 we y = 2(0)
− 1,
i.e. y =
− 1 and
−1
and
y=
−2(0) + 3,
i.e. y = 3
which confirms what we see on the graphs. In general, if we substitute x = 0 into y = mx + c we obtain
y = m (0) + c, i.e. y = c . Thus c is the y –intercept of the line defined by y = mx + c. SLOPE or GRADIENT
1 The three lines l 1 : y = x + 1, l2 : y = 2x + 1 and l 3 : y = x + 1 are drawn 2 on the same system of axes in Figure 3.1.4. What do the lines have in common and how do they differ?
−
y l2
3
l1
2 1
l3
x –3
–2
–1
1
2
3
–1
Figure 3.1.4
From Figure 3.1.4 we see that all the lines have the same y –intercept. However, the directions of the lines are different, i.e. two lines slant upwards and one slants downwards; also each line has a different “steepness”. Mathematically we use the term slope or gradient to describe the differences in steepness and direction.
108
In everyday language we speak about a gradient of 1 in 100 which means that for every 100 units (e.g. 100 metres) travelled horizontally we
rise 1 unit (e.g. 1 metre) if we travel uphill, and
drop 1 unit (e.g. 1 metre) if we travel downhill.
It is clear then that a hill with a gradient of 1 in 30 is steeper than a hill with a gradient of 1 in 100.
3.1.2 Consider the line l, defined by y = 2x + 1, drawn in Figure 3.1.5. Points P (x1 , y1 ) and Q (x2 , y2 ) lie on the line and R is the point (x2 , y1 ). y
P ( x 1 , y1 ) 1
Q ( x 2 , y2 ) R ( x 2 , y1 )
– _12
x
y = 2x +1
Figure 3.1.5
(a) Calculate the lengths of QR and PR if x 1 = 1 and x 2 = 3.
y (b) Calculate x2 2
−y −x
1 1
using the values determined in (a).
(c) Compare your answer to (b) with the coefficient of x in y = 2x + 1.
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MAT0511/003
− × ×
(a) The length of QR = y2 y1 , and the length of PR = x 2 If x 1 = 1 then y 1 = 2 1 + 1 = 3. If x 2 = 3 then y 2 = 2 3 + 1 = 7. Hence QR = 7 3 = 4 and PR = 3 1 = 2.
−
−x . 1
−
(b)
y2 x2 y (c) The value of x2 2 y = 2x + 1.
−y −x
1 1
−y
1
x1
=
4 = 2 2
−
is the same as the coefficient of x in the equation
If we assume that the points P and Q on the line l in Figure 3.1.5 are P(1, 3) and Q(3, 7), we see that as we move from P to Q there is a vertical change of 4 units and a horizontal change of 2 units. The ratio of the vertical change to the horizontal change is 42 , i.e. 2. We say that the line has a slope of 2. This gives us the steepness and direction of the line. No matter which points P and Q we take on the line,we will always find that the difference between the y –coordinates of P and Q the difference between the x –coordinates of P and Q is equal to 2 as long as we subtract the coordinates in the same order in both the numerator and denominator. Note that if P (x1 , y1 ) and Q (x2 , y2 ) are two different points on the line then
y2 x2
−y −x
1
=
1
−(y − y ) y − y −( x − x ) = x − x 1
2
1
2
1
2
1
2
.
In other words, the answer is the same regardless of which point we use as the starting point, provided we subtract in the same order in the numerator and denominator. Now we consider the line with general equation y = mx + c. Suppose two different points P(x1 , y1 ) and Q(x2 , y2 ) lie on this line. Since P and Q lie on the line the coordinates of P and Q must satisfy the equation y = mx + c. Thus we have
y1 = mx1 + c and
y2 = mx2 + c. If we subtract the first equation from the second we obtain
y2
−y
1
= mx2
− mx
1
110 i.e. we have
− y = m(x − x ) y −y m = since x =x . x −x y2
and hence
1
2
2
1
2
1
1
2
1
From this expression it is clear that the value of m is the same no matter which two points we choose on a specific line. We call m the slope of th e line. It is determined by considering vertical change in relation to horizontal change. We have the following definition. Definition 3.1.2 If two different points P (x1 , y1 ) and Q (x2 , y2 ) are on a line l , then the slope (or gradient) of l is defined by m=
y2 – y1 x2 – x1
, x2 = x 1 .
Note
We have already shown that
y2 x2
−y −x
1 1
=
y1 x1
−y −x
2
.
2
Thus the formula for the slope of a line can be given by
m=
y2 x2
−y −x
1 1
or m =
y1 x1
−y −x
2
.
2
Different authors use different words to describe the way we determine the slope m of a line. The following all describe m .
m =
y2 x2
−y −x
1
=
1
change in y vertical change rise = = change in x horizontal change run
Now try the following activity.
3.1.3 Find the slope of each of the lines given in the following three figures. For easier referencing use m 1 for the slope of l 1 , m2 for the slope of l 2 , etc.
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y l3 l2 (2,2)
l1 (4,1)
x
(0,0)
( – 1 , – 3)
Figure 3.1.6
y
l6 l5
(0,1)
x
(– 1, 0)
l4 (1, – 2)
( 5 ,– 3 )
Figure 3.1.7
112
y
l8 l7 ( – 2 , 2)
(4 , 2)
x
( – 2 ,– 4 )
Figure 3.1.8
l8 is not defined by an equation of the form y = mx + c. See Study Unit 3.1E.
l1 :
m1 =
l2 :
m2 =
l3 :
m3 =
l4 :
m4 =
l5 :
m5 =
l6 :
m6 =
l7 :
m7 =
l8 :
m8 =
−0 = 1 −0 4 −0 = 2 = 1 −0 2 −3 − 0 = − 3 = 3 −1 − 0 − 1 −2 − (−3) = −2 + 3 = − 1 1−5 −4 4 0 − (−2) 2 −1 − 1 = −2 = − 1 1 − (−2) 3 = 0−1 −1 = − 3 2−2 0 = = 0 4 − (−2) 6 2 − (−4) 6 = which is undefined. −2 − (−2) 0 1 4 2 2
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What deductions can we make about different lines and their slopes from the graphs in Activity 3.1.3?
A line with positive slope slants upwards from left to right. See Figure 3.1.6.
A line with negative slope slants downwards from left to right. See Figure 3.1.7.
A horizontal line (i.e. a line parallel to the x –axis) has a slope of zero. See Figure 3.1.8.
The slope of a vertical line (i.e. a line parallel to the y –axis) is undefined.
For positive slopes , the bigger the slope , the steeper the line . In Figure 3.1.6 compare l1 , l2 and l3 . We see that l3 is steeper than l2 , and l2 is steeper than l 1 . We have
m3 > m 2 > m 1 , since 3 > 1 >
1 4
,
i.e. magnitude of m 3 > magnitude of m 2 > magnitude of m 1 . When we compare the magnitudes of the different slopes we are actually considering the numbers
For negative slopes, the smaller the slope (i.e. the bigger the magnitude of the slope) the steeper the line . In Figure 3.1.7, l 6 is steeper than l 5 and l 5 is steeper than l 4 . We have
m6 < m 5 < m 4 , since
without their signs.
− 3 < −1 < −
1. 4
Note however that magnitude of m 6 > magnitude of m 5 > magnitude of m 4 since 3 > 1 > 14 . We summarise the conclusions of this study unit as follows.
slope of l
: y = mx + c
l
line
l horizontal l vertical
l slants upwards l slants downwards l1 steeper than l 2
y–intercept of l
↓ ↓
equation
m=0 m is undefined (i.e. a vertical line cannot be described by the equation y = mx + c) m>0 m<0 magnitude of m 1 > magnitude of m 2
114
3.1D USING TWO POINTS, OR ONE POINT AND THE SLOPE, TO DRAW A LINE In Study Unit 3.1A we consider how to draw lines using tables of values. This is not the most efficient way of drawing lines. We now look at the following two methods.
If we know that two different points lie on the line (i.e. that their coordinates satisfy the equation of the line), we can plot these points and join them by means of a straight line. We call this the two–point method. Note that to draw an accurate line it is best to plot three well–spaced points instead of just two points.
If we know that one point lies on the line (i.e that its coordinates satisfy the equation of the line) and we know the slope of the line, we can plot the point and through the point draw a line with the given slope. We call this the point–slope method.
3.1.1 THE TWO–POINT
Draw the line l : y =
METHOD
−3x + 2 using two points on the line.
SOLUTION Although we can use any two points on l , the most convenient points to use are those with one of the coordinates equal to zero. Thus we first determine the x – and y –intercepts of the line l . To obtain the x –intercept, let y = 0. Then 0 = 3x + 2, i.e. x = 23 .
−
To find the y –intercept, let x = 0. Then y = 3(0) + 2 = 2.
−
Now we have two points, namely ( 23 , 0) and ( 0, 2), which lie on l . We plot these points and draw a line through them to give the line l. See Figure 3.1.9 on the next page.
115
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2
l : y = – 3x + 2 1
–2
0
–1
2 _ 3
x 1
2
–1
–2
Figure 3.1.9
3.1.2 THE POINT–SLOPE
Draw the lines l 1 and l 2 , where
METHOD
(a) the line l 1 has slope
2 3
−
and passes through the point ( 1, 2)
(b) l2 is the line with slope
−
2 3
and (1 , 1) is a point on this line.
SOLUTION (a) We can think of a slope of
2 3
as
change in y of + 2 change in x of + 3 or as
vertical change of + 2 units . horizontal change of + 3 units
−
We begin at the point ( 1, 2) and move two units up (positive direction) and three units to the right (positive direction) and arrive at the point (2, 4).
116 We plot these two points and draw a line through them. y
6 3 (2,4) 2 (– 1 , 2)
2
–2
0
–4
x 2
4
–2
Figure 3.1.10
(b) We can write
−
2 3
as
− 2 or as 3
2
−3 . Suppose we use
− 2 . We interpret − 2 as 3
3
−
vertical change of 2 units . horizontal change of + 3 units Thus we can start at the point (1 , 1) and move two units down (i.e. in a negative direction) and three units to the right (in a positive direction) to arrive at (4, 1). We plot the two points and draw a line through them.
−
Alternatively we can use
2
−3 and interpret this as
vertical change of + 2 units . horizontal change of 3 units
−
We start at (1, 1) and move two units up and then three units to the left and arrive at ( 2, 3). Now we plot and join the points (1 , 1) and ( 2, 3). See Figure 3.1.11 on the next page.
−
−
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y 4 3 (– 2 , 3) 2
2
(1,1)
–4
x
0
–2
2
2
4 ( 4 , – 1) 3
–2
Figure 3.1.11
Now try the following activity.
3.1.4 An equation of the form ay + bx + d = 0 where a = 0 can be written as y = mx + c, where m = ba and c = da .
−
−
The equation of the line l is given by 2y + 4x
− 3 = 0.
(a) Use the x – and y –intercepts of l to draw the line l . (b) Rewrite the equation in the form y = mx + c and then draw the line l by using the slope of the line and a point through which the line passes.
(a) To obtain the x –intercept, let y = 0. Then 2(0) + 4x
− 3 = 0, i.e. x = 34 .
To obtain the y –intercept, let x = 0. Then 2y + 4(0)
− 3 = 0, i.e. y = 32 .
118 We plot the points ( 34 , 0) and (0, 32 ), and draw a line through them. y
_3 2
l
3 _ 4
0
x
Figure 3.1.12
(b) 2y + 4x
−3 = 0
⇔
2y =
−4 x + 3
⇔
y=
−2x + 32
−
The slope of l is thus 2. We know from (a) that l passes through the point ( 34 , 0). We use ( 34 , 0) and the slope of 2 to sketch the line. We interpret the slope of 2 as −12 , i.e. as
−
−
−
vertical change of 2 units . horizontal change of + 1 unit The line is sketched in Figure 3.1.13 on the next page.
119
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l ( _34 ,0)
x
0
2
1
( 7_4 , – 2)
Figure 3.1.13
3.1E RELATED PAIRS OF LINES We now look at pairs of lines which have special slope properties. Remember: Vertical lines do not have a slope, and thus
PARALLEL LINES
must be treated separately. It is clear that vertical lines are also parallel.
Two (or more) non–vertical lines are parallel if they have the same slope. Conversely, if two (or more) lines have the same slope they are parallel.
3.1.3 Suppose l 1 and l 2 are the two lines
l1 : y = 12 x + 1 and l2 : y = 12 x + 3. The two lines are parall el since they have the same slope. We draw the lines using the y–intercept of each line and the point–slope technique described in
120 Example 3.1.2. See Figure 3.1.14 below. y
1
y = _2 x + 3 y = _21x +1 2 3
1
x
0
l2 l1
Figure 3.1.14
From Figure 3.1.14 we notice that there is a vertical distance of 2 units between the two lines. Since the lines are parallel we can also obtain l 2 by shifting l 1 two units upwards, or l 1 by shifting l 2 two units downwards. This means that if we have any two non–vertical lines which are parallel, then we can obtain the one from the other by a suitable vertical shift, either upwards or downwards.
PERPENDICULAR LINES Remember: Horizontal lines have slope 0 and vertical lines have no slope, and we deal with them separately. Horizontal and vertical lines are perpendicular to each other.
Consider two lines which are not parallel to either the x–axis or y –axis and which have slopes m 1 and m 2 . If m 1 m2 = 1 then the lines are perpendicular. Conversely, if the lines are perpendicular then m 1 m2 = –1. We say that when two lines are perpendicular their slopes are 1 negative reciprocals of each other, i.e. m 2 = – . m1
×
−
×
The slope property of perpendicular lines is not as obvious as that of parallel lines. You need not be able to prove this property; you just need to apply it.
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3.1.4
−
The line l 1 is defined by the equation 2 x 3y = 6. The line l 2 passes through the point (1, 2) and is perpendicular to l 1 . Draw the two lines on the same system of axes.
SOLUTION The equation 2x 3y = 6 can be written as y = 23 x 2. Hence the slope of l 1 is 23 and the y –intercept is 2. Since l 2 is perpendicular to l 1 , the slope of l 2 is 32 .
−
Note: If m1 m2 = 1 then 2 = 1, i.e. m2 = −21 , i.e. 3 m2
m2
− =− . 3 2
−
3
− −
−
We again use the point–slope technique described in Example 3.1.2 and we obtain the following graphs.
y 5 4 3 2
(1,2)
l1
1
–5
–4
–3
–2
–1
0
x 1 2 3
–1 –2 –3 –4 –5
Figure 3.1.15
5 46 (3 , – 1)
l2
122 The following activity gives you practice in drawing lines such as the ones we have been describing.
3.1.5 The lines l 1 , l 2 and l 3 are such that
−1
l1 is defined by y = 2x
l2 is parallel to l 1 and passes through the point ( 1, 1)
l3 is perpenduclar to l 1 and passes through the point (2 , 3).
−
−
Draw the three lines on the same system of axes.
The equation of l 1 is y = 2x 1. Hence the slope of l 1 is 2 (i.e. 21 ) and the y–intercept is 1. We thus begin at 1 on the y–axis, and move 2 units
−
−
upwards and 1 unit to the right since the slope is positive. This gives us the point (1, 1) and we draw the line through (0 , 1) and (1 , 1).
We can also move 2 units downwards and 1 unit to the
left.
−
−
−
l2 is parallel to l 1 and thus has slope 2. It passes through ( 1, 1). We thus draw the line through ( 1, 1) that is parallel to l 1 .
−
We can also move 1 unit downwards and 2 units to the right.
l3 is perpendicular to l 1 and thus has slope 12 . It passes through (2 , 3). We thus begin at the point (2, 3), and move 1 unit upwards and 2 units to the left since the slope is negative. This gives us the point (0 , 2) and we draw the line through (0, 2) and (2 , 3).
−
−
−
−
The graphs are shown in Figure 3.1.16 on the next page.
−
−
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y 5 4 (0,3)
l2 l1
3 2
(– 1 , 1 )
–5
–4
–3
–2
1
–1
0
(1,1)
x 1 ( 0 , – 1)
2345
( 0 , – 2) –3
( 2 , – 3)
–4
l3
–5
Figure 3.1.16
In the last example in this study unit we consider how to draw vertical and horizontal lines.
3.1.5 On the same system of axes draw the lines l 1 and l 2 , where
−3
l1 : y =
l2 : x = 2.
124
SOLUTION
−
The equation for l 1 has the form y = mx + c, where m = 0 and c = 3. Thus l 1 is horizontal (see the summary at the end of Study Unit 3.1C), i.e. parallel to the x–axis, and has y –intercept 3. Thus we obtain l 1 by drawing a line parallel to the x –axis passing through the point (0, 3). (The equation y = 3 means that for each value of x , y = 3.)
−
−
−
Note: x = 2 cannot be written as y = mx + c.
−
We have not yet come across an equation for a straight line of the form x = c . The equation x = 2 means that x = 2 for all values of y. If we plot any two points with x –coordinate 2, for example (2 , 1) and (2, 1), and join them with a straight line, we have a straight line parallel to the y –axis which has x –intercept 2.
−
The two lines l 1 and l 2 are drawn in Figure 3.1.17. y 3
l2 2 1
(2,1)
x
0 –3
–2
–1
–1
1
2
3 ( 2, – 1)
–2 –3
l1
–4
Figure 3.1.17
Note that l 1 represents a linear function, but l 2 does not. Can you explain why?
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3.1F GRAPHS OF LINEAR FUNCTIONS WITH RESTRICTED DOMAINS We stated in Study Unit 3.1B that graphs of linear functions are straight lines.
Throughout this numbers. section we have assumed that the equations defining lines were valid for all real Suppose we have a linear function f which is defined on a domain that does not include all possible values of x for which the equation makes sense. We then say that the domain is restricted. How do we draw the graph of f in a case such as this? Look at the following example.
3.1.6 Suppose f is the linear function defined by
y = f (x) =
−3x + 2,
{ ∈ R : −1 < x ≤ 2}.
where D f = x
−
(We have already drawn the line l defined by y = 3x + 2 in Example 3.1.1 in Study Unit 3.1D.) Note that D f is a half–open, half–closed interval, with endpoints x = 1 and x = 2; 1 / D f and 2 D f .
−
− ∈ ∈ We consider the endpoints −1 and 2. Now f (−1) = −3(−1) + 2 = 5 and f (2) = −3(2) + 2 = −4. Hence R f = {y ∈ R : −4 ≤ y < 5}. We know that −4 ∈ R f because −4 = f (2) and 2 ∈ D f . However, even though f (−1) = 5, −1 ∈ / D f and hence 5 ∈ / Rf . The graph of f is shown in Figure 3.1.18.
126
y 5
(– 1, 5)
4 3 2
y = f ( x)
1
–3
–2
–1
_2 3
x 1
2
3
–1 –2 –3 –4
(2 , – 4)
Figure 3.1.18
−
We use a solid dot to indicate that the point (2 , 4) is included whereas the open dot indicates that the point ( 1, 5) is excluded.
−
Now try the following activity. STOP! By now you have read this comment many times. Have you actually been doing
the activities yourself before reading the solutions? Only you know the answer to this question, but we emphasise again that the more you try to do on your own the more you will understand!
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3.1.6 Draw the graph of the function f , where
y = f ( x) = 3 x
−3
and
{ ∈ R : x < 2 }.
Df = x
The y –intercept is
−3 and the gradient is 3.
We note that f is defined on the open interval ( namely 2. Now f (2) = 3(2) 3 = 6 3 = 3.
−
−
−∞ , 2) with only one endpoint,
The graph of f is shown in Figure 3.1.19. It extends indefinitely downwards since R f = ( , 3).
−∞
y 3
(2,3)
2
y = f ( x) 1
x –3
–2
–1
1 –1 –2 –3 –4
Figure 3.1.19
2
3
128
3.1
−
In each of questions 1 4 find the x – and y –intercepts and use them to draw the line defined by the given equation.
1. y = 3x
−1
2. 2x + y = 2 3. x + 2y = 4 4. 2x
− 4y = 6 −
In each of questions 5 8 calculate the slope, if it exists, of the line which passes through the given pair of points.
− −
5. (3, 4); ( 1, 2)
−
6. (2, 7); (2, 4)
− − 8. (−3, 4); (6, 4)
7. ( 1, 2); (1, 2)
In each of questions 9 point and has slope m .
−12 draw the line which passes through the given
−
m=4
10. (2, 3)
m=0
9. (0, 1)
11. (1, 2)
−
m=
−
12. (2, 0)
m=
2 3
1 4
13. Suppose f is the constant function defined by y = f (x) = 1 and i is the identity function defined by i (x) = x for all x R. Draw the graphs of f and i on separate systems of axes.
∈
14. On the same system of axes draw the lines l 1 , l 2 , l 3 , l 4 and l 5 where
l1 : l2 : l3 : l4 : l5 :
− −
y = 2x + 1 y = x+1 y=1 y = x+1 y = 2x + 1.
What is the slope of l 3 ?
−
−
15. The line l 1 passes through the points ( 3, 8) and ( 1, 3). The line l 2 passes through ( 1, 8) and ( 3, 2). The line l 3 passes through ( 1, 1) and (4, 1).
− − −
−
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(a) Without drawing the lines, how do you know that l 2 is parallel to l 1 ? (b) Without drawing the lines, how do you know that l 3 is perpendicular to l 2 ? 16. Find p if the line l1 , through (1, 2) and (2, 5), is parallel to the line l2 which passes through (0, 2) and (2, p).
−
− −
−
17. Find q if the line l 1 , through ( 1, 1) and (2 , 3), is perpendicular to the line l , through (2 4) and q 1 . 2
,
( , )
−
18. Suppose line l1 is defined by 2 y + x = 1. Line l2 is parallel to l1 and passes through the point (1, 2). Line l3 is perpendicular to l2 and passes through the point ( 1, 2). Draw all three lines on the same system of axes.
− −
19. Suppose l 1 and l 2 are the two lines
l1 : y =
−2 x + 3
and l2 : 2y + 4x + 1 = 0.
Explain, without drawing the lines, how you could draw l2 if you had already drawn l 1 . 20. See the comment immediately below Figure 3.1.17. Can you explain why l1 represents a function, and why l 2 does not?
−
In each of questions 21 24 the function f is defined by y = f (x) = 2 x Draw the graph of f for the given domain.
{−2, −1, 0, 1, 2} 22. D f = {x ∈ R : x > 1} 23. D f = {x ∈ R : −2 < x ≤ 1} 21. D f =
24. D f = R (i.e. the natural domain of f )
− 1.
130
3.2 FINDING EQUATIONS OF LINES
3.2A SLOPE–INTERCEPT METHOD, POINT–SLOPE METHOD AND THE TWO–POINT METHOD In Section 3.1 we concentrated on drawing lines according to information given about the gradients and points lying on the lines. We now consider how we can work backwards, and determine from any given line the equation that defines it. We can obtain the equation of any line if we know
Slope–intercept method
the slope and the y–intercept of the line (we use the slope–intercept method and obtain the slope–intercept form of the equation)
the slope of and a point on the line (we use the point–slope method and obtain the point–slope form of the equation)
any two points on the line (we use the two–point method and obtain the two–point form of the equation).
Look at Figure 3.2 .1 below. It shows a line l with slope m and y–intercept c. Suppose P (x, y) is any point on the line l . y
P (x , y ) y–c (0, c )
x –0 0
l
Figure 3.2.1
x
131
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Since (0, c) and ( x, y) are two points on l and the slope is m we have m =
y x
−c − 0.
Now
y x
−c −0
mx = y
−c
m=
⇔ ⇔
y = mx + c.
This is just one of the possible forms of the equation of a line. We call this the slope–intercept form, and this was the first form given in Study Unit 3.1B.
SLOPE–INTERCEPT FORM OF THE EQUATION OF A LINE
The equation of a line that has slope m and y –intercept c is y = mx + c.
Point–slope method
(3.2.1)
If we know the slope of a line l and one point (other than the y–intercept) that lies on the line, how do we obtain an equation that defines the line? Figure 3.2.2 is the sketch of a line l with slope m. The line goes through the point Q(x1 , y1 ). Let P (x, y) be any other point on the line. y
P (x , y) y – y1 Q (x1 , y1 )
x–x 1
0
l
Figure 3.2.2
x
132 Since P (x, y) and Q (x1 , y1 ) are points on the line l and the slope is m we have
m=
y x
i.e. we have the equation
y
−y
1
−y −x
= m (x
1 1
−x ) 1
and this is anothe r form of the equation of a line. We call this the point–slope form. POINT–SLOPE FORM OF THE EQUATION OF A LINE
The equation of a line that passes through the point Q (x1 , y1 ) and has slope m is (3.2.2) y – y1 = m (x – x1 ) .
Two–point method
If we have two points A (x1 , y1 ) and B (x2 , y2 ) on a line l then we can also find the equation of l . According to Definition 3.1.2, the slope m of l is
m=
y2 x2
By using equation (3.2.2) we obtain
y
−y
1
=
−y −x
1
− y2 x2
y1
−x
1
.
1
(x
− x ). 1
This form is known as the two–point form.
TWO–POINT FORM OF THE EQUATION OF A LINE
The equation of a line that passes through the points P (x1 , y1 ) and Q (x2 , y2 ) is y – y1 =
y2 – y1 x2 – x1
(x – x1 ).
(3.2.2)
Equation (3.2.3) can easily be simplified, and the equation of a line would not usually be left in this form. The terms in the point–slope form of the equation of a line can be rearranged to obtain the slope–intercept form. You can see this in the following activity.
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3.2.1 The line l has slope
−
4 5
−
and passes through the point ( 1, 2).
Give the equation for l in
(a) point–slope form (b) slope–intercept form.
(a) Substituting m =
y
−
4 5
−
and (x1 , y1 ) = ( 1, 2) into equation (3.2.2) we get
− 2 = − 45 (x + 1).
Point–slope form.
(b) We rearrange the terms in the point–slope form of the equation. Thus 4
y
− 2 = − 5 (x + 1)
⇔
y=
− 45 x − 45 + 2
⇔
y=
− 45 x + 65 .
Slope–intercept form.
Activity 3.2.1 shows that once we have the equation in point–slope form we can rewrite it in slope–intercept form. We can also rewrite the two–point form in point–slope form and in slope–intercept form.
134
3.2.2
−
−
Suppose the line l passes through the points (5 , 2) and ( 3, 4). Write the equation for l in two–point form, and then simplify it so that we have the equation in
(a) point–slope form (b) slope–intercept form.
−
−
Suppose we have P (x1 , y1 ) = (5, 2) and Q (x2 , y2 ) = ( 3, 4). From (3.2.3) we have 4 ( 2) y ( 2) = (x 5) . (1) 3 5
−−
−−
−
− (−2) =
− 34 .
−−
(a) Now the slope m of l is
m=
4
3
5
−−
6 = 8
−
If we simplify (1) we obtain the point–slope form of the equation, i.e. we have
y
− (−2) = − 34 (x − 5)
and the equation becomes
− 34 (x − 5). (2) Note that if we interchange P and Q , and use the point ( −3, 4), the point– y+2 =
slope form of the equation becomes
y
− 4 = − 34 (x − (−3))
y
− 4 = − 4 (x + 3).
i.e. 3
Note that (2) and (3) are equivalent, i.e. they have the same solutions. (b) If we use equation (2) we obtain
y=
− 34 x + 154 − 2
y=
− 34 x + 74 .
i.e.
(3)
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If we use equation (3) we obtain
y=
− 34 x − 94 + 4
y=
− 34 x + 74 .
i.e.
Obviously in practice we do not need to obtain the answer in more than one way.
3.2B SOME SPECIAL LINES We now return to the slope–intercept form of the equation of a line, namely
y = mx + c. We recall that the slope of a horizontal line is m = 0. Hence if l is a horizontal line (i.e. l is a line parallel to the x–axis) that passes through the point ( p, q), then its equation is y = q , which has the form y = 0x + q (see Figure 3.2.3). The special line with equation y = q is the graph of the constant function referred to in Study Unit 3.1B. A vertical line does not have a slope and thus it does not have an equation of the form (3.2.1) or (3.2.2). However, it follows from Figure 3.2.3 that if l is a vertical line (i.e. l is a line parallel to the y–axis) that passes through the point ( p, q), then its equation is x = p. y
y=q q
( p,q)
x=p
p
0
Figure 3.2.3
x
136 It has already been stated that any function f defined by
y = f (x) = mx + c is called a linear function, and the graph of every linear function is a straight line. Figure 3.2.3 shows that not all lines represent linear functions. The vertical line defined by x = p has the same x–value for every y–value, and hence does not represent a function.
HORIZONTAL AND VERTICAL LINES
The equation of the horizontal line through the point ( p, q) is
y = q. (Slope m = 0.) The equation of the vertical line through the point ( p, q) is
x = p. (It has no slope.)
3.2C THE GENERAL EQUATION OF A LINE It would be useful if we could find some general equation which would describe any line, including vertical lines. The following equations
y = 2x + 1
y
y 2=
x=
− 4 = 3(x + 1)
Slope–intercept form Point–slope form Horizontalline
−1
can be rewritten as
Verticalline
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−y+1 = 0 3x − y + 7 = 0 0x + y − 2 = 0 2x
x + 0y + 1 = 0
respectively, and each equation has the form
px + qy + r = 0 where p, q and r are constants, and where p and q are not both zero.
GENERAL EQUATION OF A LINE
Every line has an equation of the form
px + qy + r = 0,
(3.2.4)
where p, q and r are constants and p and q are not both zero. Also, every equation of the form of (3.2.4) where p and q are not both zero, is the equation of a line.
If q = 0 then (3.2.4) can be written as
y=
− pxq − qr
which is in slope–intercept form and represents a line with slope y–intercept qr .
−
The special case where p = 0 gives y = line with y –intercept qr .
−
− qr which is the equation of a horizontal
If q = 0 but p = 0 then (3.2.4) can be written as x = r . vertical line with x –intercept p
−
− qp and
− pr , which represents a
138
3.2.3 Refer to Example 3.1.4 in Study Unit 3.1E. The line l1 is defined by the equation 2x 3y = 6. The line l 2 passes through the point (1 , 2) and is perpendicular to l 1 .
−
Find the equation of l 2 in the form px + qy + r = 0.
Since 2x 3y = 6 can be written as y = 23 x 3 Thus the slope of l 2 is . 2
−
−
− 2 it follows that the slope of l
1
is 23 .
Let the equation of l 2 be y = mx + c. 3 Since the slope is we have 2
−
y=
− 32 x + c.
(1)
The line passes through the point (1 , 2). We thus substitute x = 1 and y = 2 into (1) and obtain 2=
− 32 (1) + c.
We thus have
c = 2+ 3 2 and thus
c= Hence the equation for l 2 is
y=
7 . 2
− 32 x + 72
which we can write in the form 2y + 3x
− 7 = 0.
We have said that practice is essential, but you do not need to do all the questions in the following exercise now. You may want to do the even–numbered questions now and the odd–numbered questions later.
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3.2
−
In each of questions 1 22 find a general equation of the form px + qy + r = 0 for the line l with the stated properties.
1. The slope of l is 4 and the y –intercept is 2. The y –intercept of l is 3. The slope of l is
2 3
−
3 2
− and the slope is −
1
.
4 2 . 7
−
and l passes through (2, 5).
−
4. The point ( 1, 3) lies on l and the slope of l is 7. 5. l passes through (2, 1) and (1 , 2).
−
−
6. Two points on l are ( 4, 3) and (5 , 7).
−
7. l is parallel to the x –axis and passes through (2 , 3).
−
8. l is perpendicular to the x –axis and passes through (2 , 3).
−
9. l is parallel to the y –axis and passes through (7, 10). 10. l is perpendicular to the y –axis and passes through (7, 10).
−
11. l is the x –axis. 12. l is the y –axis.
13. l passes through (0, 1) and is parallel to the line with equation y = 3x + 1.
−
14. l passes through ( 1, 1) and is parallel to the line with equation 2x + 3y 4 = 0.
−
− 16. l is parallel to the line with equation y = 2 and passes through ( −1, 0). 15. l is parallel to the line with equation x = 1 and passes through ( 1, 0).
17. l passes through (0, 1) and is perpendicular to the line with equation y = 3x + 1.
−
18. l passes through ( 1, 1) and is perpendicular to the line with equation 2x + 3y 4 = 0.
−
19. l is perpendicular to the line with equation x = 1 and passes through (2 , 3). 20. l is perpendicular to the line with equation y = 2 and passes through (2 , 3).
−
21. l passes through ( 7, 1) and is parallel to the line that passes through (5, 2) and ( 3, 3).
−
140 22. l passes through (2, 5) and is perpendicular to the line that passes through (5, 2) and ( 3, 3).
−
−
In each of questions 23 26 rewrite the equation in slope–intercept form and then give the slope and y –intercept of the line defined by the equation.
23. 2 x
3y + 4 = 0
− 24. −2x + 5y − 2 = 0 −3(x + 2) 26. y − 3 = 2(x − 1) 25. y + 7 =
−
27. Two lines l1 and l2 are perpendicular to each other and intersect at ( 1, 3). The slope of l 1 is 34 . (a) What is the slope of l 2 ? (b) Determine the y –intercept of l 1 . (c) Determine the x –intercept of l 2 . In each of questions 28 31 the graph of a linear function f is drawn. Give a formula for f and write down D f and R f .
−
28.
29. y
y
(0,1)
0 (1,0)
2
x
–3
x
0 –2
141 30.
MAT0511/003 31. y
y (– 1 , 1 )
x
0
x 0
(– 3, – 2 )
( 8,– 2 )
( 2 , – 2)
32. Equations (3.2.1), (3.2.2) and (3.2.3) respectively give the slope–intercept, point–slope and two–point forms of the equatio n of a line. The slope– intercept form involves the y –intercept. Suppose you have both the x – and the y–intercepts of a line and both are non–zero (see the sketch below). Show that x y + = 1. a b This is called the two–intercept form of the equation of a line. y
b
l a
x
142
3.3 APPLICATIONS OF LINES AND LINEAR FUNCTIONS
3.3A THE INTERSECTION OF LINES AND THE VERTICAL DISTANCE BETWEEN LINES POINT OF INTERSECTION
In Topic 2 of Book 2 we solved systems of two equations in two unknowns algebraically. In this topic we have learnt how to represent linear equations graphically. Can we now interpret the solution of a system of linear equations graphically? We answer this question in the following example.
3.3.1 Consider the following system of equations 3x x+ 4yy = = 121
−
−
.
We represent the two equations graphically in Figure 3.3.1 by drawing the lines l1 : 3x + 4y = 12 and l 2 : x y = 1. (Note that the x –intercept of l 1 is 4 and the y –intercept is 3. The x –intercept of l 2 is 1 and the y –intercept is 1.)
−
−
−
y
l2 : x – y = – 1 3
P
1
–1
l1 : 3x + 4y = 12 x
0
4
Figure 3.3.1
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The lines intersect at the point P. All the points on line l1 satisfy the equation 3x + 4y = 12 and those on line l 2 satisfy the equation x y = 1. Since P is the point of intersection of the lines it lies on both lines and hence the coordinates of P satisfy both equations. Thus the coordinates of P give the solution of
−
We find the coordinates of P in Example 3.3.2.
3x + 4y = 12 x y = 1
−
−
−
.
Although in theory we can solve a system of linear equations graphically we do not usually do so, since regardless of how accurate the sketch is we are not necessarily able to read off the exact coordinates of the point of intersection. To determine the exact coordinates of the point of intersection we need to use algebraic methods. Although we do not usually solve systems of linear equations graphically, graphs enable us to visualise cases in which systems of linear equations can have
only one solution (as in Example 3.3.1)
no solution
infinitely many solutions.
The next activity illustrates the last two situations.
3.3.1 (a) Suppose
−1 − 2y + 4 = 0 .
l1 : y = x l2 : 2x
and
Draw these two lines on the same system of axes. What can you dedu ce about the number of solutions of the following system?
2x
−
y = x 2y + 4 = 0
−1
(b) Suppose
l3 : y = 2x + 1 l4 : 4x
and
− 2y + 2 = 0 .
Draw these two lines on the same system of axes. What can you dedu ce about the number of solutions of the following system?
4x
−
y = 2x + 1 2y + 2 = 0
144 (a)
y
l2 2
l1 x
0
–2
1 –1
Figure 3.3.2
We note that the two lines hav e the same slope. We can deduce this algebraically, since the equation for l2 can be rewritten as 2 y = 2 x + 4, i.e. as y = x + 2, and thus the slope of both l 1 and l 2 is 1. Since the lines are parallel and have different y–intercepts, they never intersect. Since there are no points of intersection , the system 2x has no solution .
−
y = x 2y + 4 = 0
−1
(b) y
l4
1
– 1_2 0
l3
Figure 3.3.3
x
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Note that the two lines are the same, i.e. the one line lies on top of the other and hence all points on the lines are points of intersection of the two lines. When we rewrite the equation of l 4 , namely
Lines l3 and l4 are said to coincide, and are called coincident lines.
4x
− 2y + 2 = 0
in slope–intercept form, we obtain
y = 2x + 1 which is the equation of l 3 . Since there are infinitely many points of intersection , the system
4x
−
y = 2x + 1 2y + 2 = 0
has infinitely many solutions.
We now consider lines that are not parallel and not coincident, where there is only one point of intersection.
3.3.2 We usually find the coordinates of the point of intersection algebraically.
Consider Figure 3.3.1. We now find the coordinates of P. Since P (x, y) lies on both l 1 and l 2 , the coordinates of P must satisfy the equations 3x + 4y = 12
x
and
− y = −1.
Thus we solve the system 3x + 4y = 12 1 x y =
−
−
.
By multiplying the second equation by 3 we obtain the equivalent system 3x + 4y = 12 3x 3y = 3
−
−
.
Subtracting the first equation from the second produces the equivalent system 3x + 4y = 12 7y = 15
−
−
.
146 Thus y =
15 . 7
By substituting y =
15 7
into 3 x + 4y = 12 we obtain
3x +
60 = 12. 7
3x +
60 = 12 7
Now
⇔
3x = 12
− 607
⇔
3x =
84
− 60
⇔
x=
⇔
x=
Thus
P=
VERTICAL
DISTANCE
BE TW E EN
CORRE-
SPONDING POINTS ON
24 3
×7
8 . 7
8 15 , 7 7
.
We now see how we can obtain the vertical distance between corresponding points on two lines. Figure 3.3.4 shows two lines l 1 and l 2 , defined by y = f (x) and y = g(x) respectively. The lines intersect at P (a, b).
TWO LINES
In this case corresponding
7
y l1 : y = f ( x )
( x , g ( x ))
( x , f ( x))
points are points that have
the same first coordinates.
b
P ( a,b )
( x , g( x )) 0 a
x l2 : y = g( x )
( x , f ( x))
Figure 3.3.4
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Note that if x > a then line l 1 lies above line l 2 and if x < a then l 2 lies above l 1 . We emphasise that the reason for identifying which point lies above the other is to make sure we can identify the bigger of the two y–coordinates so that we can subtract in the correct order.
So for x > a, the point (x, f (x)) lies directly above (x, g(x)). Since f (x) is bigger than g (x), the vertical distance between the two points is given by
f (x)
− g(x).
Note that distance is non–negative and thus we must subtract the smaller number from the larger number. For x < a , the point ( x, g(x)) lies directly above the point ( x, f (x)) and thus the vertical distance between the two points is given by
g( x )
The function d can also be considered in terms of the distance formula. (See Study Unit 1.3A.) Suppose P and Q are respectively points on the graphs of g and f , so that P = (x, g(x)) and Q = (x, f (x)). If Q lies above x)If>Qg(lies x), i.e. f (xP ) , then g(x) >f (0. directly above P , then they have the same x –coordinate, and d (P, Q) = ( f (x) g(x))2 = f (x) g(x). Thus if P and Q are any two corresponding points on the two graphs then the distance between the points depends on their x– coordinate, and hence we use the notation d (x) to denote the distance, instead of d (P, Q).
−
−
−
− f (x).
From Section 2.2 of Topic 2 we know that combining two functions by the operation of subtraction creates another function. Since we know that x D f Dg we can define a new function, the difference function d , by
∈ ∩
d (x) = f (x)
− g(x)
if f (x)
≥ g(x)
d (x) = g (x)
− f (x)
if g(x)
≥ f (x).
or
The function value d (x) is the vertical distance between corresponding points on the lines defined by y = f (x) and y = g(x) for any given x such that x
∈ D f ∩ Dg .
Since the x–axis is the line defined by y = 0, we can also find the vertical distance between a point on the x–axis and the corresponding point on any other non– vertical line. In Figure 3.3.5 the line l is defined by y = f (x) and its x –intercept is a . y ( x , f ( x ))
( x ,0) ( x , 0)
0
a
( x , f ( x ))
x
l : y = f (x )
Figure 3.3.5
For x > a , l lies below the x –axis. For any x
∈ D f the vertical distance between
148 a point on l and the corresponding point on the x –axis is thus given by
− f (x) − f (x).
d (x) = 0
=
Note that f (x) < 0 when x > a since the y –coordinate of each point on the line l is negative for these values of x . f the vertical distance between a For x< l liesthe above the x –axis.point For any x xD–axis point onal, and corresponding on the is thus given by
∈
−
d (x) =
f (x) 0 = f (x).
Note that f (x) > 0 when x < a since the y –coordinate of each point on the line l is positive for these values of x .
3.3.2 Figure 3.3.6 shows the two lines
l1 : y = f (x)
and
l2 : y = g(x)
representing functions f and g . y
2
x
0
1
4
P (x , y ) –2
l1 : y = f ( x)
Figure 3.3.6
l2 : y = g( x )
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(a) Find the equation for each of the functions f and g . (b) Determine algebraically P (x, y), the point of intersection of the two lines. (c) What is the vertical distance between the two lines when x = 3? (d) If x < 85 find the value of x for which the vertical distance between corresponding points on the two lines is 9 units. (e) When x =
1 find the vertical distance between l 1 and the x –axis.
−
(f) Find the values of x for which (i) f (x) > 0
≥0 f (x) − g(x) < 0.
(ii) g(x) (iii)
(a) The slope of l 1 is 24 , i.e. 12 , and the y –intercept is
f (x) = The slope of l 2 is
−
2 1 , i.e.
1 x 2
−2. Thus
− 2.
−2, and the y–intercept is 2. Hence g(x) = −2x + 2.
(b) We solve the system 1x 2 2 2x + 2
y = y =
−
−
.
We have 1 x 2
− 2 = −2x + 2.
Hence 5 x=4 2 and thus
x= If we substitute x =
8 5
into y =
y=
8 . 5
−2x + 2 we get
− 165 + 2 = − 65 .
Hence
P=
− 8 , 5
6 5
.
150 (c) At x = 3, l1 lies above l2 . Thus the vertical distance between corresponding points on the lines is given by
d (x) = f (x)
− g(x).
Hence
− − −− − −− −
d (3) =
f (3) g(3) 1 2 (3) 2 ( 2(3) + 2) 3 2 ( 6 + 2) 2 3 2+4 2 1 3 . 2
= = = =
Thus, when x = 3 the vertical distance between the lines is 3 12 units. (d) If x < 85 then l 2 lies above l1 and the vertical distance between corresponding points on the lines is given by
d (x) = g(x)
= =
− f (x)
1
−25x + 2 − ( 2 x − 2) − 2 x + 4.
We must find the value of x such that d (x) = 9. Now
d (x) = 9
⇔
− 52 x + 4 = 9
⇔
− 52 x = 5
⇔
x=
Thus, when x =
−2.
−2 the vertical distance between the lines is 9 units.
−
(e) When x = 1 then l1 lies below the x–axis. Thus the vertical distance between a point on l1 and the corresponding point on the x–axis is given by d (x) = f (x).
−
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MAT0511/003
So
−
d ( 1) =
= = = Thus, when x = 5 units. 2 (f)
− f (−1) − ( 12 (−1) − 2) − (− 52 ) 5
.
2 1 the vertical distance between l1 and the x–axis is
−
(i) The values of x for which f (x) > 0 are the values of x for which the line l 1 lies above the x –axis. It is clear from Figure 3.3.6 that
f (x) > 0 when
x > 4.
Note that x = 4 is excluded since f (4) = 0 and we are only considering f (x) > 0.
≥
(ii) The values of x for which g (x) 0 are the values of x for which the line l 2 lies above or on the x –axis. From Figure 3.3.6 we have
g(x)
≥0
when x
≤ 1.
Here x = 1 is included since g(1) = 0 and now we are considering g(x) = 0 or g (x) > 0. (iii)
f (x)
⇔
− g( x ) < 0 f (x) < g(x).
Thus we must find the x–values for which f (x) < g(x), i.e. the x–values for which the graph of f lies below the graph of g. In other words, we must find the values of x for which l 1 lies below l 2 . Thus f (x) < g(x) when x < 85 .
A REAL–LIFE APPLICATION
We end this study unit with a discussion and an activity that relates to the manufacturing or retail business. A manufacturer is interested in the break–even point. This is the point at which manufacturing costs equal income . For example, if the cost of producing certain items is given by the linear function f , and the income derived from selling the items is given by the linear function g , then the point of intersection of the graphs of f and g is the break–even point. See Figure 3.3.7 on the next page.
152 When the graph of f lies below the graph of g, there will be a profit; when the graph of f lies above the graph of g there will be a loss. y income function g Profit Cost in rands
cost function f
P (a, b) Break−even point Loss
c
a
d
x
Number of items
Figure 3.3.7
P(a, b), the point of intersection of the graphs of f and g, is the break–even point, i.e. the point at which production cost is equal to income .
We clarify the meaning of the words profit and loss by means of this example. Suppose the costs involved in a belt–making business are: leather costs rental for a warehouse salaries for the people who are employed costs of other items purchased packaging advertising transport administrative costs. All these items make up the expenses. Some of these items stay the same, regardless of how many belts are produced, for example rental and administrative costs. These costs are called fixed costs.
153
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When the belts are sold, a certain amount is earned per belt. The total amount earned is the income. Profit = income – expenses. If income exceeds expenses, then there will be a profit. Loss = expenses – income. If income is less than expenses , then there will be a loss.
In Figure 3.3.7 you can see that for x > a the graph of the income function g lies above that of the cost function f , i.e. income is greater than cost and hence a profit is made. For x < a the graph of the cost function f lies above that of the income function g, i.e. production costs are greater than income and hence a loss is made.
−
Since d > a a profit is made when x = d . This profit is g(d ) f (d ) (i.e. the difference between the income obtained from selling d items and the cost of producing d items). If x = c then a loss is made, and the loss is equal to f (c) g(c) (the difference between the cost of producing c items and the income obtained from selling c items).
−
3.3.3 A company manufactures soccer balls. The company’s daily fixed costs (i.e. rental, phones, stationery, etc.) are R1 800. One soccer ball costs R50 to manufacture and it is sold for R80.
(a) Give an equation for the functio n f which describes the cost of producing x soccer balls per day. (b) Give an equation for the function g which gives the income obtained if x balls are sold per day. (c) Draw the graphs of f and g on the same system of axes. (d) Find the break–even point P . (e) Does the company make a profit or incur a loss if 100 balls are produ ced and sold per day? What is the profit or loss?
154 (a) Total daily cost = ( cost per item) Hence y = f (x) = 50x + 1 800.
× (number of items) + fixed costs.
(b) Total daily income = ( selling price per item) Thus y = g (x) = 80x. (c)
× (number of items).
y g
5 400
Cost in rands
f
3 600
P ( x, y)
1 800
x 20
40
60
80
100 1 20
Number of soccer balls
Figure 3.3.8
(d) In order to find P (x, y) we solve the system
y = 50x + 1 800 y = 80x
.
We obtain 50x + 1 800 = 80x i.e. we have 30x = 1 800 and hence
x = 60. Thus
y = 80
× 60 = 4 800.
The break–even point is (60, 4 800). (e) Since 100 > 60, a profit is made when 100 balls are sold. This profit is (g(100) f (100)) rands. Now
−
g(100)
− f (100)
Thus the profit is R1 200.
−
= 80(100) (50(100) + 1 800) = 8 000 5 000 1 800 = 1 200 .
−
−
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3.3B GRAPHS OF LINEAR INEQUALITIES IN TWO UNKNOWNS So far in this topic we have considered the graphical representation of linear equations as well as of the solution of a system of linear equatio ns. We now consider graphs of linear inequalities. inequalities. We also show graphically the solution of a simple system of linear
3.3.3
You will see late r why we sketch l as a dashed line in the two figures.
When we look at the sketch of any straight line in the Cartesian plane we see that the line divides the plane into two half planes. Consider Figure 3.3.9 (a) below and Figure 3.3.9(b) on the next page. The line l defined by y = 3x 2, which is represented by a dashed line, divides the plane into two halves.
−
y 6
Q( x,y ) The shadi ng in the sketc h denotes that all points above the dashed line are included,
(– 1, y )
2
P ( x , 3x – 2)
not only those that can be seen here.
l
4
x –2
–1
1 –2 –4
(– 1 ,– 5 ) –6
Figure 3.3.9(a)
2
156
y
The shadi ng in the sketc h
6
l
4
(2,4)
2
(2 , y )
denotes that all points below the dashed line are included, not only those that can be
x
seen here.
–2
–1
1
2
–2
P ( x, 3x – 2)
–4 –6
R ( x,y )
Figure 3.3.9(b)
Since l is the graph of the function defined by y = 3x coordinates of the form ( x, 3x 2).
−
− 2 all points P on l have
We now note the following with respect to Figure 3.3.9(a).
− −
−
−
The point ( 1, 5) lies on l since the coordinates satisfy y = 3x 2. Now any point which has the same x–coordinate as ( 1, 5) and lies in the shaded region has the form ( 1, y), where y > 5 (i.e. the y–coordinate of ( 1, y) is bigger than 5 since the point lies above ( 1, 5)).
−
− − − − −
−
A point Q(x, y) which has the same x–coordinate as P lies in the shaded region, i.e. lies above P , if y > 3x 2.
−
The shaded region consists of all points Q (x, y) such that y > 3x the shaded region is the graph of y > 3x 2.
−
− 2. Thus
We have indicated l by a dashed line . This means that the gra ph of the inequality y > 3x 2 excludes points on this line . If we want the graph of y 3x 2 then l must be represented by a solid line which denotes that points on this line are also included. This is how we indicate the difference between the graphs of y > 3x 2 and y 3x 2.
≥ −
−
−
≥ −
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We note the following with respect to Figure 3.3.9(b).
The point ( 2, 4) lies on l. Any point which has the sa me x –coordinate as (2, 4) and lies in the shaded region has the form (2, y) where y < 4 (i.e. the y–coordinate of ( 2, y) is smaller than 4, since the point lies below ( 2, 4)).
A point R(x, y) which has the same x–coordinate as P lies in the shaded region, i.e. lies below P , if y < 3x 2.
The shaded region consists of all points R (x, y) such that y < 3x the shaded region is the graph of y < 3x 2.
−
−
− 2. Thus
The line l is represented by a dashed line which indicates that the points on the line are excluded . If we want to show the graph of y 3x 2 then l must be represented by a solid line.
≤ −
In general we state the following.
The graph of y > mx + c is the region that lies above the line defined by
y = mx + c. The graph of y by y = mx + c.
≥ mx + c is the region that lies above or on the line defined
The graph of y < mx + c is the region that lies below the line defined by y = mx + c.
The graph of y by y = mx + c.
The graph of x > c is the region that lies to the right of the vertical line defined by x = c .
The graph of x c is the region that lies to the right of or on the vertical line defined by x = c.
The graph of x < c is the region that lies to the left of the vertical line defined by x = c .
The graph of x c is the region that lies to the left of or on the vertical line defined by x = c.
≤ mx + c is the region that lies below or on the line defined
≥
≤
158
3.3.4
− ≤
Sketch the graph of 2x + 3y 1 0. (Hint: First rewrite the inequality in the form y
≤ mx + c or y ≥ mx + c.)
We have 2x + 3y Hence
− 1 ≤ 0. 3y
and thus
y
≤ −2x + 1 ≤ − 23 x + 13 .
Thus the graph of the inequality is the region lying below or on the line defined by y = 23 x + 13 . The graph is sketched in Figure 3.3.10.
−
y
1 _ 3
_1 2
x 2x + 3y – 1 = 0
Figure 3.3.10
If you are unsure which region to shade when drawing the graph of an inequality you can use test points. For example, consider Activity 3.3.4. You may have drawn the line defined by 2 x + 3y 1 = 0 without changing the equation into the form y = 23 x + 13 . Now choose a poin t that is not on the lin e. For example, choose a point below the line, such as ( 0, 1). If you substitute the coordinates of this point into the left side of the inequality 2 x + 3y 1 0, you will obtain 2(0) + 3( 1) 1 = 4, and 4 0. The coordinates satisfy the inequality and hence you know that you must shade the region below the line.
−
−
− −
−
−
− ≤
− ≤
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If you choose, for example, the point (0 , 2) which lies above the line, and substitute the coordinates into the left side of the inequality, you will then obtain 2(0) + 3(2) 1 = 5, but 5 0. The coordinates do not satisfy the inequality and thus you cannot shade the region above the line.
−
≥
So far we have considered graphs which represent only one linear inequality. In the next example we sketch the graph of a system of linear inequalities.
3.3.4 Sketch the graph of the following system of linear inequalities.
− ≥ −4 ≤ 8 ≥ 0 ≥ 2
x y 2x + y x y
SOLUTION
We first rewrite the inequalities in the form y following equivalent system.
y y x y
≤ x+4 ≤ −2x + 8 ≥ 0 ≥ 2
We now draw the lines defined by:
(3.3.1)
≤ ... or y ≥ ... . We then have the
y = x+4 y =
−2x + 8
x = 0
y = 2. We then shade the region that is simultaneously
below or on the line defined by y = x + 4 below or on the line defined by y =
−2x + 8
to the right of or on the y –axis (the line defined by x = 0) above or on the line defined by y = 2.
(3.3.2)
160 The shaded region in Figure 3.3.11 is the set of all points that satisfy system (3.3.2) and hence system (3.3.1). Thus the shaded region is the graphical representation of the solution set of (3.3.1), and hence is the graph of the system. y x=0 x–y = – 4 or y = x + 4
8
Note that all the lines are represented as solid lines in
4
this case.
y=2
–4
0
x 4 2x + y = 8 or y = – 2 x + 8
Figure 3.3.11
The profit P in rands for x tables and y chairs is given by
P = 140x + 80y. Various methods are available to determine the point Q (x, y) in the shaded region of Figure 3.3.12 that gives the maximum profit. We apply an appropriate method to find the coordinates of Q . We then substitute these coordinates into the profit equation. This will give us the answer to the question. We do not expect you to solve such problems at this stage. We include this example as an illustration of one of the many real–life applications of graphs of linear inequalities.
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3.3C DIRECT PROPORTION AND THE USE OF LINES IN THE ANALYSIS OF EXPERIMENTAL DATA DIRECT PROPORTION
In the natural sciences we often come across direct proportion. For example, in physics we have Hooke’s Law which states that
You do not need to have studied the concepts that we use from physics, chemistry, etc. We use the m to show you further applications of straight lines.
“the force F needed to keep a spring stretched x units beyond its natural length is directly proportional to x ”.
Figure 3.3.12
We first read about direct proportion in Topic 3 of Book 1 and there is a speed– time problem dealing with direct proportion at the beginning of this book, in Study Unit 1.1A. We remind you of the definition of direct proportion.
Definition 3.3.1
If the variables x and y are related by an equation
y = cx ; where c
∈ R , c = 0
(3.3.5)
then y is directly proportional to x . The constant c is called the constant of proportionality.
162 Note
In some textbooks the phrase “y varies directly as x” is used instead of “ y is directly proportional to x”. The constant c may also be called the “constant of variation”.
The equation y = cx defines a linear function in x , and thus we can write y = f (x) = cx . The graph of this function f is a straight line with slope c and y –intercept 0, i.e. it passes through the srcin.
3.3.5 Suppose y is directly proportional to x , and that y = 20 when x = 8. (a) Find the constant of proportionality and write dow n the equation that describes this relationship. (b) Find y when x = 3.
SOLUTION (a) Since y is directly proportional to x we have
y = cx.
(3.3.6)
If we substitute x = 8 and y = 20 into (3.3.6) we have 20 = c i.e.
c=
×8
20 5 = . 8 2
Thus the constant of proportionality is
5 2
and the equation is
y = 5 x. 2 (b) When x = 3 we have y = 5 2
× 3 = 152 .
163 Note: It is not nece ssary to remember the formula for Hooke’s Law.
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The next activity relates to Hooke’s Law, which we stated earlier. We use Definition 3.3.1 to write Hooke’s Law as
F = cx where F is the force (measured in Newtons), x is the length of the spring beyond its natural length (measured in centimetres) and the constant of proportionality, c, is the spring constant.
3.3.5
For the definition of a Newton see Example 5.1.4 in Topic 5 of Book 1.
Suppose a spring has a natural length of 20 cm. Suppose a force of 30 N (Newtons) is required to hold the spring stretched 2 cm beyond its natural length. (a) Use Hooke’s Law to find the constant of proportionality c . Write down an equation to describe this relationship. (b) What force is required to keep the spr ing stretched to a length of 25 cm? Hint: Apply the method of Example 3.3.5.
(a) According to Hooke’s Law,
F = cx .
(3.3.7)
When we substitute x = 2 and F = 30 into (3.3.7) we obtain Here the derived unit for c is N.cm−1 . In physics we usually use metres for length, in which case the derived unit for the spring constant would be N.m−1 .
30 = c i.e. we have
×2
c = 15. Thus
F = 15x.
(3.3.8)
(b) If the sprin g is stretched to 25 cm then it is stretched (25 20) cm, i.e. 5 cm, beyond its natural length. Thus we substitute x = 5 into (3.3.8) and obtain F = 15 5 = 75.
−
×
Hence a force of 75 N is required to keep the spring stretched to a length of 25 cm.
164 It often happens that one quantity does not depend on just one other quantity. For example, suppose the quantities x , y and z are related by the equation
z = cxy ; where c
∈ R , c = 0.
Now z depends on two quantities x and y, and we say that z is jointly proportional to x and y , or z varies jointly as x and y .
3.3.6 You will encounter these formulas in Book 4.
A number of examples of direct and joint proportion can be found in geometry. We consider the following. (a) The circumference C of a circle of radius r is given by C = 2πr. C is directly proportional to r . 2π is the constant of proportionality. (b) The area A of a square of side x is given by A = x 2 . A is directly proportional to the square of x , or to x 2 . 1 is the constant of proportionality. (c) The area A of a triangle with length l and base b is given by A = 12 lb. A is jointly proportional to l and b . 1 is the constant of proportionality. 2
Now see whether you have understood Example 3.3.6 by trying to do the next activity.
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3.3.6 Fill in the missing words, numbers or symbols in the following. (a) The perimeter P of a square of side x is given by P = 4x. P is ................ to ................ . ....... is the constant of proportionality. (b) The area A of a circle of radius r is given by A = πr 2 . A is ................ to ................ . ....... is the constant of proportionality. (c) The volume V of a rectangular box of length l, breadth b and height h is given by V = lbh . V is ................ to ............... . ....... is the constant of proportionality.
(a) P is directly proportional to x . 4 is the constant of proportionality. (b) A is directly proportional to the square of r , or to r 2 . π is the constant of proportionality. (c) V is jointly proportional to l , b and h . 1 is the constant of proportionality.
We consider the statement given in Example 3.3.6(b), namely “the area A of a square of side x is given by A = x 2 , i.e. A is directly proportional to x 2 ”. In this formula, although A depends only on the variable x, it is not directly proportional 2
to x . A is directly proportional to x . We show this by means of two graphs. Suppose we consider several different squares, whose sides are x cm long and whose areas are A cm2 . Sides cm x 0123 4 5 6 Square of sides (x2 cm2 ) 0 1 4 9 16 25 36 Area (A cm2 ) 0 1 4 9 16 25 36 Table 3.3.1
166 Figure 3.3.13(a) shows the graph we obtain when we plot the ordered pairs (x, A) and Figure 3.3.13(b) shows the graph we obtain when we plot the ordered pairs (x2 , A). y y 40
40
Area A 30 in cm 2 20
Area A 30 in cm 2 20
10
10
x
x
2468 x in cm
10
Figure 3.3.13(a)
20 30 40 x 2 in cm 2
Figure 3.3.13(b)
From Figures 3.3.13(a) and 3.3.13(b) we see that the graph in Figure 3.3.13(b) is a straight line, which shows direct proportion. Thus A is directly proportional to x2 . The fact that the graph in Figure 3.3.13(a) is not a straight line indicates that
A is not directly proportional to x . Explicit and implicit formulas are defined in Study Unit 2.4B of Book 2.
In all the cases discussed so far we have worked with explicit formulas. Now if we have, for example, the implicit formula in v , g and h
v2 = 2gh then we say:
The square of v is jointly proportional to g and h and 2 is the constant of proportionality.
Use this comment, and the method used in Example 3.3.5, to attempt the next activity.
3.3.7 The period T of a planet is defined to be the time it takes the planet to complete one revolution around the sun, i.e. to travel around the sun once. Kepler’s Third Law of Planetary Motion states that the square of the period T is directly proportional to the cube of the planet’s average distance d from the sun.
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(a) Suppose that the period of the planet earth is approximately 365 days and that the earth’s average distance from the sun is approximately 150 million kilometres. Find the constant of proportionality and write down the equation that describes the relationship. (b) The planet Jupiter is approximately 750 million kilom etres from the sun. Find the period of Jupiter.
(a) Kepler’s Third Law states that
T 2 = cd 3 . Note the way we use scientific notation to simplify the numbers in the calculation.
If we substitute T
(3.3.9) 6
≈ 365 and d ≈ 150 × 10 into (3.3.9) we obtain (365) ≈ c(150 × 10 ) 2
i.e.
133 225 i.e. 1, 33225 i.e.
18
≈ c(3 375 000 × 10
5
)
6
18
× 10 ≈ c(3, 375 × 10 × 10 105
1, 33225 i.e.
6 3
×
c(3, 375
)
1024 )
≈ × 1, 33225 × 10 c≈ 3, 375 × 10
5
24
≈ 3, 947 × 10−
20
.
The constant of proportionality is approximately 3 , 947
T (b) We substitute d
2
≈ (3, 947 × 10−20)d 3.
× 10−
6
20
6 3
20
8 3
20
3
3
24
4
4
Now since T > 0, we have T = Hence
T
≈
1665, 14
T 2. 4
× 10
= 40, 81
2
× 10
and thus
(3.3.10)
≈ 750 × 10 into (3.3.10). We obtain T ≈ (3, 947 × 10− ) × (750 × 10 ) = (3, 947 × 10− ) × (7, 5 × 10 ) = 3, 947 × 10− × (7, 5) × 10 = 3, 947 × (7, 5) × 10 ≈ 1665, 14 × 10 . √ 2
20
= 4 081 .
Thus the period of Jupiter is approximately 4 081 days.
168 We have looked at various problems involving direct proportion. Can we use straight lines to help us establish direct proportionality between two sets of data? Suppose we have two sets of data A and B and suppose each value ( y) in A depends on a value ( x) in B. Suppose we plot the poi nts ( x, y) on a gr aph. If we can draw a straight line through these points, which also passes through the srcin, then we can write y = cx and hence say that y is directly proportional to x . We use this method to establis h or prove certain natural science laws. The following example illustrates this.
3.3.7 Hooke’s Law is stated at the beginning of this study unit.
Consider an experiment to prove Hooke’s Law for a certain spring. Suppose we determine the force (in Newtons) required to hold this spring stretched at 1 cm, 2 cm, 3 cm, 4 cm and 5 cm beyond its nat ural length. Suppose we obtai n the following results.
Length x of spring beyond
Because we can never measure completely accurately there is always a degree of error in experimental data. The points will thus not all lie on the line.
its natural length (cm)
1
2
3
4
5
Force F in Newtons (N)
8 17 24 31 40
We now plot the ordered pairs (x, F ) and we obtain the graph in Figure 3.3.15.
50 40 Force F in Newtons
X
30
X
X
20 10
X
X
12345 Length x in cm
Figure 3.3.14
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From Figure 3.3.14 we see that the experimental points lie approximately on the straight line through the srcin with equation of the form F = cx . Thus we have verified Hooke’s Law.
Example 3.3.7 shows that we can use a straight line which goes through the srcin to establish a relationship of direct proportion between two variables. This is one of the easiest relationships to establish. In experimental work it is very difficult to fit equations to curves that are not straight lines. When we plot data and obtain a graph that is not a straight line, we try to make an intelligent guess regarding the equation that represents the relationship between the variables. We do not expect from you to be able to carry out procedures such as these in this module. Now try some questions from the following exercise.
3.3 In each of questions 1–4 find the point of intersection of the lines which are defined by the given equations .
1. l1 : 3x + y = 1 l2 : x = 2 2. l1 : y = 34 x 2 l2 : 2x + y = 1
−
3. l1 : x + 4y = 3 l2 : 2x y = 4
−
4. l1 : x 1 = 0 l2 : 2y + 3 = 0
−
l1 and l2
170 5. The line l3 passes through the point of intersection of parallel to l 1 . Find the equation of l 3 if
l1 and l2 , and is
−3 −3x + 4.
l1 : y = 2x l2 : y =
−
6. The line l 3 passes through (2, 3) and the point of intersection of l 1 and l 2 , where
l1 : y = 3 l2 : 2x + y
−3
= 0.
Find the equation of l 3 . 7. The line l3 passes through the point of intersection of lines l 1 and l 2 , and is perpendicular to l 2 , where
l1 : y + 2x = 3 l2 : x
−1
= 0.
Find the equation of l 3 . 8. On the same system of axes sketch the lines l 1 , l 2 and l 3 , where
l1 : y = x l2 : y = x + 2 l3 : y =
−
1 3
x + 1.
(a) Find algebraically P(x, y), the point of intersection of l1 and l3 , and Q(x, y), the point of intersection of l 2 and l 3 . (b) How do we know that l 1 and l 2 will never intersect? Remember that in this case corresponding points are points with the same first coordinate.
(c) For any given x , determine the vertical distance between corresponding points on l 1 and l 2 . (d) When x =
−2 determine the vertical distance between l
1
and l 3 .
(e) If x > 0 find the value of x for which the vertical distance between corresponding points on l 2 and l 3 is 4 units. 9. The lines l 1 and l 2 , where l 1 : y = f (x) and l 2 : y = g(x), are sketched in Figure 3.3.15. (a) Find an equation for f and for g . (b) Determine P (x, y), the point of intersection of l 1 and l 2 . (c) What is the vertical distance between l 1 and l 2 when x = 5? (d) Determine the two points on the x –axis at which the vertical distance between them and the respective corresponding points on l2 is 6 units.
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(e) Find the values of x for which (i) f (x)
≤0
(ii) g(x) < 0 (iii) f (x)
≥ g(x). y
l2
l1 1
P ( x , y) x
0
1
3
–1
Figure 3.3.15
10. The lines l 1 and l 2 , where l 1 : y = h(x) and l 2 : y = k (x), are sketched in Figure 3.3.16. The point of intersection of the two lines is P (2, 2). y l1
P (2,2)
0
l2
x 1
Figure 3.3.16
172 (a) Find an equation for h and for k . (b) Determine the y –intercept of l 1 . (c) What is the vertical distance between the two lines when x = 5? (d) If x < 2 find the value of x for which the vertical distance between corresponding points on the two lines is 10 units. (e) Find the values of x for which (i) h x
0
(ii) h((x)) > 2
− ≤ 0.
11. A woman decides to earn some extra money by baking banana cakes at home and selling them. Her daily fixed costs are R24. It costs her R6 to bake one banana cake and she sells it for R12. (a) Give a formula for the functi on f which describes the daily cost of producing x banana cakes per day. (b) Give a formula for the fun ction g which represents her income if she sells x cakes per day. (c) Draw the graphs of f and g on the same system of axes. (d) How many cakes must she bake and sell in order to break even? (e) If her profit was R30 on a certain day, how many cakes did she bake and sell that day? 12. A small factory produces and sells T–shirts. The daily fixed costs are R800 and the cost of producing one T–shirt is R10. The T–shirts are sold for R18 each. (a) Draw the graphs representing cost and income on the same system of axes, and determine the break–even point. (b) If 80 T–shirts are produced and sold, is there a profit or loss, and what is the profit or loss? (c) If a profit of R1 200 was made on a certain day, how many T–shirts were produced and sold that day? 13. In each of the following, give the inequality represented by the graph. (a)
(b) y
y 2x – y = – 2
x y = –x
x
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(c)
(d) y
y
(1,0)
x ( 0, _ 2 )
x
14. Sketch the graph of each of the following linear inequalities. (a) x + 2 (b) y
≥0
≥ 2x
(c) x + 2y > 3 (d) x
− 2y ≤ 3
15. Sketch the graph of the following system of linear inequalities.
y y y 2x
≤ x−1 ≥ x− ≤ 4 3 ≥
3 2
16. Fill in the missing words, numbers or symbols in the following. (a) The volume V of a cube with side x is given by V = x 3 . V is ................ to ............... . ........ is the constant of proportionality. (b) The area A of an equilateral triangle√(i.e. a triangle in which all sides are equal) of side x is given by A = 43 x2 . A is ................ to ............... . ......... is the constant of proportionality.
(c) The volume of a box with length x , breadth x and height h is given by V = hx 2 . V is ................ to ................ . ......... is the constant of proportionality. 17. Write an equatio n for each of the following statements, using k as the constant of variation. (a) p is directly proportional to q . (b) r is directly proportional to the square root of s . (c) p is jointly proportional to the cube of q and the square of r . (d) The square of p is jointly proportional to s and the cube root of r .
174 18. Suppose z is directly proportional to x , and z = 30 when x = 6. (a) Find the constant of proportionality and write down the equation that describes this relationship. (b) Find z when x = 11. 19. Suppose p is jointly proportional to q and the square of r , and p = 36 when q = 3 and r = 2. Find the constant of proportionality and write down an equation that represents this relationship. 20. This question is adapted from Stewart, Redlin and Watson (see the References). The cost of printing a magazine is jointly proportional to the number of pages in the magazine and the number of magazines produced. (a) Write an equation for this joint proportion if it costs R240 000 to print 4 000 copies of a 120–page magazine. (b) How much would it cost to print 5 000 copies of a 92–page magazine?
• A linear function f is defined by y = f (x) = mx + c, where m and c are constants.
If m = 0 then f (x) = c, and f is called a constant function.
If m = 1 and c = 0 then f (x) = x and f is called the identity function.
The graph of f is a straight line l with slope m and y –intercept c . slope of l
↓ ↓ l
line
l horizontal l vertical
l slants upwards l slants downwards l1 is steeper than l 2
y–intercept of l
: y = mx + c
equation m=0 m is undefined (i.e. a vertical line cannot be described by the equation y = mx + c) m>0 m<0 magnitude of m 1 > magnitude of m 2
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• The graphs of all linear functions are straight lines, but not all straight
lines are the graphs of linear functions , e.g. a line parallel to the y –axis is not the graph of a function.
• Suppose the lines l
1
and l 2 are
l1 : y = m1 x + c1 l2 : y = m2 x + c2 .
If l1 and l2 are parallel then m1 = m 2 ; and conversely, if m1 = m 2 then l 1 and l 2 are parallel.
If l1 and l2 are perpendicular then m1 m2 = m1 m2 = 1 then l 1 and l 2 are perpendicular.
−
−1; and conversely, if
All vertical lines are parallel. All horizontal lines are parallel.
A vertical line and a horizontal line are perpendicular.
• Different forms of the equation of a
straight line
Slope–intercept form The equation of a line that has slope m and y –intercept c is
y = mx + c.
Point–slope The equationform of a line that passes through the point Q (x1 , y1 ) and has slope m is given by y y1 = m (x x1 ).
−
Two–point form The equation of a line that passes through the points Q(x2 , y2 ) is given by
y
−
−y
1=
− y2 x2
y1
−x
(x
1
P(x1 , y1 ) and
− x ). 1
General form The general form of the equation of a line is
px + qy + r = 0 where p, q and r are constants and p and q are not both zero.
The equation of the horizontal line through the point ( p, q) is given by y = q.
The equation of the vertical line through the point ( p, q) is given by
x = p.
176
• Coordinates of the point of intersection of two lines
If two lines represented by the equations y = f (x) and y = g(x) intersect at a point P , then the coordinates of P are found by solving the system
y = f (x) y = g(x)
.
Distance between (x, f (x)) and (x, g(x))
•
y l1 : y = f ( x )
( x , g ( x ))
( x , f ( x))
b
P ( a,b )
( x , g( x ))
x
0 a
l2 : y = g( x ) ( x , f ( x))
The vertical distance d (x) between corresponding points on lines l 1 and l 2 is given by d (x) = f (x) g(x) when x > a
−
and
d (x) = g(x)
− f (x)
when x < a.
• Distance between (x, f (x)) and (x, 0) y ( x , f ( x ))
( x ,0) ( x , 0)
0
a
( x , f ( x ))
x
l : y = f (x )
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The vertical distance d (x) between corresponding points on the line l and the x –axis is given by
d (x) = 0
− f (x) = − f (x)
when x > a
and
d (x) = f (x)
− 0 = f (x)
when x < a.
Break–even point, profit and loss
•
y income function g Profit Cost in rands
cost function f
P (a, b) Break−even point Loss
c
a
x
d Number of items
The break–even point is the point of intersection of the graphs of the cost function f and the income function g , i.e. the point at which cost = income.
When g (x) > f (x) a profit is made and profit = g (x)
When g (x) < f (x) a loss is made and loss = f (x)
− f (x).
− g(x).
• Graphs of linear inequalities Inequality y > mx + c y mx + c y < mx + c y mx + c x>c x c x
≥ ≤ ≥ ≤
Graph region above the line defined by y = mx + c region above or on the line defined by y = mx + c region below the line defined by y = mx + c region below or on the line defined by y = mx + c region to the right of the line defined by x = c region to the right of or on the line defined by x = c region to the left of the line defined by x = c region to the left of or on the line defined by x = c
178
• Direct and Joint Proportion Term directly proportional
Examples y = cx y is directly proportional to x c is the constant of proportionality
T2 = 4 2
× 10−
20 3
d
3
T is directly proportional to d 4 10−20 is the constant of proportionality
×
jointly proportional
z = cxy z is jointly proportional to x and y c is the constant of proportionality v2 = 2gh v2 is jointly proportional to g and h 2 is the constant of proportionality
CHECKLIST Now check that you can do the following.
SECTION 3.1
1. Draw a line using a table of v alues. Activity 3.1.1 2. Find the slope of a line using the coordinates of two points on the line. Activity 3.1.3 3. Draw a line using two points (we usua lly use the x – and y –intercepts). Example 3.1.1; Activity 3.1.4(a) 4. Draw a line using one point on the line and the slope of the line. Example 3.1.2; Activity 3.1.4(b) 5. Draw a line which is parallel or perpendicular to a given line. Examples 3.1.3, 3.1.4; Activity 3.1.5 6. Draw vertical and horizontal lines. Example 3.1.5
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7. Draw graphs of linear functions with restricted domains. Activity 3.1.6
SECTION 3.2
1. Find the
slope intercept form of the equation of a line point slope form of the equation of a line two point form of the equation of a line general equation of a line.
−− −
Activities 3.2.1, 3.2.2, 3.2.3; Exercise 3.2
SECTION 3.3
1. Find the point of intersection of two lines. Example 3.3.2 (and discussion directly after it); Activity 3.3.2 2. Find the vertica l distance between corresponding points on two lines, or between a point on a line and the corresponding point on the horizontal axis. Activity 3.3.2 3. Apply your knowledge of linear functions and straight lines to simple problems involving linear cost and income functions, break–even points, profit and loss. Activity 3.3.3 4. Sketch the graphs of linear inequalities. Example 3.3.3; Activity 3.3.4 5. Sketch the graph of the solution set of a system of linear inequalities. Example 3.3.4 6. Recognise equations which show direct or joint proportion and apply these concepts to various real–life situations. Examples 3.3.5, 3.3.6, 3.3.7; Activities 3.3.5, 3.3.6, 3.3.7
180
PARABOLAS OUTCOMES After studying this topic you should be able to do the following. SECTION 4.1: Characteristics of Parabolas defined by y = a(x
− h)2 + k
Recognise the role of a with respect to the parabola defined by y = ax 2 .
Recognise the role of k with respect to the parabola defined by y = ax2 + k.
Recognise the role of h with respect to the parabola defined by y = a (x
Determine various characteristics of a parabola defined by
y = a (x
− h)2 + k.
Know the characteristics of parabolas defined by y = f (x) = a(x
− h)2.
− h)2 + k.
SECTION 4.2: Sketching Parabolas
− h)2 + k.
Sketch parabolas defined by y = a (x
Sketch parabolas defined by y = ax 2 + bx + c.
Use the discriminant to determine the number of times a parabola cuts the x–axis. Know the characteristics of parabolas defined by y = ax 2 + bx + c.
SECTION 4.3: Finding the Equation of Parabola
Find the equation of a parabola if you are given three unrelated points on the parabola.
Find the equation of a parabola if you are given the vertex and another point on the parabola.
Find the equation of a parabola if you are given the x –intercepts and one other point on the parabola.
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SECTION 4.4: Using Parabolas and Quadratic Functions
Use a rough sketch of an appropriate parabola to solve a quadratic inequality.
px + q Solve a rational inequality of the form rx + s > 0 (where > can be replaced by <, or ) by using a rough sketch of the parabola defined by y = ( px + q)(rx + s).
≥ ≤
Find maximum and minimum values of quadratic functions in certain applied problems.
182
4.1 CHARACTERISTICS OF PARABOLAS Polynomials are dealt with in Topic 1 of Book 2.
In the last topic we dealt with lines. Many lines are graphs of linear functions. Remember that a linear function is defined by a polynomial expression of degree 1. We now consider functions which are define d by certain polynomial
expressions of degree 2. These are called quadratic functions and their graphs are called parabolas.
Definition 4.1.1 A function f defined by
f (x) = ax 2 + bx + c; a, b, c,
∈R
and a = 0,
is called a quadratic function in x . The graph of a quadratic function is called a parabola.
Note that in the above definition we require a = 0. Why is this necessary? Well, if a = 0 then f will reduce to a linear function. Can you explain why this is so?
From now on, when we deal with quadratic equations or functions, we assume a = 0 without stating this each time.
We have already come across examples of parabolas. See, for example, Figure 2.3.1, in Study Unit 2.3B of Book 2, as well as Figures 1.2.17 and 1.3.1 in Topic 1 of this book. We do this by means of completing the square.
In Study Unit 1.2C of Book 2 we showed that we can write ax2 + bx + c in the form a (x h)2 + k. In this section we now consider parabolas whose equations are written in the form y = a(x h)2 + k. The constants a, h and k relate to certain characteristics of parabolas. In the next section we look at parabolas defined by the general equation y = ax 2 + bx + c.
−
−
Since quadratic polynomials (i.e. polynomial expressions of degree 2) are defined for all real numbers, the domain of a quadratic function is R. We can thus sketch the graph of a quadratic function by plotting some points that lie on the graph and joining them with a smooth curve.
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In the next three study units we use tables of values to find separate points, which we then plot and join, to sketch three different types of parabolas. We consider parabolas defined by
y = ax 2
y = ax 2 + k
y = a (x
h)2 + k.
−
The sketches of these parabolas illustrate the roles of a, h and k in determining the shape and position of the parabola with respect to the x – and y –axes.
4.1A PARABOLAS DEFINED BY y = ax 2 In this study unit we consider parabolas defined by y = ax 2 . We look at some of their characteristics and consider how different values of a affect the shape of the parabolas.
4.1.1 Consider the parabolas defined by the following equations. (b) y = 12 x2
(a) y = x 2
(c) y = 2x2
Set up one table of values for all three cases, using integer values of x from 3 to 3.
−
For each case plot the points and join them by means of a smooth curve.
Draw all three graphs on the same system of axes.
SOLUTION x
−3 −2 −1
y = x2
9
y = 12 x2
9 2
y = 2x 2
18
4
0 1 2
1 0 1 4 2
8
1 2
0
2 02
3
9 1 2
2
8 18
9 2
184
y y = 2x 2
18
16 The arms of the parab olas
y = x2
14
open upwards. Note that the value of a influences how
12
close together the arms of the parabola are.
10
8
y = 12 x2
6
4
2
x
−4 −3 −2 −1
0
1
2
3
4
Figure 4.1.1
Before we discuss the properties of the parabolas shown in Figure 4.1.1, try the following activity.
4.1.1 On the same system of axes sketch the parabolas defined by the following equations. (a) y =
− x2
(b) y =
− 12 x2
(c) y =
−2x2
Set up one table of values for all three cases, using integer values of x from 3 to 3.
−
For each case plot the points and join them by means of a smooth curve.
185
x
−x2 y = − 12 x2 y = −2x2 y=
MAT0511/003
−3 −9 − 92 −18
−2 −4 −2 −8
−4 −3 −2 −1
The arms of the parabolas open downwards. Once again note that the value
of a influences the extent
to which the arms of the parabola are close together
−1 0 1 2 3 −1 0 −1 −4 −9 − 12 0 − 12 −2 − 92 −2 0 −2 −8 −18 y 0
1
2
3
4
−2 −4 −6 −8 −10
x
y=
− 12 x2
y=
−x2
12
or far apart from each other.
−−14 −16 −18
y=
−2x2
Figure 4.1.2
STANDARD PARABOLA
If f denotes a function then the graph of f is symmetric with respect to the y –axis if
f ( x) = f (x). In this case
− −
f ( x) = ( x)2 = x2 = f (x).
−
We first consider the parabola defined by y = f (x) = x 2 which is sketched in Figure 4.1.1. This is known as the standard parabola . It has the following characteristics.
It is symmetric with respect to the y –axis, i.e. the line defined by x = 0. We call this line the axis of symmetry of the parabola. This means that if we fold the sketch of the parabola along the y –axis, then the left arm of the parabola lies directly on top of the right arm. Symmetry with respect to the y –axis also means that each point on the left arm of the parabola has
186 a corresponding point (with respect to the y–axis) lying on the right arm. For example, ( 2, 4) lies on the left arm of the stand ard parabola. The corresponding point (with respect to the y –axis) is (2, 4), which lies on the right arm. See Figure 4.1.3.
−
y y = x2
( 2, 4)
−
(0, 0)
(2, 4)
x
Figure 4.1.3
The arms of the parabola open upwards ; we say the parabola is concave up. If we look at the graph of y = x 2 from left to right we see that the curve “falls” to the srcin and then “rises”. Mathematically we say that the function f decreases on the interval ( , 0) and increases on the interval (0, ).
−∞
∞
The parabola has a turning point , called the vertex. This is the po int of intersection of the parabola and its axis of symmetry. In this case the vertex is the point ( 0, 0).
As x increases the function value f (x) decreases to 0 and then increases, and thus 0 is the minimum value of the function. We know this even without looking at the graph. Algebraically we have x 2 0 for all x R and thus f (x) has a minimum value of 0.
≥
D f = R and R f = [0,
∈
).
∞ PARABOLAS DEFINED 2
BY y = ax , a > 0
From Figure 4.1.1 we see that the parabolas defined by y = 12 x2 and y = 2x2 have the same characteristics as the standard parabola. We also note the following.
The parabola defined by y = 2 x2 is narrower than the standard parabola, i.e. its arms are closer together than the arms of the standard parabola. The parabola defined by y = 12 x2 is wider than the standard parabola.
187 PARABOLA DEFINED BY y =
−x
2
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From Figure 4.1.2 we see that the parabola defined by y = x2 is a reflection of the standard parabola in the x–axis. In other words we can obta in the new parabola by “turning” the standard parabola “upside–down”. We use the phrase “reflection in the x–axis” to suggest that the x–axis behaves as a mirror, and “reflects” one graph to give another.
−
4.1.2 Write down characteristics, similar to those given for the standard parabola, for the parabola defined by y = f (x) = x2 .
−
The parabola is symmetric with respect to the y –axis.
The arms of the parabola open downwards, i.e. the parabola is concave down.
As we move from left to right the curve rises to ( 0, 0) and then falls, i.e. the function f increases on ( , 0] and decreases on [ 0, ).
The vertex is the point ( 0, 0).
As x increases the function value f (x) increases to 0 and then decreases, and thus 0 is the maximum value of f (x). Algebraically, since x 2 0 for all x R, we have x2 0 for all x R and thus f (x) has a maximum value of 0.
−∞
∈
PARABOLAS DEFINED 2
BY y = ax , a < 0
− ≤
D f = R and R f = (
∞ ∈
≥
−∞, 0].
− 12 x2 and y = −2x2 2 have the same characteristics as the parabola defined by y x . We also note − = the following. In Figure 4.1.2 we note that the parabolas defined by y =
The parabola defined by y = by y = x2 .
−2x2 is narrower than the parabola defined
The parabola defined by y = y = x2 .
− 12 x2 is wider than the parabola defined by
−
−
We can now state the general role of a .
188
THE ROLE OF a WITH RESPECT TO THE PARABOLA DEFINED BY y = ax 2
The sign of a determines whether the parabola opens upwards or downwards. If a > 0, it opens upwards. If a < 0, it opens downwards.
In T opic 1 of Boo k 1 we use the term “magnitude” for “numerical value”.
The numerical value of a determines how close together or how far apart the arms of a parabola are. The bigger the numerical value of a, the narrower the parabola becomes. The smaller the numerical value of a , the wider the parabola becomes.
Table 4.1.1
4.1B PARABOLAS DEFINED BY y = ax 2 + k We investigate the relationship between parabolas defined by y = ax2 + k and the parabola defined by y = ax 2 for a fixed value of a . In other words we cons ider a specific value of a and different values of k . PARABOLAS DEFINED
We first consider an example in which a = 1.
BY y = x2 + k
4.1.2 (a) On the same system of axes sketch the parabolas defined as follows. (i) y = f (x) = x 2 (ii) y = g(x) = x 2 + 2 (iii) y = h(x) = x 2
−3
Set up one table of values for all three cases, using integer values of x from 3 to 3, plot the points and join them by means of smooth curves.
− (b)
(i) What is the relationship between the graphs of g and f ? (ii) What is the relationship between the graphs of h and f ?
189
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SOLUTION (a)
−3 −2 −1
x y = x2
9
y = x2 + 2 1 1 y = x2
−3
4 6
6
0
1
0
3
2
1 23
14 9 3 6 11
−2 −3 −2
1
1
6
y 12
10
y = g(x) = x 2 + 2
8
y = f (x) = x 2
6
4
y = h (x) = x 2
2
x
−4 −3 −2 −1
0
3
−
−2 −4
1
2
3
4
Figure 4.1.4
(b)
Algebraically, x2 0 and hence x2 + 2 2, i.e. we have g (x) 2.
≥
≥
≥
(i) The graph of g is the same of fg except that it has been shifted 2 units . The graphsasofthat have the same characteristics, upwards f and except that the vertex of the graph of g is no longer (0, 0).
Dg = R and the graph of g – has the y –axis as its axis of symmetry – is concave up, and – has vertex ( 0, 2). Thus g (x) has a minimum value of 2 and hence R g = [ 2,
∞).
190
We know that x2 thus x2 3 3. h(x) 3.
− ≥− ≥−
(ii) The graph of h is the same as that of f except that it has been shifted 3 units downwards. Thus Dh = R and the graph of h has the y –axis as its axis of symmetry, is concave up and has vertex (0, 3). The graph of h has its lowest point when x = 0, and thus the minimum value of h (x) is 3 and hence R h = [ 3, ).
−
≥ 0 and
Hence
PARABOLAS DEFINED
−
− ∞
Now try the following activity in which a = 1.
BY y = ax2 + k, a = 1
4.1.3 (a) On the same system of axes sketch the parabolas defined as follows.
− 32 x2 (ii) y = q(x) = − 32 x2 + 1 (iii) y = r (x) = − 32 x2 − 2 (i) y = p (x) =
Set up one table of values for all three cases, using integer values of x from 3 to 3.
−
(b) State the characteristics of the graphs of q and r in relation to those of the graph of p.
(a)
−3 − 272 y = − 32 x2 y = − 32 x2 + 1 − 25 2 y = − 32 x2 − 2 − 31 2 x
−2 −6 −5 −8
−1 0 1 2 3 − 32 0 − 32 −6 − 272 − 12 1 − 12 −5 − 252 − 72 −2 − 72 −8 − 312
191
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y 4
−4 −3 −2 −1
2 0
2
3
4
1
2
−−4 −6 −8 −10 −12 −14 −16
x
y = q(x) =
− 32 x2 + 1 y = p(x) = − 32 x2
y = r (x) =
− 32 x2 − 2
Figure 4.1.5
q(x) thus
3 2 2
≤ 1 since − x ≤ 0 and − x + 1 ≤ 1. 3 2 2
r(x) 2 since and hence 32 x2
≤−
3 2 2
− x ≤0 − − 2 ≤ −2.
(b) The graph of q is the same as that of p except that it has been shifted 1 unit upwards. Thus D q = R; the axis of symmetry of the graph is the y –axis, it is concave down and the vertex is ( 0, 1). Thus q (x) has a maximum value of 1, and hence R q = ( , 1].
−∞
The graph of r is the same as that of p except that it has been shifted 2 units downwards. Thus Dr = R; the axis of symmetry of the graph is the y–axis, it is concave down and ( 0, 2) is its turning point. Hence r (x) has a maximum value of 2, and hence R r = ( , 2].
−
−
−∞ −
In the following table we summarise the effect of k on parabolas defined by y = ax 2 + k for a fixed value of a .
192
THE ROLE OF k WITH RESPECT TO THE PARABOLA DEFINED BY y = ax 2 + k
For a fixed value of a the parabola defined by y = ax 2 + k has the same shape as the parabola defined by y = ax 2 . It is obtained by shifting the parabola defined by y = ax 2
−k > 0 when k < 0. −3, then −k = −(−3) = 3. Note:
For example, if k =
k units upwards if k > 0
−k (i.e. the numerical value of k) units downwards if k < 0.
The vertex of the parabola defined by y = ax 2 + k is (0, k ). The vertex (0, k ) is
above the x –axis if k > 0
at the srcin if k = 0
below the x –axis if k < 0.
Table 4.1.2
We often mentioned thethe importance ofbetween using mathematical terminology and have notation correctly. Note difference the expressions lowest (or highest) point and the smallest (or biggest) value. Consider this terminology in relation to parabolas defined by y = ax 2 or y = ax 2 + k. Consider the parabola defined by y = f (x) = ax 2 + k, where a > 0 and k < 0. (See Figure 4.1.6.)
y
f
x
A(0, k)
Figure 4.1.6
In Figure 4.1.6 we note that
all points (x, y) on the graph satisfy the equation y = f (x)
the lowest point on the graph of the function f is the vertex, A
the smallest value of f (x) is k , the y –coordinate of the vertex
193
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the smallest value of f (x) occurs when x = 0, which is the x –coordinate of the vertex.
We have just discussed the case where a > 0 and k < 0. You may find it helpful to formulate similar statements for the case a < 0.
4.1C PARABOLAS DEFINED BY y = a (x
− h)2 + k
In the previous two study units we looked at parabolas defined by y = ax 2 and y = ax 2 + k. We now consider parabolas defined by y = a(x h)2 + k.
−
We determine the effect of h by keeping the values of a and k fixed, and then considering different values of h . PARABOLAS DEFINED
h)2
BY y = ( x
In the next example and activity we assume a = 1 and k = 0, i.e. we consider parabolas defined by y = ( x h)2 .
−
−
4.1.3 (a) On the same system of axes sketch the parabolas defined by (i) y = f (x) = x 2
(ii) y = g(x) = ( x
− 1)2
by setting up one table of values for both functions, using integer values of x from 3 to 4.
−
(b) What is the relationship between the graphs of g and f ?
SOLUTION (a)
−3 −2 −1 0 1 2 3 4 −4 −3 −2 −1 0 1 2 3
x x
−1
y = x2 y = (x
− 1)2
9
4
1
0 1 4 9 16
16
9
4
1 0 1 4
9
194
y 16
14 12 10
8
y = f (x) = x2
6
4
y = g (x) = ( x
− 1)2
2
−4 −3 −2 −1
0
1
x 2
3
4
Figure 4.1.7
(b) Both functions have the same domain and range, and both graphs have the same shape. However, relative to the graph of f , the axis of symmetry of the graph of g, and hence also its vertex, have shifted 1 unit to the right . Thus the axis of symmetry of the graph of g is the line defined by x = 1 and the vertex is the point (1, 0).
In Example 4.1.3 we have h > 0, and the axis of symmetry of the parabola defined by y = (x h)2 lies to the right of the y–axis and is the line defined by x = h. The vertex is the point ( h, 0). In the next activity we consider the axis of symmetry when h < 0.
−
195
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4.1.4 (a) On the same system of axes sketch the parabolas defined by (i) y = f (x) = x 2
(ii) y = l (x) = ( x + 2)2
by setting up one table of values for both functions, using integer values of x from 5 to 3.
−
(b) Give the equation of the axis of symmetry and the coordinates of the vertex of the graph of l . (c) From the graphs in (a) describe the relation ship between the graphs of l and f .
(a)
−5 −4 −3 −2 −1 0 1 −3 −2 −1 0 1 2 3 4
x x+2 y = x2
25 1 6
y = ( x + 2)2
9
4
4
2
3
5
9
4
1 0 1
9
1
0
1 4 9 16 25
y 14 12
y = f (x) = x 2
10
8 6
y = l (x) =
( x + 2)2
4
2
x
−5 −4 −3 −2 −1
0
1
Figure 4.1.8
2
3
4
196 (b) The axis of symmetry of the graph of l is the line defined by x = the vertex is ( 2, 0).
−
−2 and
(c) The graph of l , i.e. the parabola defined by y = (x + 2)2 , has the same shape as the graph of f , i.e. the parabola defined by y = x2 , but it is 2 units further to the left than the graph of f .
In Activity 4.1.4 we may write the equation y = (x + 2)2 as y = (x ( 2))2 , and the axis of symmetry is the line defined by x = 2. Thus, when h < 0, the axis of symmetry of the parabola defined by y = ( x h)2 is the line defined by x = h and it lies to the left of the y–axis . The vertex is the point ( h, 0).
− −
PARABOLAS DEFINED BY y = a(x
2
− h) , a = 1
−−
We have considered the effect of the sign of h for parabolas defined by y = a (x h)2 , where a = 1. Now, in the next activity, we consider the effect of different values of h , when a = 1.
−
4.1.5 (a) On the same system of axes sketch the graphs of
−2x2 (ii) y = g(x) = −2(x − 2)2 (iii) y = h(x) = −2(x + 1)2 (i) y = f (x) =
by setting up one table of values for all the functions, using integer values of x from 3 to 4.
−
(b) Describe the relationship between the graphs of (i) g and f (ii) h and f .
197
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(a)
x
(x
− 2)
(x + 1)
−2x2 y = −2(x − 2)2 y = −2(x + 1)2
y=
−3 −5 −2 −18 −50 −8
−2 −4 −1 −8 −32 −2
−4 −3 −2 −1 y = h ( x) =
−2(x + 1)2
3
4
4
5
−2 0 −2 −8 −18 −32 −18 −8 −2 0 −2 −8 0 −2 −8 −18 −32 −50 1
−2 −4 −6 −8 −10 −12 −14 −16
0 1 2
y 0
−1 0 1 2 3 −3 −2 −1 0 1 2
3
2
4
5
x
y = g (x) =
−2(x − 2)2
y = f (x) =
−2x2
Figure 4.1.9
(b)
(i) The graph of g is obtained by shifting the graph of right.
f 2 units to the
(ii) The graph of h is obtained by shifting the graph of f 1 unit to the left.
198 In the following table we summarise the effect of h on parabolas defined by y = a (x h)2 .
−
THE ROLE OF h WITH RESPECT TO THE PARABOLA DEFINED BY y = a (x h)2
−
2 2 a (x For a fixed valueasofthea parabola the parabola defined y= h) has the same shape defined by y by = ax . It is obtained 2 by shifting the parabola defined by y = ax
−
Note:
−h > 0 when h < 0
h units to the right if h > 0
−h (i.e. the numerical value of h) units to the left if h < 0.
The axis of symmetry of the parabola is the line defined by x = h. The vertex is the point ( h, 0) and it is
to the right of the y –axis if h > 0
at the srcin if h = 0
to the left of the y –axis if h < 0.
Table 4.1.3
In the examples and activities that we have considered so far we have looked in turn at the effects of a, h and k on parabolas defined by y = a (x h)2 + k. We have the following.
−
When we speak of the value of a (or h or k ) we mean both the sign and numerical value of a (or h or k ).
The value of a determines the shape of the parabola.
The value of k determines the vertical shift with respect to the parabola defined by y = ax 2 . The value of k is the y –coordinate of the vertex of the parabola and is the maximum or minimum value of the function.
The value of h determines the horizontal shift with respect to the parabola defined by y = ax 2 . The equation of the line which is the axis of symmetry of the parabola is x = h, and h is the x –coordinate of the vertex .
In the next example we use these properties to predict the characteristics of two specific parabolas defined by y = a (x h)2 + k, where a = 0, h = 0 and k = 0. In Activity 4.1.6 you can sketch the parabolas using tables of values, and see whether the predictions are correct.
−
199
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4.1.4 For each of the functions f and g defined by (a) f (x) = 2(x
− 1)2 + 3
(b)
g (x) =
− 12 (x + 2)2 − 1
(i) state whether the graph is concave up or down (ii) if the graph is concave up state whether it is narrower or wider than the standard parabola; if it is concave down state if it is narrower or wider than the parabola defined by y = x2 (iii) give the axis of symmetry of the graph
−
(iv) give the vertex of the graph (v) give the maximum or minimum value of the function (vi) give the domain and range of the function.
SOLUTION (a) When we compare f (x) = 2(x a = 2, h = 1 and k = 3.
− 1)2 + 3 with y = a(x − h)2 + k we see that
(i) Since a = 2, i.e. a > 0, it follows that the parabola is concave up. (ii) The parabola is narrower than the standard parabola since a = 2, i.e. a > 1. (iii) The axis of symmetry is the line defin ed by x = h, i.e. x = 1. (iv) The vertex is the point (h, k), i.e. ( 1, 3). (v) Since the graph is conc ave up it has a lowest point. Hence the minimum value of f (x) is k , i.e. 3. (vi) D f = R and R f = [3,
∞). (b) When we compare g (x) = − 12 (x + 2)2 − 1 with y = a (x − h)2 + k we have a = − 12 , h = −2 (since ( x + 2) = ( x − (−2))) and k = −1. (i) Since a = − 12 , i.e. a < 0, it follows that the parabola is concave down. x2 since the
(ii) The parabola is wider than the parabola defined by y =
magnitude of a is which is less than 1. (iii) The axis of symmetry is the line defin ed by x = h, i.e. x =
−
1 2,
(iv) The vertex is the point (h, k), i.e. ( 2, 1).
− −
−2.
(v) Since the graph is concave down the graph has a highest point. Hence the maximum value of g (x) is k , i.e. 1. (vi) Dg = R and R g = (
−∞, −1].
−
200
4.1.6 Sketch the parabolas defined by (a) y = f (x) = 2(x
− 1)2 + 3
(b)
y = g(x) =
− 12 (x + 2) − 1 −2 to 4 for (a); and
by setting up tables of values using integer values of x from from 5 to 1 for (b).
−
Confirm that the graphs have the properties that have been deduced in Example 4.1.4.
(a)
−2 −1 0 1 2 3 4 −3 −2 −1 0 1 2 3
x
(x
− 1) 1)2 + 3
y = 2(x
21
11
5 3 5 1
1 21
− y 12
10 8
y = f (x) = 2(x
6
− 1)2 + 3
4
(1, 3)
2
−3 −2 −1
x 0
1
Figure 4.1.10
2
3
4
201
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(b)
−5 −4 −3 −2 −1 0 1 −3 −2 −1 0 1 2 3 (x + 2) −3 − 32 −1 − 32 −3 − 112 y = − 12 (x + 2)2 − 1 − 11 2 x
y 1
x
−5 −4 −3 −2 −1
0
−1 −2 −3 −4
1
2
y = g(x) =
− 12 (x + 2)2 − 1
5
−
Figure 4.1.11
Figures 4.1.10 and 4.1.11 show us that the parabolas have the properties deduced in Example 4.1.4.
We now know the effects ofa , h and k on the characteristics of parabolas defined by y = a (x h)2 + k. Thus we can now sketc h parabolas using these cha racteristics, and we no longer need to use tables of values. In addition to these characteristics, we use the y –intercept and x –intercept(s) (if any exist) to help us sketch the graphs.
−
202 Study Figure 4.1.4 in Study Unit 4.1B. Each of the parabolas has a but the parabola defined by
y = h (x) = x 2
2
y = f (x) = x has only one x –intercept
y = g (x) = x 2 + 2 has no x –intercepts.
y –intercept,
− 3 has two x–intercepts
We now consider this in more detail. Remember that the y –intercept is obtained by substituting x = 0 and the x –intercept(s) is (are) found by substituting y = 0 into the equation that defines the parabola. Thus by substituting x = 0 into each of the equations above we see that the y–intercepts of the graphs of h , f and g are 3, 0 and 2, respectively.
−
For the x –intercept(s) we substitute y = 0 into each of the equations. For h we obtain 0 = x2 i.e. we have
−3
2
and thus
x =3
√ ± 3. √ √ Hence the x –intercepts of the graph of h are − 3 and 3. x=
For f we obtain
x2 = 0 i.e. we have
x = 0. Thus the graph of f has only one x –intercept, namely 0. For g we get 0 = x2 + 2 i.e. we have
x2 =
−2
which has no real solution. Thus the graph of g has no x –intercept.
203
MAT0511/003
We now look at the general case, i.e. the parabola defined by y = a(x
− h) 2 + k .
When we substitute x = 0 into the equation we obtain
y = a ( h) 2 + k
−
= ah 2 + k. Thus the y –intercept is ah 2 + k. When we substitute y = 0 into the equation we obtain 0 = a (x
− h)2 + k.
Now 0 = a(x
⇔
a(x
− h)2 = −k
⇔
(x
− h)2 = −ak
(x
⇔
− h) =
since a = 0
− ± − k a
x=h
⇔
− h)2 + k
±
k a
−k ≥ 0
if
a
if
−k ≥ 0. a
Thus we have the following.
−
k < 0. There are two x –intercepts if ak > 0, i.e. if a
There is only one x –intercept if ak = 0, i.e. if k = 0.
k > 0. There is no x –intercept if ak < 0, i.e if a
−
−
In the following table we summarise the characteristics of parabolas defined by
y = a (x
− h) 2 + k .
204
CHARACTERISTICS OF PARABOLAS DEFINED BY y = f (x) = a(x h)2 + k
−
Note: It is very impo rtant
The domain and range of f are
=
R
Rf
=
[k, (
∞) −∞, k]
if a > 0 if a < 0.
concave up if a > 0 concave down if a < 0.
The parabola is
The bigger the numerical value of a, the narrower the parabola becomes. The smaller the numerical value of a, the wider the parabola becomes.
The axis of symmetry is the line defined by x = h .
The vertex is (h, k ).
When a > 0, f (x) has a minimum value, namely k . When a < 0, f (x) has a maximum value, namely k .
that students know and will be able to apply these cha-
Df
racteristics in problems or exercises.
The y –intercept is ah 2 + k, or substitute x = 0 in y = a(x h)2 + k.
−
The x–intercepts (if any) are h substitute y = 0 in y = a(x
− −ak and h + −ak , or
− h)2 + k.
Table 4.1.4
Try one more activity before going on to the next section.
4.1.7 Suppose f is the function defined by y = f (x) =
−2(x + 1)2 + 8. Determine
(a) D f and R f (b) the y –intercept and x –intercept(s) of the graph (c) the axis of symmetry of the graph
205
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(d) the turning point (vertex) of the graph (state whether it is the lowest or the highest point of the graph).
(a) Since f is defined for all x
and hence
∈ R we have D f = R. Also (x + 1)2 ≥ 0 −2(x + 1)2 ≤ 0
so that
Thus R f = (
−2(x + 1)2 + 8 ≤ 8.
−∞, 8].
(b) When we substitute x = 0 into the equation we obtain
y=
−2(1)2 + 8 = −2 + 8 = 6.
Thus the y –intercept is 6. When we substitute y = 0 into the equation we obtain 0 =
−2(x + 1)2 + 8.
Now 0=
−2(x + 1)2 + 8
⇔
2(x + 1)2 = 8
⇔
(x + 1)2 = 4
⇔
x+1 =
⇔
x= x=
⇔
−3
√ ± 4 −1 ± 2
or x = 1.
−3 and 1. (c) When we compare y = −2(x + 1)2 + 8 with y = a (x − h)2 + k we have Thus the x –intercepts are
a=
2, h =
−
1 and k = 8.
−
The axis of symmetry is the line defined by x = h, i.e. the line defined by
x=
−1.
(d) The turning point (vertex) of the parabola is ( h, k), i.e. ( 1, 8). It is the highest point of the graph since the parabola is concave down. This follows from the fact that a = 2, i.e. a < 0.
−
−
206 We have seen that the axis of symmetry of a parabola defined by y = ax 2 + k is the line defined by x = 0. We can also show algebraically that, for example, the axis of symmetry of the parabola defined by y = 3(x 2)2 + 1 is the line defined by x = 2. We proceed as follows.
−
We write y = 3(x 2)2 + 1 as y = 3z2 + 1, where z = x 2. The axis of symmetry of the parabola defined by y = 3z2 + 1 is the line defined by z = 0.
−
−
Now z = 0 is the same as x 22 = 0, i.e. x = 2. Hence the axis of symmetry of the parabola defined by y = 3z + 1, i.e. by y = 3(x 2)2 + 1, is the line defined by x = 2.
−
−
4.1 1. Use the method described just after Activity 4.1.7 to show that the equation of the axis of symmetry of the parabola defined by y = a (x h)2 + k is x = h.
−
2
2. State how you would shift the graph of y = 3x in order to obtain the graphs defined by the following equations. (a) y = 3(x
− 1)2 − 2
(b) y = 3(x + 1)2 + 2
− 1)2 + 2 (d) y = 3(x + 1)2 − 2 (e) y = 3(x − h)2 + k for h < 0 and k > 0 (c) y = 3(x
3. For each of the functions f and g defined by (a) f (x) = ( x + 3)2 (b) g(x) =
−3(x − 2)2 + 4
determine (i) the domain and range of the function (ii) the y –intercept and x –intercept(s) (if any) of the graph (iii) the equation of the axis of symmet ry of the graph (iv) the vertex of the graph (state whether it is the lowest or the highest point of the graph).
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4.2 SKETCHING PARABOLAS
4.2A SKETCHING PARABOLAS DEFINED BY y = a (x
− h) 2 + k
In the previous section we noted several properties of parabolas defined by y = a(x h)2 + k. In this section we use these properties to sketch such parabolas.
−
4.2.1 Sketch the parabola defined by
y = f (x) =
−2(x + 1)2 + 8
without using a table of values.
SOLUTION In Activity 4.1.7 we found that
D f = R, R f = (
−∞, 8]
the y –intercept is 6
the x –intercepts are
the parabola is concave down
the axis of symmetry is the line defined by x =
the vertex is ( 1, 8), which is the highest point of the graph.
−3 and 1
−1
−
We use this information to sketch the parabola in Figure 4.2.1.
208 In order to sketch the parabola in Figure 4.2.1 below, we
mark off the x –intercepts
mark off the y –intercept
draw the axis of symmetry
plot the vertex
join the points so that the parabola looks symmetrical with respect to the axis of symmetry.
y
( 1, 8)
−
Note that the axis of sym-
6
metry lies half–way between the x –intercepts. This is true
y=
for any parabola which cuts
−2(x + 1)2 + 8
the x –axis twice, as we will prove later, just after Table 4.2.2.
−3
−1
Figure 4.2.1
Now try the following activity.
1
x
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4.2.1 Without using a table of values sketch the parabola defined by
y = f (x) = 13 (x
− 3)2 + 1.
Show all the relevant calculations.
Domain and range f is defined for all x R and hence D f = R . Since (x 3)2 0 we have 13 (x 3)2 0, i.e. f (x) 1. Thus R f = [1, ).
≥
−
≥
∈
∞
−
1 3 (x
≥
− 3)2 + 1 ≥ 1 and hence
y– and x –intercepts Substitution of x = 0 into the equation gives y = 13 ( 3)2 + 1 = 3 + 1 = 4.
−
Hence the y –intercept is 4. If y = 0 then the equation becomes 0 = 13 (x which gives 1 ( 3 x
We thus have
− 3)2 + 1
− 3)2 = −1.
− 3)2 = −3 which has no solutions since (x − 3)2 ≥ 0. Thus there are no x –intercepts. (x
Axis of symmetry Comparing y = 13 (x 3)2 + 1 with y = a (x h)2 + k we have a = 13 , h = 3 and k = 1. Thus the axis of symmetry is the line with equation x = 3.
−
−
Vertex (turning point) The vertex is ( h, k), i.e. (3, 1), and it is the lowest point of the graph since the parabola is concave up (because a = 13 , i.e. a > 0).
The parabola is sketched in Figure 4.2.2.
210
y 4
(6, 4)
y = 13 (x
− 3)2 + 1
(3, 1) x 3
6
Figure 4.2.2
You will notice that in this case we can obtain only two of the points on the parabola directly from the equation, since there are no x–intercepts. In order to sketch the curve better we use symmetry to obtain another point on the parabola. We know that (0, 4) is a point on the left arm of the parabola since 4 is the y–intercept. This point lies 3 units to the left of the axis of symmetry. Thus the point on the right arm of the parabola that corresponds to (0, 4) lies 3 units to the right of the axis of symmetry and is hence the point (6, 4). We plot this point and use it when we sketch the parabola.
We now move on to parabolas defined by y = ax 2 + bx + c. We sketch them by applying the techniques we have used so far.
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4.2B SKETCHING PARABOLAS DEFINED BY y = ax 2 + bx + c At the beginning of Section 4.1 we defined a quadratic function in x as a function f that is defined by
f (x) = ax 2 + bx + c;
where a = 0.
Quadratic functions are normally expre ssed in this form. We now consider the different techniques that we can use to sketch parabolas defined in this way.
4.2.2 We consider the function f defined by
y = f (x) = 3x2
− 12x + 9.
(1)
We can obtain the information we require to sketch the graph of f by
h)2 + k by completing the square,
Refer to Study Unit 1.2C of
rewriting the equation as y = a (x
Book 2 if you have forgotten how to complete the square.
and then proceeding as we did in the previous example and activity.
−
We do this now.
y = f (x) = 3x2
− 12x + 9 3(x − 4x) + 9 3(x2 − 4x + 22 − 22 ) + 9 3(x − 2)2 + 3(−4) + 9 3(x − 2)2 − 12 + 9 3(x − 2)2 − 3 2
= = = = = Domain and range Df = R Since (x 2)2 0 we have f (x) = 3(x
− ≥
− 2)2 − 3 ≥ −3 and hence R f = [−3, ∞).
y– and x –intercepts By substituting x = 0 into (2) we obtain y = 3( 2)2
− − 3 = 12 − 3 = 9.
Thus the y –intercept is 9.
(2)
212 By substituting y = 0 into (2) we obtain 0 = 3(x
− 2)2 − 3.
Now 0 = 3(x
⇔
3(x
⇔ ⇔
(x
− 2)2 − 3
− 2)2 = 3 2
− 2) = 1 (x − 2) = ±1
⇔
x=2
⇔
x = 1 or x = 3.
±1
Thus the x –intercepts are 1 and 3. f (x) = a (x h)2 + k where a = 3, h = 2 and k = 3.
−
−
The axis of symmetry The axis of symmetry is the line defined by x = 2. The vertex
−
The vertex is ( 2, 3) and it is the lowest point of the parabola since the graph is concave up (because a = 3, i.e. a > 0). The graph is sketched in Figure 4.2.3.
y 9
y = 3x 2
1
3
2
(2, 3)
Figure 4.2.3
−
− 12x + 9
x
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In Example 4.2.2 we used equation (2) to find the y – and x –intercepts, the axis of symmetry, the vertex and R f . We can also find the y – and x –intercepts by using equation (1). By substituting x = 0 into (1) we obtain
y = 3(0)2
− 12(0) + 9 = 9,
i.e. the y –intercept is 9. By substituting y = 0 into (1) we obtain 0 = 3x2
− 12x + 9.
Now 0 = 3x 2
x2
⇔ ⇔ ⇔
(x
− 12x + 9
− 4x + 30=
Divideby3.
− 1)(x − 3) = 0 x
−1 = 0
⇔
or x
−3 = 0
x = 1 or x = 3.
Thus the x –intercepts are 1 and 3.
Take care not to make the following common mistake. Do not write, for example,
y = 3x 2 as
− 12x + 9
x2
y
=
4x
−
3.
+
(1) 2
( )
Equation (2) has been obtained by dividing only the RHS of (1) by 3. This is not valid.
See point 4 of Table 2.1.2, Study Unit 2.1A of Book 2.
Remember that we produce an equivalent equation (i.e. an equation having the same solution set as the srcinal equation) by dividing both sides of the srcinal equation by the same number.
214 It is clear that (1) and (2) represent different parabolas. The coefficient of x2 in both equations is positive and thus both parabolas are concave up. However, in (1) the coefficient of x2 is 3, and in (2) it is 1. The parabola defined by (1) is thus narrower than the parabola defined by (2). Another point to note is that the y–intercepts of the two parabolas are different. If we substitute x = 0 into (1) and into (2) then we see that the y –intercepts of the parabolas defined by (1) and (2) are respectively 9 and 3. However, when we replacey by 0 we can write 3 x2 as
− 12x + 9 = 0
x2 0
÷3 = 0
− 4x + 3 = 0
(3) (4)
because both sides have been divided by 3. Hence equations (3) and (4) are equivalent. We can thus obtain the x –intercepts of the parabola defined by (1) by solving either (3) or (4). Similarly we can obtain the x –intercepts of the parabola defined by (2) by solving either (4) or (3). Hence the two parabola s have the same x –intercepts. Please take note of the difference between
y = 3x 2
− 12x + 9
and 3 x2
− 12x + 9 = 0
and, in general, between
y = ax 2 + bx + c and
ax2 + bx + c = 0.
In the previous example the x–intercepts of the graph of y = 3 x2 1 and 3. We note that
y = 3 x2
12x + 9 = 3(x2 4x + 3) = 3(x 1)(x 3)
− − − −
and that both x = 1 and x = 3 are the solutions of 0 = 3( x
− 1)(x − 3).
− 12x + 9 are
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In general, if
y = f (x) = a(x
− r1)(x − r2)
then the x –intercepts of the graph of f are r 1 and r 2 . Conversely, if r 1 and r 2 are the x–intercepts of the graph of a quadratic function f , then the function f can be defined by
f (x) = a (x
− r1)(x − r2).
(4.2.1)
Equation (4.2.1) is thus a third possible form of the equation which defines a quadratic function. We now return to a parabola defined by the general equation
y = ax 2 + bx + c
(4.2.2)
and consider its y – and x –intercepts. If we substitute x = 0 into (4.2.2) we obtain
y = a(0) + b(0) + c = c. Hence the y –intercept is c. If we substitute y = 0 into (4.2.2) we obtain the equation
ax2 + bx + c = 0. You may need to revise “nature of roots” just before Example 2.3.6 in Study Unit 2.3A of Book 2.
(4.2.3)
Do you remember that in Study Unit 2.3A of Book 2 we use the discriminant ∆ to determine whether equation (4.2.3) has two solutions, only one solution or no solutions? What does this mean in terms of the parabola defined by (4.2.2)? Well, the solutions of (4.2.3) give us the x –intercepts of the parabola and hence the discriminant tells us whether, and how often, the parabola cuts the x–axis. The following table summarises the role of the discriminant with respect to the x–intercept(s) of the parabola.
THE ROLE OF ∆ = b2
− 4ac
If ∆ > 0, the parabola cuts the x –axis twice, i.e. the parabola has two x –intercepts. (Equation (4.2.3) has two different roots.)
If ∆ = 0, the parabola touches the x –axis once, i.e. the parabola has one x –intercept. (Equation (4.2.3) has only one root.)
If ∆ < 0, the parabola does not cut or touch the axis, i.e. the parabola has no x –intercept. (Equation (4.2.3) has no real roots.)
Table 4.2.1
216 Diagrammatically we have the following cases.
If ∆ > 0 then there are two x–intercepts, i.e. the parabola cuts the x–axis twice.
y
y
x
0
If ∆ = 0 then there is one x –intercept, i.e. the parabola touches the x –axis once.
y
0
x
0
y
x
0
x
If ∆ < 0 then there are no x–intercepts, i.e. the parabola does not cut or touch the x –axis.
y
0
y
x
0
x
Look again at Example 4.1.2 in Study Unit 4.1B. Figure 4.1.4 shows three parabolas: one cuts the x–axis twice, one touches the x–axis and one does not cut or touch the x –axis. In the next example we determine the discriminant for each of the three functions whose graphs are sketched in Figure 4.1.4.
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4.2.3 For f (x) = x 2 , a = 1, b = 0 and c = 0, and hence ∆
= b2
− 4ac = 0 − 4(1)(0) = 0.
For g (x) = x2 + 2, a = 1, b = 0 and c = 2, and hence
= b2
− 4ac = 0 − 4(1)(2) = −8 < 0. For h (x) = x2 − 3, a = 1, b = 0 and c = −3, and hence 2 ∆ = b − 4ac = 0 − 4(1)(−3) = 12 > 0. ∆
From Figure 4.1.4 we see that
If you have forgotten the quadratic formula, see the paragraphs just after Activity 2.3.4 in Study Unit 2.3C of Book 2.
the graph of f touches the x –axis and we have ∆ = 0
the graph of g does not cut or touch the x –axis and we have ∆ < 0
the graph of h cuts the x –axis twice and we have ∆ > 0.
If y = ax 2 + bx + c, then the quadratic formula
x=
−b ±
√2 b 2a
− 4ac
gives us the x –intercepts. Consider the general quadratic equation that defines a function f , namely
y = f (x) = ax 2 + bx + c. We now use completion of the square to rewrite this equation in the form
y = a(x
− h)2 + k,
where h and k can be expressed in terms of a , b and c . This will enable us to obtain general formulas for the axis of symmetry and the vertex of the parabola, and for the minimum or maximum value of f (x). Having formulas to use means that we do not have to complete the square every time we want to draw a parabola. However, understanding the method of completing the square means you do not need to panic if you forget the formulas!
218 We have
y = ax2 + bx + c b = a x2 + x + c a
− − − − − b x+ a
b 2a
= a x2 + b x +
b 2a
= a x2 +
a
b 2a
2
= a x+
b 2a
2
= a x+
b 2a
2
= a x+
2
2
b 2a
2
a
b 2a
+c
2
+c
ab2 +c 4a2 b2 4a
+c +
4ac b2 . 4a
− h)2 + k we have −b 2 + 4ac − b2 x−
If we write this in the form y = a(x
y=a
where h =
2a
4a
−b and k = 4ac − b2 . 2a
4a
From this, if f is defined by
f (x) = ax 2 + bx + c then we conclude the following.
(h, k) =
b 4ac b2 , 2a 4a
−
parabola.
−
,
or =
− − b ,f 2a
b 2a
is the vertex of the
The equation of the axis of symmetry of the parabola is x =
−b .
4ac b2 If a > 0 then the minimum value of f (x) is , or f 4a
2a b . 2a
If a < 0 then the maximum value of f (x) is
−
4ac b2 , or f 4a
−
− −
b . 2a
We summarise the characteristics of a parabola defined by y = ax 2 + bx + c in the table on the next page.
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CHARACTERISTICS OF A PARABOLA DEFINED BY y = f (x) = ax 2 + bx + c 2
We rewrite y = ax2 + bx + c as y = a Note: (Alternatively)
f
Rf =
− − − b 2a
∞
; f
;
∞
b 2a
b2 .
+ 4ac
4a
− −
−
The domain and range of f are
Df
if a > 0
x
b 2a
Rf
= =
if a < 0
R
− ∞ −∞ − 4ac b2 , 4a
4ac b2 , 4a
if a > 0 if a < 0.
concave up if a > 0 concave down if a < 0.
The parabola is
The bigger the numerical value of a, the narrower the parabola becomes. The smaller the numerical value of a, the wider the parabola becomes.
The axis of symmetry is the line defined by x =
The vertex is
−
b , 4ac b2 2a 4a
−
−b . 2a
− − or =
b ,f 2a
b 2a
.
2 When a > 0, f (x) has a minimum value, namely 4ac b 4a b or f . 2a 2 When a < 0, f (x) has a maximum value, namely 4ac b 4a b or f . 2a
−
− −
−
The y –intercept is c .
The x –intercepts (if any) are
−b −
√2 b 2a
− 4ac
and
Table 4.2.2
−b +
√2 b 2a
− 4ac .
220 We mentioned earlier (see the margin note next to Figure 4.2.1) that when a parabola has two x –intercepts, the axis of symmetry lies half–way between these intercepts. For interest we now show that this is true in general. If the parabola defined by y = ax 2 + bx + c cuts the x–axis twice then it passes through the points
See the midpoint formula, equation (1.3.3) in Study Unit 1.3A of Topic 1.
− −
b2 2a
b
−
− 4ac , 0
− 4ac , 0
.
The point which lies half–way between these points is
− −√ − − b
b2 4ac 2a
+
√
b+ b2 4ac 2a
2
Note: It is not important to remember these simplified formulas.
b2 2a
b+
and
If we simplify the expression we have
−
,
0
.
− −
2b , 0 , 4a
i.e. we have
b , 0 . 2a
−
Now the axis of symmetry is the vertical line which is defined by x = b . It is 2a b , thus parallel to the y –axis and passes through the point 2a 0 .
−
We have thus shown that the axis of symmetry lies half–way between the x–intercepts. Now use the properties given in Table 4.2.2 to do the following activity.
4.2.2 Without using a table of values sketch the graph of g , where g is defined by
y = g(x) = Show all the appropriate calculations.
−x2 − 3x + 1.
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Note: (Alternatively)
Rg = But
− − ∞
b 2a
; f
When we compare y = b = 3 and c = 1.
−
−x2 − 3x + 1 with y = ax 2 + bx + c we have a = −1,
Domain and range
−b = 2a
3 . 2
Dg =
−
− − − − − b 2a
f
= = = =
4ac b2 4a 4 9 , 4
−∞ − −∞ − −
Rg =
Hence,
=
R
3 2
=
2
= (
−∞
3 +1 2 9 9 + +1 4 2 9 + 18 + 4 4 13 1 =3 . 4 4 3
− −
Hence, Rg =
−
1 , 34]
y– and x –intercepts The y –intercept is 1. To find the x –intercept we substitute y = 0 into the equation. We obtain
−x2 − 3x + 1 = 0 which we cannot factorise. Thus we need to use the quadratic formula. We have
−b ± √b2 − 4ac √ 2a 3± 9+4 = √−2 3 ± 13 = −2 . −3 − √13 −3 + √13 x =
1 . 4
− ∞, 3
,
Thus the x –intercepts are
and
2
2
.
Axis of symmetry The axis of symmetry is the line defined by
y=
−b = −(−3) = − 3 . 2a
2
2
−
Vertex The vertex is
−
b 2a
,
4ac b2 4a
−
3 4( 1)(1) ( 3)2 , 2 4( 1) 3 4 9 , 2 4 3 1 , 3 . 2 4
− − − − − − − − − − = = =
222 Note: (Alternative) Vertex is
− − b ; 2a
f
b 2a
Do not try to learn the formula for the y –coordinate of the vertex. You can find it by substituting x = b into the equation that defines the function. If we 2a do this we have
−
2 − −23 − 3 −23
y =
+1
− 94 + 92 + 1
=
= 13 4 = 3
1 . 4
The vertex is the highest point since the parabola is concave down (because a = 1, i.e. a < 0).
−
The graph is sketched in Figure 4.2.4.
y
(
−
3 1 ,34) 2
1−3+√13
−3−√13 2
2
− 32 y=
x
−x2 − 3x + 1
Figure 4.2.4
Note that the x–intercepts are irrational numbers. However, we can find their approximate values and thus know roughly where to plot them. Using a calculator we find
√13 ≈ 3, 61. We thus have x
so that
x
≈ 3 ±−32, 61
≈ 6−, 612
or x
≈ −−0,261
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i.e.
x
≈ −3, 31
or x
≈ 0, 31.
It is not always necessary to use all the information given in Table 4.2.2. For example, if ax 2 + bx + c factorises, we use the factorised form instead of the
quadratic formula to solve ax 2 + bx + c = 0 and find the x –intercepts. Now practise sketching a few parabolas by doing some questions from Exercise 4.2.
4.2 1. On separate systems of axes, without using tables of values, sketch the parabolas defined by (a) y = 4(x + 1)2 + 1 (b) y = 4(x + 1)2 (c) y = 4(x + 1)2
1
− 12 x2 + x −+ 12 (e) y = 3x2 − 5x − 2.
(d) y =
2. Use the discriminant to determine whether the parabola with equation y = 2x2 3x + 2 cuts the x –axis.
−
3. Consider the graph of f , where
y = f (x) =
−6 + 5x − x2
and complete the following statements (select the appropriate word where applicable). (a) The graph of f is concave up/down. (b) The axis of symmetry is the line defined by .................. . (c) The maximum/minimum value of y is
1 4
.
(d) The graph of f cuts the x –axis in ............ place(s). (e) The graph of y = p + 5x p = ......... .
− x2 touches the x –axis at only one point if
224
4.3 FINDING THE EQUATION OF A PARABOLA So far in this topic we have considered characteristics of parabolas, and the way in which these charac teristics enable us to sketch the parabolas. In this section we now find the equations of parabolas, given sufficient information about them.
4.3A THREE FORMS OF THE EQUATION OF A PARABOLA We have considered three different forms of quadratic equations that define parabolas. They are
If the graph touches the
x–axis at only one point,
y = ax 2 + bx + c; a, b, c (general form)
∈ R and0 a =
(4.3.1)
y = a (x h)2 + k; a, h, k R and 0a = ((h, k ) is the vertex of the parabola)
−
∈
(4.3.2)
R and0a = y = a (x r1 )(x r2 ); a, r1 , r2 (r1 and r 2 are the x –intercepts of the parabola).
−
−
∈
(4.3.3)
then r1 = r2 .
If we want to find the equation that defines a particular parabola, the information we have determines which equation is more convenient to use. We may use
equation (4.3.1) if we are given 3 unrelated points that lie on the parabola
equation (4.3.2) if we know the vertex and one other point on the parabola
equation (4.3.3) if we are given the x –intercept(s) and another point on the parabola.
The information may be specified or supplied on a graph. Study the following example.
4.3.1 (a) A parabola passes through the points (1, 2), ( 1, 4) and (3, 8). Find the equation of the parabola.
−
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(b) Consider the parabola in the sketch below. Determine its equation.
y 6 5
y = f (x)
4 3 2 1
x
−1
1
−1 −2 −3
3
2
4
5
(2, 3)
−
Figure 4.3.1
(c) Consider the parabola in the sketch below. What is its equation?
y
8 7 6
y = f (x)
5 4 3 2 1
x
−1
0
1
Figure 4.3.2
2
3
226
SOLUTION (a) We are given three points on the parabola but we are not given any special information about them. Thus we make use of equation (4.3.1), namely y = ax2 + bx + c . We know that the points lie on the parabola and hence the coordinates of the points must satisfy the parabo la’s equation. When we substitute the coordinates of the different points into the equation we obtain 2 = a+b+c 4=a
−b+c
8 = 9 a + 3 b + c.
(1) (2) (3)
We subtract (2) from (1) and obtain
−2 = 2 b
i.e. we have
b= When we substitute b =
−1.
(4)
−1 into (1) we obtain 2=a
and hence
c=3
1+c
− − a.
(5)
We substitute (4) and (5) into (3) and obtain 8 = 9a i.e. we have
−3+3−a
8 = 8a and thus
a = 1. It follows that
c=3
− 1 = 2.
Thus the equation for the parabola is
y = x2
− x + 2.
(b) We are given the vertex and one other point on the parabola (since the y–intercept is 1 we know that (0, 1) lies on the graph) and thus we use equation (4.3.2), namely y = a (x - h)2 + k.
We have
y = a(x
− 2)2 − 3.
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The coordinates of the point (0, 1) satisfy the equation of the parabola and we thus have 1 = a( 2)2
− − 3.
Now
1 = a( 2)2
− −3
1 = 4a
⇔ ⇔
4a = 4
3
−
a = 1.
⇔
The parabola thus has the equation
y = (x
− 2)2 − 3
y = x2
− 4x + 1.
which can be rewritten as
(c) When we look at the parabola we can identify3 points on the parabola. We may use equation (4.3.1). However, notice that we have been given the two x –intercepts and thus we can also use equation (4.3.3), namely y = a(x – r1 )(x – r2 ).
The x –intercepts are
i.e. we have
−1 and 3 and thus the equation becomes y = a (x − (−1))(x − 3) y = a (x + 1)(x
− 3).
We substitute the coordinates ( 0, 6) into the last equation and obtain 6 = a (1)( 3)
−
i.e. we have 6= and thus
−3a
a= Thus the parabola has the equation
y=
2.
−
−2(x + 1)(x − 3)
which can be rewritten as
y=
−2x2 + 4x + 6.
228 In the following activity you need to find the equations of two parabolas which have been sketched on the same system of axes.
4.3.1 Write down the equations of the functions f and g whose graphs are the parabolas sketched in Figure 4.3.3.
y
5
(1, 5)
4 3
y = g ( x)
2 1
x
−1 y = f (x)
0
1
2
3
−1 −2 −3
Figure 4.3.3
The sketch shows that the vertex of the graph of f is ( 1, 5). Knowing the vertex suggests that we should use the form
h)2 + k
y = a (x
−
to find the equation for f . We therefore have
y = a (x
− 1)2 + 5.
Now the graph of f cuts the y–axis at 3, so the point (0, 3) lies on the graph of f , and hence the coordinates ( 0, 3) satisfy the equation that defines the graph.
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Hence 3 = a( 1)2 + 5
−
i.e. we have
a=
−2.
The equation that defines f is thus
y =
= = =
−2(x − 1)2 + 5 −2(x2 − 2x + 1) + 5 −2x2 + 4x − 2 + 5 −2x2 + 4x + 3.
At first glance it would appear that all we know about the graph of g is that it cuts the y–axis at 2, and that the value of a should be positive. However, we notice that the graph of g cuts the x–axis at the same points as the graph of f , whose equation we now know. If we find these x –intercepts then we can use
−
y = a (x
− r1)(x − r2)
to find the equation for g . To find the x–intercepts of the graph of f we substitute y = 0 into the equation y = 2x2 + 4x + 3. We obtain
−
−2x2 + 4x + 3 = 0. You can also make use of the quadratic formula.
Since this does not factorise we substitute y = 0 into y = We have
Now
−2(x − 1)2 + 5 = 0. −2(x − 1)2 + 5 = 0
⇔
2(x
− 1)2 = 5
⇔
(x
− 1)2 = 52
⇔ ⇔
x
±
−1 = ± x=1
5 2
5 . 2
−2(x − 1)2 + 5 instead.
230
− − − − − − − − − − − − 5 2
Thus the x–intercepts of the graph of g are 1 + equation that defines g is
y = a x
5 2
1+
= a x2
1+
= a x2
2x + 1
= a x2
2x
3 2
x
5 2
x
and hence the
5 2
1
5 2
1
5 2
and 1
x+ 1+
5 2
1
5 2
5 2
.
The y –intercept is 2 and hence ( 0, 2) lies on the graph. When we substitute x = 0 and y = 2 into the previous equation we obtain
−
−
−
−2 = − 32 a i.e. we have
a=
4 . 3
Thus the equation that defines g is
− 2x − 32
4 2 x 3
− 83 x − 2 .
y= 4 3 which we rewrite as
y=
x2
If a parabola touches the x –axis at only one point then it does not matter whether we use equation (4.3.3) or (4.3.2). Suppose a parabola touches the x–axis at ( p, 0). If we su bstitute r1 = p and
r2 = p into equation (4.3.3) we obtain y = a (x
− p)(x − p)
and thus
y = a (x
− p)2.
This equation is now in the form of (4.3.2).
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Since the parabola touches the x –axis at p, i.e. at the point ( p, 0), this point must be the vertex of the parabola. If we substitute h = p and k = 0 into equation (4.3.2) we get y = a(x p)2
−
which can be rewritten as
y = a(x
− p)(x − p).
This equation is now in the form of (4.3.3). Hence we have shown that if a parabola touches the x–axis in only one point then equation (4.3.3) reduces to equation (4.3.2) and vice versa. Now try some of the questions from the next exercise.
4.3 1. In each of the following cases, find the equation of the parabola which has the given properties. (a) The vertex is ( 2, 6) and the parabola passes through (1, 3).
−
−
−2 and 5 and the y–intercept is −5. − − (d) The x –intercepts are −4 and 2 and the maximum y –value is 6. (Hint: (b) The x –intercepts are
(c) It passes through the points (0, 0), ( 1, 2) and ( 2, 22).
the axis of symmetry lies half–way between the two x –intercepts.)
(e) The x –intercepts are 0 and 4 and the graph passes through ( 1, 1).
− −
2. Each of the following figures shows a parabola. Find the equation of each parabola. (a)
(b)
y
y
−
( 1, 4)
3
x x 2
( 3, 4)
− −
232 (c)
(d)
y
y
x
−1
(1,
− 52 )
4
x 1
(e)
y 9
(3, 1 12 )
(6, 3)
x
5
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4.4 USING PARABOLAS AND QUADRATIC FUNCTIONS
4.4A SOLVING QUADRATIC AND SIMPLE RATIONAL INEQUALITIES In Study Unit 2.3C of Book 2 we solve quadratic inequalities by using the following inequality properties, namely Remember that A and B rep-
if AB > 0 then A > 0 and B > 0 or A < 0 and B < 0 or if AB < 0 then A > 0 and B < 0 or A < 0 and B > 0;
resent algebraic expressions.
or by using the split–point method. We can also use parabo las to solve such inequalities. Quadratic inequalities
Solve the inequality
Note: The
4.4.1
three
2 x2
alternative
methods to solve quadratic inequalities are by using: I.
Inequality properties
II.
Split-point method
III.
Rough sketch of
− 5x − 3 > 0,
using a rough sketch of the parabola defined by y = 2x2
− 5x − 3.
SOLUTION We consider the parabola defined by
parabola
y = 2x 2
− 5x − 3.
We obtain the x –intercepts of the parabola by solving 2x2
5x
− −
We have
(2x + 1)(x and thus
− 3) = 0
2x + 1 = 0 or x Therefore
x=
− 12
− 3 = 0.
or x = 3.
3 = 0.
234 Thus the parabola cuts the x –axis at a = 2, i.e. a > 0.
− 12 and 3, and it is concave up since
A rough sketch of the parabola is shown in Figure 4.4.1. Note that we need not find the y –intercept or vertex as these are not relevant to the problem.
y
Note: The
three
2
y = 2x
alternative
methods to solve quadratic inequalities are by using
− 12
− 5x − 3
x 3
0
I. Inequality properties II. Split-point method III. Rough
sketch
of
parabola
Figure 4.4.1
Now we want to find values of x such that 2x2
− 5x − 3 > 0 i.e. such that the graph of y = 2x2 − 5x − 3 lies above the x –axis. From the sketch we see that the parabola lies above the x –axis when x < 12 or x > 3. Thus the solution of 2 x2 5x 3 > 0 is x < 12 or x > 3. You can check this answer by using any of the methods mentioned at the beginning of the study unit.
− −
−
−
You can practise the technique used in Example 4.4.1 in the next activity.
4.4.1 Solve the inequality 2+x
− x2 ≥ 0
by using a rough sketch of the parabola defined by y = 2 + x
− x2 .
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We rewrite the equation of the parabola as
y= We solve
−x2 + x + 2.
−x2 + x + 2 = 0 to find the x–intercepts.
Now
x2 + x + 2 = 0
−
x2
⇔ (x
⇔
−x−2 = 0
− 2)(x + 1) = 0 x=
⇔ Thus the x –intercepts are a = 1, i.e. a < 0.
−
−1
or x = 2.
−1 and 2 and the parabola is concave down since
We sketch the parabola defined by y =
−x2 + x + 2 in Figure 4.4.2.
y
x
−1
0
2
y=
−x2 + x + 2
Figure 4.4.2
The values of x for which 2+x
x2
0
− ≥ are the values of x for which the graph of y = −x2 + x + 2 lies above or on the
x–axis, i.e. values of x such that
−1 ≤ x ≤ 2. Thus the solution of 2 + x − x2 ≥ 0 is −1 ≤ x ≤ 2.
236 In Activity 4.4.1 we can also reason as follows. The solution of the inequality 2+x
− x2 ≥ 0
Remember the inequality sign reverses direction when we multiply both sides by a
is the same as that of
negative number.
We now consider the parabola defined by y = x 2 x 2. It also has x –intercepts of 1 and 2 but it is concave up since in this case a = 1, i.e. a > 0. Figure 4.4.3 shows a rough sketch of the parabola defined by y = x2 x 2.
−2 − x + x2 ≤ 0. − −
−
− −
y
y = x2
−x−2
x
−1
0
2
Figure 4.4.3
Now the values of x that satisfy
− 2 − x + x2 ≤ 0 are the values of x for which the parabola in Figure 4.4.3 lies below or on the x–axis, i.e. 1 x 2.
− ≤ ≤ Thus the solution of −2 − x + x2 ≤ 0, i.e. of 2 + x − x2 ≥ 0, is −1 ≤ x ≤ 2 which is the same as the solution we obtained in Activity 4.4.1.
Rational inequalities
Can we use a parabola to solve a simple rational inequality of the form
px + q rx + s
p, q, r, s
< 0;
≤
where < can be replaced by , > or look at the following example.
∈R
and
p = 0, r = 0
≥? STOP and THINK for a while and then
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4.4.2 Solve the inequality
− ≤0
x 3 2x + 1
(1)
by using a rough sketch of an appropriate parabola.
SOLUTION Do you remember why 2x + 1 = 0?
−
We first notice that x 3 is undefined if 2 x + 1 = 0, i.e. if x = 2x + 1 x = 12 cannot form part of the answer.
−
− 12 and hence
an inequality by a positive
Secondly we note that for x = 12 , 2x + 1 can be either positive or negative, depending on the values of x, but that (2x + 1)2 is always positive. (Can you explain why?) Thus, for x = 12 we can multiply both sides of (1) by (2x + 1)2 and obtain (2x + 1)2 (x 3) 0. 2x + 1
number does not change the
After cancelling we obtain
− −
Multiplying both sides of
− ≤
direction of the inequality
(2x + 1)(x
sign.
− 3) ≤ 0.
− ≤ 0 we solve (2x + 1)(x − 3) ≤ 0 for x = − 12 .
Thus instead of solving x 3 2x + 1
Consider the parabola defined by y = (2x + 1)(x
− 3), i.e. by y = 2x2 − 5x − 3.
The x–intercepts are 12 and 3 and the par abola is conc ave up. Figure 4.4.4 shows a rough sketch of this parabola.
−
y
− 12
0
3
x
y = ( 2x + 1)(x
Figure 4.4.4
− 3)
238 From Figure 4.4.4 we see that the parabola lies below or on the x –axis if 1 x 3. Thus the solu tion of (2x + 1)(x 3) 0 is 12 x 3, but the 2 solution of x 3 0 is 12 < x 3 since we must exclude x = 12 . 2x + 1
− ≤ ≤
− ≤
−
− ≤
≤
− ≤ ≤ −
We can use a technique similar to the one illustrated in Example 4.4.2 to solve some other more complicated rational inequalities, but this involves drawing graphs of polynomials of degree 3 or more. These graphs are beyond the scope of this module.
4.4B FINDING MAXIMUM AND MINIMUM VALUES In certain applied problems we need to find the maximum or minimum value of a quadratic function, value you of thethat function variablefunction at which the or minimum occurs.and Wethe remind if a quadratic written in f is maximum the form y = f (x) = ax 2 + bx + c then the vertex of the corresponding parabola is
−
b 2a
4ac b2 , 4a
−
− − or
b, f 2a
b 2a
and thus
the minimum function value (when a > 0) or the maximum function value (when a < 0)
occurs when x =
−b , and y = 4ac − b2 , or y = f −b 2a
4a
2a
If a quadratic function is written in the form
y = f (x) = a(x
.
− h)2 + k
then the vertex of the corresponding parabola is ( h, k) and thus the minimum function value (when a > 0) or the maximum function value (when a < 0)
occurs when x = h, and this value is y = k .
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We apply these comments in the next few problems.
4.4.3 Have a look again at Activity 2.3.7 in Study Unit 2.3B of Book 2. Here we are given “When Simon kicks a rugby ball into the air the height ground after t seconds is given by
h = 15t
If you continue to study mathematics you will learn a quicker method to solve problems such as these.
h (in metres) above the
− 5t 2.”
Since we now know how to find the maximum value of a quadratic function and the value of the function variable at which the maximum occurs we can determine algebraically the maximum height that the ball reaches and the time it takes to reach this height. We have
− 5t 2 −5(t 2 − 3t ) −5 t 2 − 3t +
h = 15t
= = = =
− − − − − − − 5 t
3 2
2
3 2
2
5 t
3 2
5
+ 11
2
3 2
2
9 4
1 . 4
Thus the ball reaches its maximum height, i.e. 11 14 metres, after 1 12 seconds. In Activity 2.3.7 of Book 2 we were given the height of 11 14 metres, and found that the ball reached this height after 32 seconds. Since we found no other time at which the ball reached this height we deduced that this was the maximum height that the ball reac hed. We have now shown this to be the case. Look again at Figure 2.3.1 in Study Unit 2.3B of Book 2, which shows the path of the ball.
4.4.4 A farmer wants to enclose a rectangular section of land bordered by a river, to prevent cattle getting into his maize fields. He has 4 000 m of fencing. If he does not fence the land along the river, what is the biggest area of land that he can enclose?
240
SOLUTION We first represent the problem by means of a diagram. River
xm
xm (4 000 – 2 x ) m
Figure 4.4.5
We let the length of fencing on the two equal sides be x m. Since there is a length of 4 000 m available, the third side will be ( 4 000 2x) m long.
−
We know that in a rectangle, area = length
See Book 4 for area.
× breadth
and hence we let
A = x(4 000
= 4 000 x
− 2 x) − 2 x2
where A represents area in square metres. This equation for the area represe nts a function whose graph is a parabola. We sketch this in Figure 4.4.6.
Area A of land (m2 )
A = 4 000 x
− 2 x2
x 0
1 000
000 2
Length of side (m) Figure 4.4.6
241 The function has a maximum value since a < 0, and the value is given by the second coordinate of the vertex.
MAT0511/003
The maximum value of A is given by
A= where a =
4 ac b2 4a
−
−2, b = 4 000 and c = 0.
Thus Note that the axis of symme-
A =
try is defined by x = 1 000; if we substitute x = 1 000 into
A = 4 000 x
− 2x
2
we will
= 2 000 000 .
also find the maximum value of A .
−16 −0008 000
The maximum area that can be enclosed is thus 2 000 000 m 2 .
Try the following activity, which involves a minimum value.
4.4.2 Suppose the difference between two integers is 20. Find these two intege rs so that their product is as small as possible.
Let the smaller number be x . Then the larger number is x + 20. Their product P is given by Note that we can also de-
P = x(x + 20)
note the two numbers by a,
a
= x2 + 20x = x2 + 20x + (10)2 = (x + 10)2 100
− 20; or by b − 10, b + 10,
etc. We will always find the
−
same answer.
which is of the form P = a (x
− (10)2
− h)2 + k where a = 1, h = −10 and k = −100.
Since a > 0 the function P has a minimum value. Thus the minimum product is 100 and this occurs when x = 10. The one number is 10 and hence the other number is x + 20, i.e. 10 + 20, i.e. 10. Thus the two numbers are 10 and 10.
−
−
−
−
−
242
4.4 1. Use a rough sketch of an appropriate parabola to solve each of the following inequalities. (a) x2 + 5x + 6 2
≥0
(b) 3 + x 2x > 0 (c) x 3 0 x+2 (c) 3x + 1 0 1 x
− −≥ − ≤
2. The sum of two inte gers is 20. Find these two inte gers such tha t their product is a maximum. What is their product? 3. A house owner has 44 m of fencing, and wants to enclose a rectang ular vegetable garden so that one side is formed by a neighbour’s fence. What is the largest area he can enclose with the length of fencing available? 4. A ball is thrown into the air and its heig ht h (in metres) above the ground after t seconds is given by 5t 2 .
h = 20t
−
What is the maximum height reached by the ball and how long does it take to reach this height? 5. It is estimated that 20 000 people will atten d a soccer match when the tickets cost R20 each. It is predicted that for each R1 increase in the price of a ticket the attendance will drop by 500 people. (a) What should the tick ets cost for the organisers to make the most money? (b) How many people should attend for the organisers to make the most money and how much will they make? (Hint: If the ticket price increases by x rands, then the attendance will be 20 000 500x.)
−
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• A function f defined by f (x) = ax 2 + bx + c; a, b, c
R
and a = 0
∈
is called a quadratic function in x .
• The graph of a quadratic function is called a parabola. • The equation y = ax2 + bx + c can be rewritten in the form y = a(x − h)2 + k by completing the square.
• The parabola defined by y = x2 is called the standard parabola. • Characteristics of parabolas defined by y = f (x) = a(x − h)2 + k
Domain and range
Df
=
Rf
=
R
[k,
)
if a > 0
(
, k]
if a < 0
−∞∞
concave up if a > 0 concave down if a < 0.
The parabola is
The bigger the numerical value of a, the narrower the parabola becomes. The smaller the numerical value of a, the wider the parabola becomes.
The axis of symmetry is the line defined by x = h.
The vertex is (h, k ).
The y –intercept is ah 2 + k.
The x –intercepts are h
− −ak and h + −ak .
When a > 0 the function f has a minimum value ; the minimum value of f (x) is k . When a < 0 the function f has a maximum value ; the maximum value of f (x) is k .
• Number of times a parabola defined by y = ax2 + bx + c cuts the x–axis
If ∆ > 0, the parabola cuts the x –axis twice.
If ∆ = 0, the parabola touches the x –axis once.
If ∆ < 0, the parabola does not cut or touch the x –axis.
244
• Characteristics of a parabola defined by y = f (x) = ax2 + bx + c
Note: (Alternatively)
y
2
= ax + bx + c
− −
2
− − − b 2a
= a x
+f
b 2a
2
y = ax + bx + c = a x
f
Rf =
b 2a
;
; f
b 2a
∞
+
4 ac b2 4a
−
if a > 0
− − − ∞
Domain and range
2
b 2a
Df Rf
if a < 0
=
R
=
2
if a > 0 [ 4ac4−a b , ) 4 ac−b2 ( , 4a ] if a < 0
−∞
∞
concave up if a > 0 concave down if a < 0.
The parabola is
The bigger the numerical value of a, the narrower the parabola becomes. The smaller the numerical value of a, the wider the parabola becomes. The axis of symmetry is the line defined by x = b . 2a
−
2ab
−
The vertex is
The y –intercept is c .
The x –intercepts are
4ac4 b2 , or a
,
−
2ab
f
2ab
− − − −− − − − − − −
,
.
b2 4ac and b + b2 4ac . 2a 2a When a > 0 the function f has a minimum value ; the minimum 2 b value of f (x) is 4ac b or f 4a 2a When a < 0 the function f has a maximum value ; the maximum 2 b . value of f (x) is 4ac b or f 4a 2a b
• Three ways of finding the equation of a parabola
Given three unrelated points on a parabola: use the equation
y = ax 2 + bx + c.
Given the turning point ( h, k ) and another point on the parabola: use the equation y = a (x h)2 + k.
−
Given the x–intercepts r1 and r2 and another point on the parabola: use the equation y = a (x r1 )(x r2 ).
−
−
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• Quadratic inequalities can be solved by using sketches of parabolas. +q • Rational inequalities of the form px = 0, r = 0 (where > can rx + s > 0, p be replaced by < , ≥ or ≤) can be solved by sketching the parabola defined by y = ( px + q)(rx + s).
246
CHECKLIST Now check that you can do the following. SECTION 4.1
1. Recognise the role of a with respect to the parabola defined by y = ax 2 . Table 4.1.1 2. Recognise the role of k with respect to the parabola defined by y = ax2 + k. Table 4.1.2 3. Recognise the role of h with respect to the parabola defined by y = a (x Table 4.1.3
− h)2.
4. Determine various characteristics of a parabola defined by y = a (x h)2 + k. Example 4.1.4; Activity 4.1.7
−
5. Know the characteristics of parabolas defined by y = f (x) = a(x Table 4.1.4
− h)2 + k.
SECTION 4.2
1. Sketch parabolas defined by y = a (x Example 4.2.1; Activity 4.2.1
− h)2 + k.
2. Sketch parabolas defined by y = ax 2 + bx + c. Example 4.2.2; Activity 4.2.2 3. Use the discriminant to determine the number of times a parabola cuts the x–axis. Example 4.2.3 4. Know the characteristics of parabolas defined by y = ax 2 + bx + c. Table 4.2.2 SECTION 4.3
1. Find the equation of a parabola if you are given thre e unrelated points on the parabola. Example 4.3.1(a) 2. Find the equation of a parabola if you are given the verte x and another point on the parabola. Example 4.3.1(b); Activity 4.3.1
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3. Find the equation of a parabol a if you are given the x –intercepts and one other point on the parabola. Example 4.3.1(c); Activity 4.3.1
SECTION 4.4
1. Use a rough sketch of an appropriate parabola to solve a quadratic inequality. Example 4.4.1; Activity 4.4.1
px + q 2. Solve a rational inequality of the form rx + s > 0 (where > can be replaced by <, or ) by using a rough sketch of the parabola defined by y = ( px + q)(rx + s). Example 4.4.2
≥ ≤
3. Find maximum and minimum values of quadratic functions in certain applied problems. Examples 4.4.3, 4.4.4; Activity 4.4.2
248
HYPERBOLAS OUTCOMES After studying this topic you should be able to do the following. SECTION 5.1: Characteristics of Hyperbolas
Sketch hyperbolas defined by y =
k x
using a table of values.
Know the properties of hyperbolas defined by y = kx for k > 0 and k < 0. This includes being able to find the coordinates of the two points on the hyperbola closest to the srcin and the distance of these points from the srcin.
Sketch hyperbolas without using a table of values.
Find the equation of a hyperbola given one point on the hyperbola.
Find the equation of a hyperbola given the distance from the origin to either of the two points on the hyperbola closest to the srcin, and some fact that will determine the sign of k in the equation y = kx .
SECTION 5.2: Inverse Proportion
Know and apply the definitions of inverse proportion, direct and inverse proportion, and joint and inverse proportion.
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5.1 CHARACTERISTICS OF HYPERBOLAS
5.1A y =k, k > 0 x
HYPERBOLAS DEFINED BY
In Example 1.1.2 of Study Unit 1.1A we represented graphically the time it would take Moses and a varying number of his friends to fold notices and seal them in envelopes. We assumed that Moses and each of his friends worked at the same rate and that he could complete the task in 8 hours if he worked alone. In tabular form we had the following information. Number of people (n) 1 2 3 4
Time taken in hours (t ) 8 4 2 23 2
Table 5.1.1
The time taken was found by dividing 8 hours by the number of people carrying out the task, i.e. 8 t = , where n N and 1 n 4. n The graph defined by this equation is given in Figure 5.1.1 on the next page. This graph represents a function in n .
∈
Do you know why t = 8n defines a function? If not, revise functions in Topic 2.
≤ ≤
t 10 8
6 4
2
n 1
2
3
4
Figure 5.1.1
250 The graph in Figure 5.1.1 consists of points which we cannot join as it makes no sense to have, for example, 3 13 people. We now look at graphs similar to the one in Figure 5.1.1, in which we can join the dots. We consider functions defined by y=
k x
can also be written
y=
as xy = k.
Rational expressions are dealt with in Topic 1 of Book 2.
k x
where k is a non–zero constant and x is any non–zero real number. A function of this form is called a rational function since kx is a rational expression. Since x can be any non–zero real number we can obtain the graph of such a function by plotting a few points and joining them by means of a smooth curve that follows the pattern formed by the dots. The graph of the function y = kx is called a rectangular hyperbola, but we shall just refer to it as a hyperbola.
Since k = 0 we may have k > 0 or k < 0. We first draw the graph of y = k > 0, and investigate characteristics of this hyperbola. THE HYPERBOLA
for
5.1.1
DEFINED BY 8
y=
k x
x
We draw the graph of the function f , defined by
y = f ( x) =
8 . x
We set up the following table of values.
x
−16 −8 −4 −2 −1 −
y
−
1 2
1 2
−
1 4
−1 −2 −4 −8 −16 −32
0
1 4
1 2
12
48
unde- 32 16 8 4 2 1 fined
16
1 2
Table 5.1.2
Note that when x = 0, y is undefined since we cannot divide by zero. Once we have drawn the graph we shall investigate what happens to y as x approaches 0.
251
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We plot the points and join them as shown in Figure 5.1.2.
y 20 16
y= 8 x
12 8
−20 −16 −12 −8 −4
4
x
4
y = 8x
−4 −8 −12 −16 −20
8
12
16
20
Figure 5.1.2
Note the following characteristics of the graph of y = 8x .
R
−{0} = {x ∈ R : x = 0}
The graph consists of two separate parts, called branches, one in the first quadrant and the other in the third quadrant.
D f and R f consist of all real numbers except 0, i.e. D f = R f = R
−{0}.
There is no x –intercept since
8 x
There is no y –intercept since
8 x
From the branch in the first quadrant we see that as x approaches zero from the right (i.e. x gets closer and closer to zero but remains bigger than zero) the values of y , which are all positive, become larger and larger, i.e. y increases without bound. We say that
= 0 has no solution. is undefined for x = 0.
y approaches infinity as x approaches 0 from the right or
y tends to infinity as x tends to zero from the right.
252 Mathematically we write this phrase as
y
→∞
as x
→ 0+.
As x assumes larger and larger positive values, we see that the values of y get closer and closer to zero but remain positive. We can write this mathematically as 0+ as x . y
→
→∞
From the branch in the third quadrant we see that as x approaches zero from the left, the values of y, which are all negative, remain negative but become numerically bigger and bigger. We say that y decreases without bound and we write y as x 0− .
→ −∞
→
When x is negative and decreases without bound, i.e. when the numerical values of x become larger and larger, we see that y again gets closer and closer to zero, but remains negative. We write
y
→ 0−
as x
→ −∞ .
Note The + or associated with the zero indicates the direction from which we approach zero. The + indicates that we are approaching zero from the positive side of the horizontal (in the case of x ) or vertical (in the case of y) axes. Similarly, the indicates that we are approaching zero from the negative side of the horizontal or vertical axes. See Figure 5.1.3.
−
−
y
0
x x
0
0–
y
0
–
Figure 5.1.3
+
0
+
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Consider Figure 5.1.4 which shows the hyperbola defined by gether with the lines defined by y = x and y = x.
−
y y=
−x
8 x
to-
y=x
P
y= 8 x
x
O y= 8 x
y=
Q
Figure 5.1.4
The branches of the hyperbola are symmetric with respect to the line defined by y = – x , i.e. if we fold Figure 5.1.4 along the line defined by y = x we find that the branch in the third quadrant lies directly on top of the branch in the first quadrant.
−
We leave it as an exercise to show algebraically that if (a, b) lies on a hyperbola then (b, a) also lies on the hyperbola.
Each branch is symmetric with respect to the line defined by y = x, i.e. if we fold Figure 5.1.4 along this line we see that half of each branch lies directly on top of the other half. We also see from Figure 5.1.4 that if the point (a, b) lies on the hyperbola, then so does the point (b, a). This is true for all graphs that are symmetric with respect to the line y = x. For example we see from Table 5.1.2 that both ( 1, 8) and ( 8, 1) lie on the branch in the 1 first quadrant, and 16, 12 and , 16 lie on the branch in the third 2 quadrant.
−
−
−
−
The points on the hyperbola which lie closest to the srcin O are P and Q , which are the points of intersection of the hyperbola and the line defined by y = x (see Figure 5.1.4). We find the coordinates of these points by solving simultaneously the equations
y=x
(1)
and
y=
8 . x
When we substitute (1) into (2) we obtain
x=
8 . x
(2)
254
Thus
x2 = 8 and hence
x=
±
√
8.
√8 then y = √8 and if x = √8 then y = √8. Since P is in the first quadrant we − have P = √8−, √8 , and since Q is in √ √ the third quadrant we have Q = − 8, − 8 . If x =
See equation (1.3.2) in Study Unit 1.3A of Topic 1.
From the distance formula we find that the distance d (O, P), from P to the srcin O is
d ( O, P ) =
=
√ √ − + − 8
0
2
8
0
2
units
√8 + 8 units
= 4 units. Similarly, d (O, Q) = 4 units. It is also obvious from the symmetry of the graph that P and Q are the same distance from O . HYPERBOLAS DEFINED BY k y= ,k>0 x
Hyperbolas defined by y = xk for k > 0 all have the same characteristics as the hyperbola defined by y = 8x . Consider Figure 5.1.5 which shows a hyperbola defined by y = kx for some k > 0 together with the line defined by y = x .
y
y=x
P
O y = kx
Q
Figure 5.1.5
y = kx
x
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To find the coordinates of P and Q, when we do not have a specific value of k, we solve simultaneously
y=x
(3)
and
y=
k . x
(4)
We substitute (3) into (4) and obtain
x=
k x
i.e. we have
x2 = k and hence
√k is a real number sinc e √ k > 0. By definition k > 0.
x=
±
√
k.
√ √ √ √ If x = k then y = k, and if x = − k then y = − k. Since P is in the first quadrant we must have x > 0 and y > 0. Thus P =
√ √ k,
k .
Since Q is in the third quadrant we must have x < 0 and y < 0 and consequently
Q= When k = 8 we found that the distance was 4 units. Note that 4=
√16 = √2 × 8,
−√
k,
−√ k
. The distance from P to the srcin O is
d (O, P) =
√ d (O, P) = d (O, Q) = 2k where k = 8.
k
2
0
=
√k + k units
=
√
i.e. we have the distance in terms of the formula
√ √ − + − k
0
2
units
2k units
which is also the distance from Q to the srcin. Remember that P and Q are the points on the hyperbola that are closest to the src in. Since the dist ance of P and Q from the srcin is given by 2k units it follows that as k increases so the points on the hyperbola that are closest to the srcin move farther away from the srcin.
√
5.1.1 (a) On the same system of axes give a rough sketch of the hyperbolas defined by y = 1x and y = 2x . Also include the line defined by y = x . Let R denote the point of intersection of the hyperbola defined by y = 1x and the line defined by y = x in the first quadrant. Let S denote the point of intersection of the hyperbola defined by y = 2x and the line defined by y = x in the third quadrant. (b) Determine the coordinates of R and the distance from R to the srcin.
256 (c) Determine the coordinates of S and the distance from S to the srcin. (d) Suppose T (x, 13 ) lies on the hyperbola defined by y = 2x . Determine the x–coordinate of T .
−
(a)
y
y= 1 x
R
y=x
y= 2 x
x
O y= 1 x
S
y= 2 x
Figure 5.1.6
(b) We substitute y = x into y =
1 x
and obtain
x=
1 x
i.e. we have
x2 = 1 and thus
x= Since R is in the first quadrant we have R = ( 1, 1) . The distance from R to the srcin is
√
1.
±
x = 1 and hence y = 1. Thus
12 + 12 units, i.e.
√2 units.
257 (c) We substitute y = x into y =
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and obtain
x=
2 x
i.e. we have
x2 = 2 and thus
√ x=
2.
±
Since S is in the third quadrant we have x = Thus S
√ = −
√ 2, − 2 .
The distance from S to the srcin is
√ i.e. 4 units, i.e. 2 units.
−√2 and hence y = −√2.
√ √ − + − 2
2
(d) In order to find the x –coordinate of T we substitute y = tion y = 2x . We obtain
− 13 = 2x i.e. we have
x = 21
−
and hence
x=
3
−2 × 3 −
= 6. Thus T =
1 . 3
− − 6,
2
2
−
1 3
units,
into the equa-
258
5.1B HYPERBOLAS DEFINED BY
y=
k , k>0 x
In this study unit we investigate characteristics of hyperbolas defined by y = where k < 0.
k x
5.1.2 THE HYPERBOLA DEFINED BY 4 y= x
Consider the function f defined by
−
Note that
k=
−
4 x
=
−4 , so that x
y = f ( x) =
− 4x .
(a) Complete the following table of values.
−4, i.e. k < 0. x
− 8 −4 −2 − 1 −
1 2
0
1 2
1248
y
(b) Plot the points obtained from the table in (a) and join them in a suitable way to obtain the graph of f .
(a)
x
y
−8 −4 − 2 −1 − 1 2
1
2
4
8
1 2
0
undefined
1 2
1248
−8 − 4 −2 −1 −
1 2
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(b)
y y=
− 4x
8
6 4
2
2
4
6
8
x
−8 −6 − 4 −2
−2 −4 −6 −8
y=
− 4x
Figure 5.1.7
Study Figure 5.1.7 carefully and then try the following activity.
5.1.3 Copy Figure 5.1.7 without the dots shown on the branches of the hyperbola. Draw the lines defined by y = x and y = x. Let P denote the point in the second quadrant where the line defined by y = x cuts the hyperbola; let Q denote the corresponding point in the fourth quadrant.
− −
For the hyperbola defined by y = f (x) = 4x write down characteristics similar to those given for the hyperbola defined by y = 8x .
−
260
y=
y
− 4x
y=x
P
O
x y=
Q
y=
− 4x
−x
Figure 5.1.8
The hyperbola has two branches, one in the second quadrant and the other in the fourth quadrant.
Df = Rf = R
There is no x –intercept since
−{0}. 4
− x = 0 has no solution. There is no y –intercept since − x is undefined for x = 0. 4
In the second quadrant
y
→∞
as x
→ 0− ,
i.e. y tends to infinity as x tends to zero from the left; and
y
→ 0+
as x
→ −∞ ,
i.e. y tends to zero but remains positive, as x tends to
−∞
.
In the fourth quadrant
y i.e. y tends to
−∞
→ −∞
as x
→ 0+ ,
as x approaches zero from the right; and
y
→ 0−
as x
→∞,
i.e. y tends to zero but remains negative, as x tends to
∞.
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The branches are symmetric with respect to the line defined by y = x .
Each branch is symmetric with respect to the line defined by y = – x .
The points on the hyperbola closest to the srcin are P and Q (see Figure 5.1.8), which are the points of intersection of the hyperbola and the line defined by y = x. We find the coordinates of these points by solving simultaneously the equations
−
y=
−x
y=
− 4x .
and
If we substitute y =
4
−x into y = − x we obtain −x = − 4x
i.e. we obtain
x2 = 4 and hence
x=
If x = 2 then y =
−
±2.
−2, and if x = −2 then y = 2.
−
Thus P = ( 2, 2) ( P is in the second quadrant) and Q = (2, 2) ( Q is in the fourth quadrant).
The distance from P to the srcin O is
d (O, P) =
=
√
( 2)2 + (2)2 units
−
8 units
√
= 2 2 units. Similarly, either by symmetry or by the distance formula, the distance of Q from the srcin is 2 2 units.
√
262 HYPERBOLAS DEFINED BY k y= ,k<0 x
Hyperbolas defined by y = kx for any k < 0 have the same characteristics as the hyperbola defined by y = 4x .
−
y=
Consider Figure 5.1.9 which shows the hyperbola defined by k < 0, and the line defined by y = x.
−
k x,
for some
y
y = kx
k<0
P O
x y = kx
Q
y=
−x
Figure 5.1.9
We find the coordinates of P and Q for any negative k ously the equations
y=
−x
y=
k . x
and
When we substitute y =
∈ R by solving simultane-
−x into y = kx we obtain −x = kx
i.e. we obtain
x2 = Remember that k < 0, hence
−k > 0 and √−√k is defined. By definition −k > 0 for all k < 0. Note t hat −√−k = √k.
and hence
If x =
−k √ x = ± −k.
√−k then y = −√−k and if x = −√−k then y = √−k.
263 Since P is in the second quadrant we must have x < 0 and y > 0. Since Q is in the fourth quadrant we have x > 0 and y < 0.
Hence P =
−√−
k,
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√−k and Q = √−k, −√−k.
The distance from P (and from Q ) to the srcin O is
d ( O, P ) =
=
−√ −k
+ √ −
−
−
2
k
2
units
( k) + ( k) units
√
−2k units.
=
Hence it follows that as the magnitude of k increases, so the two points on the hyperbola that are closest to the srcin move farther away from the srcin.
5.1.4 Suppose V and W are points in the second quadrant. V is the point closest to the srcin on the hyperbola defined by y = 1x ; W is the corresponding point on the hyperbola defined by y = 2x . Calculate the distances from V and W to the srcin.
−
−
The distance from the srcin to either of the two closest points on the hyperbola defined by y = kx , where k < 0, is 2k units.
√−
Thus the distance from V to the srcin is the distance from W to the srcin is i.e. 2 units.
√ √
(− ) (− ) (− ) (− ) 2 2
1 units, i.e. 2 units; 2 units, i.e. 4 units,
In Activity 5.1.4 we found that
d (O, W ) = 2 units
>
√2 units = d (O, V )
and hence W lies farther away from the srcin than V . We expect this since the magnitude of 2 (i.e. 2) is greater than the magnitude of 1 (i.e. 1).
−
−
264
5.1C FINDING EQUATIONS OF HYPERBOLAS We can find the equation of a hyperbola if we know
one point that lies on the hyperbola or
the distance OP, where O is the srcin and hyperbola closest to O
P is one of the points on the
and some other fact from which we can deduce whether the constant equation y = kx is positive or negative.
k in the
5.1.2
−
(a) Suppose (2, 3) lies on a hyperbola. Find the equation of the hyperbola. (b) Suppose a hyperbola is defined by 2xy p = 0 where p < 0. If the shortest distance from the srcin to a point on the hyperbola is 3 units, find the value of p and write the equation that defines the hyperbola in the form y = kx .
−
SOLUTION (a) Suppose the equation of the hyperbola is y = kx . Since (2, 3) lies on the hyperbola, x = 2 and y = 3 must satisfy the equation y = kx . After substitution we obtain k 3= 2 and thus
−
−
−
k=
−6 .
6 x.
Hence the equation of the hyperbola is y =
−
(b) We rewrite the equation in the form y = kx . Since 2xy we have
−p=0 2xy = p
i.e. we have
y=
p . 2x
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This equation is in the form
y=
k p , where k = . x 2
Since p < 0, it follows that k < 0. When k < 0, we know that the shortest distance from the srcin to a point on the hyperbola defined by y = kx is 2k units. This distance is 3 units.
√−
Now
√−2 k = 3
− =
⇔
Note that 9 2x .
−
− 92 x
is the same as
2
p 2
3
⇔
√− p = 3
⇔
−p = 9
⇔
p=
Thus p =
−9. 9 2
−9 and the equation of the hyperbola is y = − x .
Now try the following activity.
5.1.5 Suppose the shortest distance from the srcin to a point on a hyperbola is
2 3
unit.
If the branches of the hyperbola lie in the first and third quadrants, find the equation of the hyperbola.
266 The equation of the hyperbola is y = kx . Since the branches lie in the first and third quadrants, we have k > 0. Thus the shortest distance from the srcin to a point on the hyperbola is 2k units.
√
Thus
√
2k =
2 3
i.e. we have 2k =
4 9
k=
2 . 9
and hence
Note that 2 9x .
2 9
x
is the same as
Thus the equation is y =
2 9x
(or 9 y = 2x ).
5.1 1. Write each of the following equations in the form y = kx . (a) xy =
−3
(b) 4 xy = 7
(c) xy + 1 = 0 (d)
−5xy + 3 = 0
(e) mxy + n = 0, m = 0, n = 0 2. Show algebraically that if the point ( a, b) lies on the hyperbola defined by y = kx then the point ( b, a) also lies on the hyperbola. 3. On separate systems of axes give rough sketches of the hyperbolas defined by (a)
xy =10
(b) 4
xy + 9 = 0.
In each case indicate one of the points on the hyperbola closest to the srcin and give the distance from this point to the srcin. 4. On the same system of axes give rough sketches of the hyperbolas defined by 3 10 y= and y = . x x
−
−
267
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5. The point ( a, b) lies on a hyperbola. Find the equation of the hyperbola in terms of a and b . 6. The shortest distance from the srcin to a point on a hyperbola is 43 units. If the two branches of the hyperbola lie in the second and fourth quadrants find the equation of the hyperbola.
7. Suppose a hyperbola is defined by mxy + 2 = 0, m = 0, and the shortest 1 distance from the srcin to a point on the hyperbola is unit. If the two 2 branches of the hyperbola lie in the first and third quadrants, find the value of m .
268
5.2 INVERSE PROPORTION
5.2A INVERSE PROPORTION Just as direct proportion occurs frequently in the natural sciences, so does inverse proportion. We first mentioned inverse proportion in Study Unit 3.2A of Book 1, and we remind you of its definition.
Definition 5.2.1 If the variables x and y are related by an equation
y=
c , where c x
∈ R and c = 0,
(5.2.1)
then y is inversely proportional to x . The constant c is called the constant of proportionality.
Note
Equation (5.2.1) is the same as
xy = c.
(5.2.2)
Some authors use the phrase “ y varies inversely as x” instead of “ y is inversely proportional to x ”. The constant c may also be referred to as “the constant of variation”. We know that y = xc defines a function and that the graph of this function is a hyperbola. In a real–life situation c and x are usually both positive and in such a case the graph then consists only of the branch of the hyperbola that lies in the first quadrant.
269
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5.2.1 Suppose z is inversely proportional to t , and that z = 7 when t = 4.
(a) Find the constant of proportionality and write dow n the equation that describes this relationship. (b) Find z if t = 10.
SOLUTION (a) Since z is inversely proportional to t we have
z=
c . t
(1)
When we substitute t = 4 and z = 7 into (1) we obtain 7=
c 4
and hence
c = 28. Thus the constant of proportionality is 28 and the equation is z = (b) When t = 10 we have z =
28 10
28 t .
= 2, 8.
Now try the following activity in which we consider the relationship between the pressure and the volume of a gas.
5.2.1 You are not expected to be familiar with such laws. We use them as examples.
Pressure is measured in Pascals, which is abbreviated by Pa, and 1 Pa = 1 N.m −2 , i.e. 1 Pascal is 1 Newton per square metre.
Boyle’s Law states that the pressure P exerted by a gas at constant temperature is inversely proportional to the volume V in which it is enclosed.
(a) Write Boyle’s Law as an equation, where c is the constant of proportionality. (b) Suppose a sample of air occupies 0 , 08 m3 and its pressure is 1, 01 105 Pa. What will the pressure be if this sample is compressed to 0 , 048 m3 and the temperature remains constant?
×
270
(a) P =
c or PV = c. V
(b) Suppose the new pressure is x Pa. Then
and
1, 01
5
× 10 × 0, 08 = c x
× 0, 048 = c.
x
× 0, 048 =
x=
Hence
i.e. we have
5
1, 01
× 10 × 0, 08
1, 01
× 10 × 0, 08
5
0, 048 5
≈ 1, 68 × 10 . Thus the new pressure is approximately 1 , 68 × 10 Pa. 5
We mentioned in Study Unit 3.3C that one variable need not just depend on one other variable. For example, if the variables x , y and z are related by the equation
z=
c ; c xy
∈ R and c = 0
then we say that z is inversely proportional to x and y . If
cy ; c R and c = 0 x then we say that z is directly proportional to y and inversely proportional to x. Sometimes the word “directly” is omitted. z=
∈
5.2.2 Write down the equation that expresses the following statements. Use c as the constant of proportionality.
(a) p is inversely proportional to the square root of s . Look at Study Unit 3.3C if you have forgotton what “jointly proportional” means.
(b) z is jointly proportional to x and the cube of y , and inversely proportional to t .
271
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(c) The resistance R of a wire is directly proportional to its length l and inversely proportional to the square of its diameter d .
SOLUTION (a) p =
√cs
(b)
z=
cxy3 t
(c)
R=
cl d2
Now see whether you have understood this example by trying to do the following activity.
272
5.2.2 (a) Write down the equation that expresses the fact that “the pressure P of a sample of gas is directly proportional to the temperature T and inversely proportional to the volume V ”. (b) Suppose 2, F = G m1 m r2 where G is the constant of proportionality. Write down, in words, the relationship which is expressed by this equation.
(a) P =
cT V
(b) F is jointly proportional to m1 and m2 and inversely proportional to the square of r .
5.2 1. Suppose r is inversely proportional to t , and that r = 11 when t = 2. (a) Find the constant of proportionality c and write down the equation that describes this relationship. (b) Find t when r = 2. 2. Write an equatio n for each of the following statements, using k as the constant of proportionality. (a) p is inversely proportional to q and r . (b) z is directly proportional to the square of x and inversely proportional to the cube of y . (c) l is jointly proportional to m and the square root of n , and inversely proportional to p and q . 3. If p is jointly proportional to q and r , state the relationship between q and the other two variables.
273
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4. Write down the relationship expressed by the following equations, where k is the constant of proportionality.
kx2 y kR (b) Q = ST kxz2 (c) y = r (a) z =
5. This question is adapted from Stewart, Redlin and Watson (see the References). The resistance R of a wire varies directly as its length l and inversely as the square of its diameter d . (a) A wire 1 , 2 m long and 0,005 m in diameter has a resistance of 140 Ω (ohms). Write an equation for this relationship and find the constant of proportionality. (b) Find the resistance of a wire made of the same materia l that is 2 , 8 m long and has a diameter of 0 , 007 m.
• The graph of the function f defined by y = f ( x) =
k x
where k is a non–zero constant and x is a non–zero real number is called a rectangular hyperbola or a hyperbola for short.
274
• Hyperbolas defined by y = kx have two branches, in either the first and third quadrants, or in the second and fourth quadrants, depending on whether k > 0 or k < 0.
y
k> 0
y
y = kx
k< 0
y = kx
x
x
For k > 0:
The branches lie in the first and third quadrants.
The branches are symmetric with respect to the line defined by y =
Each branch is symmetric with respect to the line defined by y = x.
The two points on the hyperbola which lie closest to the srcin are the points of intersection of the hyperbola and the line defined by y = x .
√ k , √k
The coordinates of these points are
and
√k , √k . − 2k units.
−√
The distance from either of these points to the srcin is
− x.
As k increases so the two points on the hyperbola closest to the srcin move farther away from the srcin.
For k < 0:
The branches lie in the second and fourth quadrants.
The branches are symmetric with respect to the line defined by y = x.
−x .
Each branch is symmetric with respect to the line defined by y =
The two points on the hyperbola which lie closest to the srcin are the points of intersection of the hyperbola and the line defined by y = x. The coordinates of the se points are k, k and k, k . The distance from either of these points to the srcin
−√−
√−− −√−
√−
is 2k units. As the magnitude of k increases so the two points on the hyperbola closest to the srcin move farther away from the srcin.
√−
• The equation of a hyperbola can be found if we know By “sign of k” we mean whether k is positive or negative.
one point on the hyperbola
the distance from the srcin to either of the two points clos est to the srcin, and some fact which will determine the sign of k in the equation y = kx .
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• Inverse Proportion Term
inversely proportional
Examples
c x y is inversely proportional to x c is the constant of proportionality y=
c z = xy z is inversely proportional to x and y c is the constant of proportionality
directly and inversely proportional
jointly and inversely proportional
cx y z is directly proportional to x and inversely proportional to y c is the constant of proportionality z=
cxy t z is jointly proportional to x and y and inversely proportional to t c is the constant of proportionality z=
CHECKLIST Now check that you can do the following. SECTION 5.1
1. Sketch hyperbolas defined by y = Example 5.1.1, Activity 5.1.2
k x
using a table of values.
2. Know the properties of hyperbolas defined by y = kx for k > 0 and k < 0. This includes being able to find the coordinates of the two points on the hyperbola closest to the srcin and the distance of these points from the srcin. Discussion after Example 5.1.1; Activity 5.1.3 and discussion after Activity 5.1.3
276 3. Sketch hyperbolas without using a table of values. Activity 5.1.1; Exercise 5.1(3 and 4) 4. Find the equation of a hyperbola given one point on the hyperbola. Example 5.1.2(a) 5. Find the equation of a hyperbola given the distance from the origin to either of the two points on the hyperbola closest to the srcin, and some fact that will determine the sign of k in the equation y = kx . Example 5.1.2(b); Activity 5.1.5
SECTION 5.2
1. Know and apply the definitions of inverse proportion, direct and inverse proportion, and joint and inverse proportion. Examples 5.2.1, 5.2.2; Activities 5.2.1, 5.2.2, 5.2.3
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COMBINATIONS OF GRAPHS OUTCOMES After studying this topic you should be able to do the following.
Interpret various combinations of graphs , including lines, parabolas, circles and hyperbolas.
6.1 GRAPHS, GRAPHS AND MORE GRAPHS
6.1A GRAPHICAL REPRESENTATION OF SYSTEMS OF LINEAR AND QUADRATIC EQUATIONS In Study Unit 3.3A we looked at the graphical representation of a system of two linear equations in two unknowns. From these graphs we were able to see why such systems can have
See Study Unit 2.5B in Book 2.
only one solution
no solution, or
infinitely many solutions.
We now look at the graphical representation of a system consisting of a linear and a quadratic equation. We first remind you that a quadratic equation in two unknowns is an equation of the form
278
ax2 + bxy + cy2 + dx + ey + f0 =
(6.1.1)
where a , b, c, d , e, f R and where the constants a , b and c are not simultaneously zero. In this study unit we consider quadratic equations of the form
∈
All these equations can be rearranged so that the equations have the same form as equation (6.1.1).
y = ax 2 + bx + c, the graph of which is a parabola
(x
xy = k , the graph of which is a hyperbola.
h)2 + (y
−
k)2 = r 2 , the graph of which is a circle
−
In Study Unit 2.5B of Book 2 we solved algebraically systems consisting of one linear and one quadratic equation. We also stated that such systems usually have two solutions. Graphically we shall now see that these systems can have
two solutions
only one solution, or
no solution.
6.1.1 (a) On the same system of axes , sketch the graph of the quadrati c function f defined by y = f ( x) = x2 4x + 3
−
as well as the lines
l1 : y = x
−1 − 3 14 y = x − 5.
l2 : y = x l3 :
(b) What can you deduce from (a) about the numb er of solutions of each of the following systems of equations? (i) (ii) (iii)
y=x 1 y = x 2 4x + 3
− − y = x − 3 14 y = x 2 − 4x + 3 y = x −5 y = x 2 − 4x + 3
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SOLUTION (a) For the graph of f we have – the x –intercepts are 1 and 3 – the y –intercept is 3 – the vertex is ( 2, 1)
−
– the graph is concave up. For l 1 , the x – and y –intercepts are respectively 1 and
−1. For l 2 , the x – and and −3 14 . For l 3 , the x – and y –intercepts are respectively 5 and −5. y –intercepts are respectively 3 14
We sketch the graphs below.
y
l1
f
l2
3
l3 2
3
x
3 14
1
−1
5
−3 14 −5 Figure 6.1.1
(b) From Figure 6.1.1 we see the following. (i) The line l 1 cuts the graph of f twice, i.e. the system
y=x 1 y = x 2 4x + 3
− −
has two solutions.
280 (ii) The line l 2 touches the graph of f at one point, i.e. the system
y = x 3 14 y = x 2 4x + 3
− −
has only one solution. (iii) The line l 3 does not cut the graph of f at all, i.e. the system
y x 5 = x 2 4x + 3 y=
−−
has no solution.
See Exercise 6.1.
We do not usually solve systems of linear equations graphically; neither do we usually solve systems consisting of one linear and one quadratic equation graphically. We leave it as an exercise for you to solve algebraically each of the systems given in Example 6.1.1(b) in order to confirm the visual representation. Now try the following activity which involves circles and straight lines.
6.1.1 (a) On the same system of axes sketch the circle defined by
x2
− 2x + y2 − 3 = 0
as well as the lines
l1 : y = l2 : y =
−x −x − 3
l3 : y = 2.
(b) What can you deduce from (a) about the numb er of solutions of each of the following systems of equations? (i) (ii) (iii)
y= x 2 x + y2 3 = 0
−
x2
−
x2
− 2 x + y2 − 3 = 0
− y = −x − 3 x2 − 2 x + y2 − 3 = 0 y=2
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(a) By completing the square we can rewrite the equation of the circle as
(x
− 1)2 + y2 = 4.
Thus the circle has centre (1, 0) and radius 2. For the line l1 , the x– and y–intercepts are 0, i.e. the line passes through the srcin. The line has slope 1.
−
For the line l 2 , the x – and y –intercepts are both
−3.
The line l 3 is parallel to the x –axis and its y –intercept is 2. The graphs are sketched below.
y
l3
2
(x (1, 0)
−3
−1
3
l1
−3 l2
Figure 6.1.2
(b) From Figure 6.1.2 we see the following. (i) Line l 1 intersects the circle twice, i.e. the system
x2
−
y= x 2x + y2 3 = 0
−
−
has two solutions. (ii) Line l 2 does not cut the circle, i.e. the system
x2 has no solution.
−
y= x 3 2x + y2 3 = 0
−− −
− 1)2 + y2 = 4 x
282 (iii) Line l 3 touches the circle once, i.e. the system
x2
− 2x + y2 −
y=2 3=0
has only one solution.
See Exercise 6.1.
Again we leave it as an exercise for you to solve each of the systems of equations given in Activity 6.1.1 algebraically. We also leave it as an exercise for you to show graphically that a system of the form
y = mx + c xy = k
can have no solution, only one solution or two solutions.
6.1B INTERPRETING COMBINATIONS OF GRAPHS This study unit contains a number of examples and activities in which we deduce various facts from different combinations of graphs.
6.1.2 In the sketch on the next page, the circle (with centre the srcin) and the parabola cut each other at points A and B, and B is the vertex of the parabola. The line joining P (a point on the upper semi–circle) and Q (a point on the parabola) is parallel to the y –axis.
(a) Write down (i) the coordinates of B (ii) the equation of the circle (iii) the equation of the upper semi–circle, i.e. the semi–circle that lies on or above the x –axis. (b) Determine the equation of the parabola. (c) Determine the length of PQ if the x –coordinate of P and Q is 1 12 .
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y
(0, 3) A
P
Q
x
B
Figure 6.1.3
SOLUTION (a)
(i) Since (0, 3) lies on the circle, the radius of the circle is 3. Hence B is the point (3, 0) . (ii) The equation of the circle is
x2 + y2 = (3)2 i.e.
x2 + y2 = 9 . (iii) If
x2 + y2 = 9 then
± 9 − x2 . For the upper semi–circle we have y ≥ 0 and thus the equation of this semi–circle is y = 9 − x2 . y=
(b) The parabola has (3, 0) as its ve rtex. If we write the equation of the parabola in the form
y = a (x
− h)2 + k,
y = a (x
− 3)2 + 0.
then we have
284 The point (0, 3) also lies on the parabola, and thus 3 = a (0 i.e. we obtain
− 3)2
3 = 9a and hence
a=
1 . 3
Hence the equation of the parabola is
y=
1 (x 3
− 3)2
y=
1 2 x 3
− 2x + 3.
i.e. we have
(c) For 0 x 3 the upper semi–circle lies on or above the parabola and the vertical distance d (x) between any two corresponding points on these graphs is given by
≤ ≤
See Vertical Distance between Corresponding Points on Two Lines in Study Unit 3.3A.
− −
d ( x) =
x2
9
=
9
1 2 x 3
− 2x + 3
− x2 − 13 x2 + 2x − 3.
Now the length of PQ is the distance between P and Q when x = 1 12 , i.e. d 1 12 , and
− − − −− − − −
d 1
1 2
=
9
=
9
= = =
2
3 2
9 4
36
1 3
3 2
3 +3 4
3
9
3 4
4
27 4 3 3 2
√
− 34
− 34
= 6 3
√ 4− 3 √ 3 2 3−1 = √4 3 2 3−1 Thus the length of PQ is
4
.
units.
2
+2
3 2
3
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6.1.3 Figure 6.1.4 shows the graphs of f and g . The graph of f is the parabola which cuts the x –axis at 1 and 3, and the y –axis at 6. The graph of g is a line which passes through (0, 2) and (3, 0). The graphs of f and g intersect at (3, 0) and at ( p, q). M and N are points that lie on the parabola and line respectively, and the line joining them is parallel to the y –axis.
− −
y 8
M
6
4
y = f (x) 2
−2 −1
1
2 N
3
4
5
x
−2 y = g(x)
( p, q )
Figure 6.1.4
(a) Find the equation that defines f . (b) Find the equation that defines g . (c) Find p and q . (d) Calculate the length of MN if MN passes through (2, 0). (e) If p < x < 3 determine the maximum length of MN and the value of x at which this occurs. (f) Find the values of x for which (i) f ( x) < g (x) (ii) f ( x) g (x)
≤ 0.
286
SOLUTION (a) Since the graph of f cuts the x –axis at x = 1 and x = 3 the equation of f has the form y = a (x + 1) (x 3) .
− −
If we substitute x = 0 and y = 6 into the equation we obtain 6 = a (0 + 1) (0 i.e. we obtain 6=
−3a
a=
−2.
and hence
3)
−
Thus the equation that defines f is
y=
−2 (x + 1) (x − 3)
y=
−2x2 + 4x + 6.
which can be rewritten as
(b) The graph of g passes through the points (3, 0) and (0, 2). Hence the slope of the line is
0 ( 2) , 3 0
−− −
Since the y –intercept is
i.e.
−
2 . 3
−2, the equation that defines g is 2 y = x − 2. 3
(c) ( p, q) is one of the points of intersection of the graphs of f and g and hence the coordinates p and q must satisfy the system 2 x 2 3 y = 2x2 + 4x + 6
y=
−
−
.
We now solve this system. By substituting the first equation into the second equation we obtain 23 x 2 = 2x2 + 4x + 6.
−
−
Now
⇔ ⇔
6x2
2 x 3
− 2 = −2x2 + 4x + 6
2x
− 6 = −6x2 + 12x + 18
− 10x − 24 = 0
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⇔
− 5x − 12 = 0
(3x + 4) (x
⇔
x=
⇔ 4
Since p < 0 we have p = 4 y = 23 x 2. Thus q = 23 3
− 43
−
or x = 3.
. We find q by substituting x = 2 = 89 2 = 2 89 .
− − − 3
− 3) = 0
− −
−
4
into
−3
(d) For 43 < x < 3 the graph of f lies above the graph of g . Thus the vertical distance between corresponding points on these graphs is given by
−
d (x) =
f ( x)
=
2
= =
−2x
− g (x) + 4x + 6
−
− 2 x 3
2
−2x2 + 4x + 6 − 23 x + 2 −2x2 + 103 x + 8.
Since MN passes through ( 2, 0) the length of MN is given by d (2). Now
d (2) =
=
−2 (2)2 + 103 (2) + 8 −8 + 203 + 8
20 3 2 = 6 . 3
=
Thus the length of MN is 6 23 units. (e) From (d) we see that for function d , where
− 43 < x < 3 the length of
d (x) =
MN is given by the
−2x2 + 103 x + 8.
Thus we see that the length of MN is given by a quadratic function in x. (Do not confuse the quadratic function d with the quadratic function f .) As x varies from 43 to 3 so the length of MN changes from very short to longer. At some stage it becomes shorter again. Thus the length varies and at some point it will be a maximum. We know that the maximum (or minimum) value of any quadratic function is the y–coordinate of the vertex of its graph. Thus we determine the vertex of the graph of d . We have
−
288
d (x) =
= = = = =
−2x2 + 103 x + 8
− − − − − − − − − − −2
x2
2 x2
5 x +8 3 5 x+ 3
5 6
2
2 x
5 6
2
2 x
5 6
2
2 x
2
5 6
+2
25 36
+1
7 +8 18
+9
7 . 18
5 6
2
+8
+8
7 Hence the vertex of the graph of d is 56 , 9 18 . Thus the maximum length 7 of MN is 9 18 units and this occurs when x = 56 .
(f)
(i) f ( x) < g (x) when the graph of f lies below the graph of g , i.e. when the parabola lies below the straight line. Thus f ( x) < g (x) for x < (ii)
− 43 or x > 3. f ( x) g (x)
≤0
when
f ( x) g (x) < 0 or f ( x) g (x) = 0 i.e. when
( f ( x) < 0 and g (x) > 0) or ( f ( x) > 0 and g (x) < 0) or when
f ( x) = 0 or g (x) = 0 Thus f (x)g(x)
≤ 0 when
the graph of f lies below and the graph of g lies above the x–axis, or
the graph of f lies above and the graph of g lie s below the x–axis,
or the graph of f cuts the x –axis, or
the graph of g cuts the x –axis.
Hence f ( x) g (x) 0 when x > 3 or i.e. when x 1.
≥−
≤
−1 < x < 3 or x = −1 or x = 3,
Now see whether you can do the following activity.
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6.1.2 Figure 6.1.5 shows a circle (centred at the srcin O), a line (the graph of the function f ) and a parabola (the graph of the function g ).
y
g
D
A
P
E
C
B 5
x
R
O
f
Q
Figure 6.1.5
The line and circle intersect at A and B. The parabola and circle inte rsect at D and E , and C is the turning point of the parabola. The line PQ which joins the point P on the parabola and the point Q on the lower semi–circle is parallel to the y –axis and lies to the left of the y –axis. AC is also parallel to the y –axis. (a) Find the equation of the circle. (b) If A = ( v, 3) find the value of v . (c) Find the equation that defines f . (d) Find the equation that defines g . (e) Determine the length of PQ if the length of OR is 2 units. (f) If a hyperbola defined by y =
k x
passes through A find the value of k .
(g) Use the graphs to find the values of x for which
√
25
− x2 + 13 x − 53 > 0.
290 (a) The circle has centre at the srcin and radius 5 units (the distanc e from B to O is 5 units). Thus the equation is
x 2 + y2 = 5 2 i.e.
x2 + y2 = 25. (b) A is a point on the circle and thus the coordinates of
A must satisfy the
equation of the circle. We substitute x = v and y = 3 into the equation of the circle.
v2 + 32 = 25 v2 = 16
⇔
v=
⇔
±4.
Now since A lies in the second quadrant it follows that v < 0 and hence v = 4.
−
See Study Unit 3.2A.
(c) The line passes through A ( 4, 3) and B (5, 0) and thus by the two–point formula the equation that defines f is
−
(0
3)
(y
− 3) = (5 −−(−4)) (x − (−4)) .
(y
− 3) = (5(−0 −(−34))) (x − (−4))
Now
⇔
y
− 3 = − 39 (x + 4)
⇔
y
− 3 = − 3x − 43
⇔
y=
− 3x + 53 .
Thus f is defined by y =
− 13 x + 53 .
(d) Since the line AC is parallel to the y –axis the points A and C have the same x–coordinate, namely 4. C lies on the x –axis, hence the y –coordinate of C is 0. The vertex of the parabola is thus C ( 4, 0), and the equation that defines g has the form
−
−
y = a (x which simplifies to
− (−4))2 + 0
y = a (x + 4)2 .
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Since the parabola passes through D (0, 5) we substitute x = 0 and y = 5 into y = a(x + 4)2 . We obtain 5 = a (4)2 i.e. we obtain 5 = 16a and hence 5 a = 16 . Thus the equation that defines g is
y=
5 (x + 4)2 16
y=
5 2 5 x + x + 5. 16 2
which can be written as
(e) The equation of the circle is
x2 + y2 = 25. which we can write as
y=
25
±
− x2 .
Thus the equation of the lower semi–circle, i.e. the semi–circle that lies below or on the x –axis, is
y=
−
25
− x2 .
Now, for 5 x 0 the graph of g lies above the lower semi–circle and thus the vertical distance between corresponding points on these graphs is given by
− ≤ ≤
d (x) =
5 2 5 x + x+5 16 2
d (x) =
5 2 5 x + x+5+ 16 2
− − 25
− −
i.e. we have
25
x2
x2 .
Since the distance from R to O is 2 units, and R lies on the negative x–axis, the x –coordinate of R is 2. Thus the length of PQ is d ( 2). Now
−
d ( 2) =
−
= =
5 5 ( 2)2 + ( 2) + 5 + 16 2 5 5 + 5 + 21 4 5 + 21. 4
Thus the length of PQ is
−
−
√
−
√
5 + 4
√
21
units.
− 25 − (−2)2
292 (f) The equation of a hyperbola is
y=
k . x
Since A is a point on the hyperbola, the coordinates of A must satisfy the equation. If we substitute x = 4 and y = 3 into the equation we obtain
−
k
3=
.
−4
Thus
k=
−12.
(g) We rewrite
as
We manipulate the given inequality so that it shows more clearly a relationship between the graphs. Note that y = 25 x2 is the equation of the upper semi– circle.
√
− −
x 3
25
x2 +
25
x2 >
− 53 > 0
− 3x + 53 .
Thus we must determine the values of x for which the upper semi–circle lies above the line defined by y = 13 x + 53 . From Figure 6.1.5 we see that this is so for 4 < x < 5. Thus the inequality holds for
−
−
−
4 < x < 5.
−
6.1 1. Determine algebraically the solutions (if they exist) of each of the systems of equations given in Example 6.1.1(b). 2. Determine algebraically the solutions (if they exist) of each of the systems of equations given in Activity 6.1.1(b). 3.
(a) On the same system of axes sketch the hyperbola defined by xy = 4 and the lines
l1 : y = x l2 : y = l3 : y =
−x −x + 4.
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(b) Use the graphs in (a) to state the number of solutions that each of the following systems of equations has. (i) (ii) (iii)
−
y=x xy = 4
y= x xy = 4 y=
−x + 4
xy = 4 (c) Solve algebraically each of the systems given in (b).
4. Figure 6.1.6 shows one branch of a hyperbola defined by y = kx , and a straight line. The hyperbola and line intersect at the points M and L .
y
N
9
R
O
x
L
Q
P
M
−9
Figure 6.1.6
The point (2, 4) lies on the hyperbola. R is a point on the x–axis and P is a point on the line so that RP is parallel to the y–axis and cuts the hyperbola at Q .
−
(a) Find the equation of the straight line through M and L . (b) Find the equation of the hyperbola. (c) Calculate the length of NM if N is a point on the x–axis and NM is parallel to the y –axis. (d) Calculate the length of QP if the length of OR is 3 units. (e) Calculate the length of RQ if the length of RP is 4 units.
294 5. Figure 6.1.7 shows the graphs of the functions f and g, both parabolas. P and Q are the points of intersection of the two parabolas. M is a point on the graph of f and N is a point on the graph of g such that MN is perpendicular to the x –axis and lies between P and Q .
y
6
M
P
f
g
−2
−1
3
−3
Q
x
N
Figure 6.1.7
(a) Find the equations that define f and g . (b) Find the coordinates of P . (c) If the length of MN is 4 units find the x –coordinate of M and N . (d) The length of MN will change as x takes on different values. Calculate the maximum length that MN can have and the value of x at which this occurs. (e) Find the values of x for which (i) f ( x) g (x) > 0 (ii) 2 x2 3x 9 < 0.
− −
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6. Figure 6.1.8 shows the graphs of the relation h, which is a circle, with centre at the srcin; the function f , which is a parabola; and the function g, which is a straight line. The parabola and line intersect at P and Q . M is a point on the line and N is a point on the lower semi–circle such that MN is parallel to the y –axis. MN cuts the upper semi–circle at S and the parabola at T when 2 < x < 0.
−
y
P
M
f
h
S
−2
R
Q 2
x
T
g
N
Figure 6.1.8
(a) Find the equations that define h , f and g . (b) Find the coordinates of P . (c) Calculate the length of SN when R = ( 1, 0).
−
(d) Calculate the maximum length of M T as x varies between and the value of x at which this occurs. (e) Give the values of x for which
− 4
x2 + x
≥ 2.
−4 and 2
296
• Graphically a system consisting of one linear and one quadratic equation in two unknowns has
two solutions one solution, or
no solutions.
• Graphical solution of certain inequalities How to interpret the table below:
For example, the first row tells us that the solution of the inequality f ( x) > g (x) is the set of all values of x for which the graph of f lies above the graph of g . Inequality
Solution Values of x for which
f ( x) > g (x)
the graph of f lies above the graph of g
f ( x)
the graph of f lies on or above the graph of g
≥ g (x)
f ( x) < g (x)
the graph of f lies below the graph of g
f ( x)
the graph of f lies on or below the graph of g
≤ g (x)
f ( x) g (x) > 0 both graphs lie above or both graphs lie below the the x –axis f ( x) g (x)
≥0
both graphs lie above or both graphs lie below the x–axis or at least one of the graphs cuts the x –axis
f ( x) g (x) < 0 one graph lies above the x –axis and the other lies below the x –axis f ( x) g (x)
≤0
one graph lies above the x –axis and the other lies below the x –axis, or at least one of the graphs cuts the x –axis.
297
•
•
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Vertical distance d (x) between corresponding points on the graphs of the functions f and g
d (x) = f ( x)
d (x) = g (x)
− g (x) − f ( x)
when f ( x)
If the function d is a quadratic function then the maximum vertical distance between corresponding points on the two graphs is given by the y –coordinate of the vertex of the graph of d .
when g (x)
≥ g (x). ≥ f (x).
Vertical distance d (x) between corresponding points on the graph of f and the x –axis
d (x) = f ( x) when f ( x)
d (x) =
− f ( x)
≥ 0. ≤ 0.
when f ( x)
CHECKLIST Check that you can do the following.
1. Interpret various combinations of graphs, including lines, parabola s, circles and hyperbolas. Examples 6.1.1, 6.1.2, 6.1.3; Activities 6.1.1, 6.1.2, Exercise 6.1
298
STATISTICS OUTCOMES After studying this topic you should be able to do the following. SECTION 7.1: Data Collection and Organisation
Use the terminology of statistics: data (discrete and continuous), population, sample, tally, class, class limit, class width, class boundary, class midpoint, frequency, frequency distribution, grouped frequency distribution.
Organise data in tables in a meaningful way.
SECTION 7.2: Using Graphs to Represent Data
Use pie graphs to repres ent data.
Use histograms and frequency polygons to represent data (this includes application of the terminology class midpoint and class boundary).
Use stem–and–leaf plots to represent data.
Use line graphs to repre sent data.
SECTION 7.3: Some Statistical Measurements
Determine the arithmetic mean of a set of data (grouped or raw, with or
without a calculator). Determine the median, the mode(s) (if any) and the mean deviation of a set of data.
SECTION 7.4: Probability
Apply the terminolo gy of probability: outcome, sample space, event.
Calculate the empirical probability of a given event.
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7.1 DATA COLLECTION AND ORGANISATION
7.1A COLLECTING DATA Data is a word we use to describe information that can be quantified in some way. Note that the word “data” is plural; we thus say “the data are ... ” and not “the data is ...”.
In our society today we depend a great deal on the answers we get from the collection and analysis of data. These activities form part of the subject we call statistics. Apart from collecting and analysi ng data, we also have vari ous possible graphical representations of the data. Statistical graphs are not the same as the graphs you have dealt with so far, which represent ordered pairs in the Carte sian plan e. When we use the word grap h in this topic, we mean a visualisation of given information. Think of the example given at the beginning of this book, in which we looked at a batsman’s shots, and considered a frequency table and graph relating to a specific cricket match. When the captain of the opposing team decides where to place fielders on the field, he takes into account the likelihood of the batsman batting in a particular direction. During a whole season’s cricket matches people are able to record information such as that shown in Table 1.1.1 or Figure 1.1.3 of Study Unit 1.1A. A captain, with experience, will know how to make use of this information. In more complicated situatio ns, such as the analysis of population trends, highly qualified people, called statisticians, obtain and analyse huge amounts of data, using a variety of statistical tools in order to interpret their findings and make decisions for the future. Statistics can then be considered as a way of using past knowledge to make future plans. It is thus extremely useful because it provides us with a guide as to what is happening in our soc iety. For example, in the Application of Topic 4, Book 2, we looked at a table showing the numbers of HIV–positive mothers in different years. Many other items of information also need to be obtained and analysed so that possible conseque nces can be predicted and strategies planned to deal with the expected consequences . In today’s information age it is essential that we learn how to collect, organise and analyse data. The idea that this would become reality was already evide nt many years ago, when H.G. Wells (1866 – 1946) said “Statistical thinking will one day be as necessary for efficient citizenship as the ability to read and write”. The word “statistics” is derived from “state–istics”, i.e. data collected by states to find out information such as how many farms there were from which taxes could be collected, or how many young men there were who could be drafted into an army. Now statistical data is common in our everyday lives, as we can see from the article below, taken from the 28 November 1998 issue of the Pretoria News.
300 World population now: 6 billion. In 2050: 9–10 billion Babies born each second: 5 Nations unable to feed their people: 82 World population growth: 98% in poorest countries By 2020, food production increase needed: 70% (800 million tons of grain) Forests destroyed each year: 16 million hectares People going to bed hungry each night: 800 million People earning less than $1 a day: 1,3 billion Children dying from malnutrition: 400 every 15 minutes – like a jumbo jet crashing every quarter of an hour Refugees in 1970: 3 million; 1998: 28 million Women make up 60% of the world’s farmers Table 7.1.1
Statistical data are sometimes easy to control, but this is not always the case. Suppose you were to set up a physics experiment in a laboratory, to test the effect of heat on different types of metal wire. You can easily control the data–collection process for this experiment, since the temperatures applied to the different types of wires, and the lengths of the wires, can be measured at any given time. Once you have decided on the variables being measured, you can make certain that other variables are exclud ed. In this way accurate resul ts can be obtained and conclusions can be reached with some certainty. However, in many statistical problems it is extremely difficult to control the data. If we want to find out how successful an advertising campaign has been, such as a campaign to persuade people to use washing powder “Clean”, it is difficult to control and analyse the data accurately. For example, we cannot find out exactly – how many people use washing po wder and thus may have bough t “Clean” (i.e. the target market) – how many people actually bought “Clean” – how many people bought “Clea n” direct ly as a result of the adver tising campaign. Each of these items can be affected by other factors in ways we cannot measure exactly, and it is thus difficult to obtain measurements so precise that we can be completely certain what they mea n. Statistics gives us the tools we need to be reasonably sure that our conclusions are correct, so that we can make the best decisions in any given situation. At this stage we will not define what we mean by being reasonably sure. In statistics we use a measure called the “level of confidence” as a measure of “being sure”. We will not discuss this furt her, since this topi c is an introduction to statistics and not an in–depth study of particular statistica l concepts.
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While statistics provides useful information, we also need to be aware that statistical data can be misinterpreted, and can mislead people who do not understand data–collection procedures , simple data analysis, or the graphical interpretation of information. Consider Example 7.1.1.
7.1.1 Suppose that a report states that in region X, in 1998, “the unemployment rate was 6,1% for females and 14,5% for males”. A newspaper reports this as follows. “During 1998, the ratio of male to female unemployed rose further to approximately 2,4.” What do you think is wrong with the newspaper report?
SOLUTION 14, 5 ≈ 2, 4 6, 1 However, 14,5 and 6,1 represent percentages, and not actual numbers. Suppose there are 2 500 employable males and 4 000 employable females in region X. Then 14 , 5% × 2 500 ≈ 363 and 6 , 1% × 4 000 = 244. Thus the ratio of unemployed males to females is in fact approximately 363 244 , i.e. approximately 1,5, which is significantly different from the stated figure, which is 2,4. This is another good reason to have understood ratios and percentages when you studied them in Book 1!
Consider now the second item of information in Table 7.1.1. Babies born each second : 5 It is unlikely that researchers obtained data from all possible hospitals, clinics, maternity homes, etc., throughout the world. Instead of investi gating every possible case, smaller groups can be investigated. These groups provide va luable information, provided they are representative of the whole population. We use the word population to mean all possible members of a particular group. In this case, the populatio n consists of all newborn babies. This is an example of a very large, but finite, population. The technique of creating smaller, manageable, representative groups from which we can obtain meaningful information is called sampling. There are many methods used to determine appropriate samples of suitable size; once again these are beyond the scope of this module. When we make conclusions about a population as a whole on the basis of a certain sample, we say that the conclusions are inferred from the analysis of
302 the sample. The part of statistics dealing with the conditions under which such inferences are valid is called inductive statistics or statistical inference. When we only describe and analyse a certain group without making any inferences about a larger group, we are dealing with deductive or descriptive statistics.
7.1B ORGANISING DATA
“Tally” is a word we use for “keep a score”, or “keep count”.
Each time we have a collection of 5 items, we draw a line through the group of 4 tally marks. In this way we count each “crossed out” group as 5. Raw data Class Class frequency Frequency distribution
In order to analyse the information obtained from the sample, we first need to record the information. A useful record– keeping tech nique is to keep a tally. We write | || || | |||| ||||
the first time we obtain a particular response when we have two such responses when we have three such responses when we have four such responses when we have five such responses.
We continue this process, and in the end we have something like this: |||| |||| |||| || . This represents 17: three lots of five, and two more. Data are numbe rs. For the ti me being we consid er only inte gers. Before we organise the collected data in any way, we refer to the data as raw data. We can arrange the raw data in a variety of ways, for example in ascending or descending order. It is often useful to arrange the data into categories, which we call classes, and determine how many items there are in each category or class. The number of items in a class is called the class frequency . When we arr ange the da ta in a table showing the classes with their corresponding frequencies, we have a frequency table, which we also call a frequency distribution. We illustrate the meaning of these words in an example.
7.1.2 Suppose a government wants to find out more about education. They are interested to find out – how many families in a given commu nity have children of primary school age (i.e. 6 years old to 13 years old) – what the monthly income of each family is
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– how many children of primar y school age there are in each family – how many children of primar y school age in each family atte nd school.
Please note that these are arbitrary numbers, chosen for the purpose of this example.
Suppose 124 families in a particular communit y were chosen, and asked to complete a questionna ire. The sample size in this case is thus 124, and the statisticians would need to decide whether meaningful decisions about the community as a whole could be made on the basis of this sample. The information can then be arranged into a table. Look at the first column in Table 7.1.2. In this column
If a variable can theoreti-
I represents income per family, per month, in rands Nf represents number of families in each income group Np represents number of primary school age children N p+ represents number of primary school age children attending school N p× represents number of primary school age children not at tending school.
cally assume any real value between two given values, we say it is a
continu-
ous variable; otherwise it is
called a discrete variable. Data which can be represented by discrete or continuous variables are referred to as discrete or continuous data, respectively.
In this example we assume that the ages of the children were their ages in years at the time of the survey, so that, for example, two children whose ages were 7 years and 3 months, and 7 years and 10 months, respectively, would both have been classified as 7 years old. This classification provides discrete data.
< 501– 1 001– 1 501– 2 001– 2 501– 3 001– 3 501– 500 1 000 1 500 2 000 2 500 3 000 3 500 4 000
I
Total Nf Np N p+ N p×
12 27 72 1 35 37 121 35 14
26 130 107 23
28 140 128 12
14 56 46 10
13 42 36 6
3 11 11 0
1 124 3 589 3 489 0 100
Table 7.1.2
From the table we see that in all columns (except the columns representing the families earning R3 001 or more per month) there is a difference between the number of children who could be attending school ( N ), and the number actually p attending school (N p+ ). Of the 124 families selected for the survey, there are 589 children of primary school age 489 of the primary school age children go to school 100 of the primary school age children do not go to school. Suppose we now want to analyse this group further, and find out the ages of the child ren not attending school. There is now only one item of information we want to recor d: the ages of the childr en not attend ing schoo l. We cannot obtain this information from Table 7.1.2, and we thus need to carry out further
304 investigations to obtain this information. We need to find out how many children in each age category are not attending school, i.e. how many 6–year olds, how many 7–year olds, etc., up to the 13–year olds, do not go to school. Suppose we let x represent the ages of the children, and n the number of children in each age group who do not go to school. Every x value thus represents a class and the number n the class frequency. Table 7.1.3 shows the distribution of the different frequencies, i.e. the number of children in each age group who do not attend school. x 6 7 8 9 10 11 12 13 n 1 5 8 6 4 7 17 19 24
Table 7.1.3
Since the total of the N p× row in Table 7.1.2 is 100, the total of the numbers in the n row in Table 7.1.3 must also be 100.
When we record information such as this, we have no difficulty in assigning exact values to the variables x and n , since both sets of data are given as discrete data. However, suppose we wanted to record the heights of a group of childr en. If we find a height of 131,2 cm, and another height of 131,3 cm, it is theoretically possible to find several more measur ements between the two. Depending on the accuracy of our measuring equipment, we could have, for example, children whose heights are 131,23 cm and 131,24 cm. Measurements such as these represent continuous data. When we have potentially continuous measurements it becomes more difficult to set out the information in a frequency table such as Table 7.1.3. Also, there are often too many measurements for it to be practical to list each item separately. We then need to arrange the data a little differently, and we use a grouped fre-
See Section 7.2, Example 7.2.3.
quency distribution, in which we consider classes with specific class widths and class limits . Later on, when we draw histograms, we need to introduce the concept of class boundaries as well.
Look at the next two examples.
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7.1.3 Consider the measurement of the midday temperature in Johannesbur g each day during February of 1998. Temperature is also a continuous variable (the number of decimal places we use to express the measurements will depend on the accuracy and sophisticat ion of the measuring equipme nt we use). Suppose we obtain the following measurem ents, correct to one decimal place.
Day
1 2 3 4 5 6 7 8 9 10 11 12 13 14
Temperature (in ◦ C, Day Temperature (in ◦ C, to one decimal place) to one decimal place) 25,4 15 26,4 26,3 16 24,5 24,9 17 27,2 28,2 18 28,4 30,9 19 28,3 30,8 20 28,5 31,3 21 29,4 30,2 22 30,2 29,7 23 31,4 28,6 24 33,2 27,0 25 33,4 25,9 26 33,5 25,8 27 33,6 26,1 28 32,1
Table 7.1.4
There are too many differe nt values in this table for it to make sense for us to list each specific temperat ure in a class of its own. We organise the data as follows.
Look for the lowest and highest measurements (in this case the lowest is 24, 5◦ C and the highest is 33 , 6◦ C).
Choose class limits and determine the classes. Suppose we ch oose to ◦
group the data into classes so that each class has a class width of 1 C (i.e. each successive class represents a change in temperature of one degree Celsius). Thus, if the first class begins at 24 , 5◦ C (the lowest temperature), it ends at 25 , 4◦ C, and hence it makes sense to choose the last class with limits of 33, 5◦ C and 34 , 4◦ C because the highest temperature falls into this class. There are thus ten classes.
Count the total number of times a temperat ure occurs in each class. These totals are the class frequencies. We denote the frequency by n .
Set up a frequency table which summarises the above information.
306
Classes 24,5– (◦ C) 25,4 n 35
Check: As you would expect, the total of all the numbers in the n row is 28, since there were 28 temperatures recorded in February , 1998. 25,5– 26,5– 27,5– 28,5– 29,5– 30,5– 31,5– 32,5– 33,5– 26,4 27,4 28,4 29,4 30,4 31,4 32,4 33,4 34,4 2 3 33 41 22
Table 7.1.5
7.1.4 Suppose we have the following measurements (rounded to the nearest centimetre) of the heights of 100 students.
152 156 161 162 163 161 165 161 160 166 165 162 169 161 172 161 167 157 174
163 163 167 159 162 166 165 168 163
165 169 164 166 166 175 165 161 162
163 164 157 168 162 162 170 158 173
164 153 172 165 157 156 172 161 170
164 166 158 170 165 170 162 168 159
158 166 161 153 167 170 163 159 158
161 161 167 169 167 163 158 162 168
154 159 169 164 158 162 158 163 174
168 162 166 162 152 172 161 168 159
Table 7.1.6
The shortest stude nt’s height is 152 cm; the height of the tallest student is
We choose to group the measurements into classes whose widths are all 5 cm, where the first class begins at 151 cm; the last clas s will then end at 175 cm.
We count the number of times each measurement occurs in each class. We record these frequencies in Table 7.1.7.
We check that the total of the class frequencie s is 100.
175 cm.
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Classes (in cm) 151–155 156 – 160 161 – 165 166 – 170 171–175
Class frequencies 5 18 42 27 8
Table 7.1.7 Note
– Since the width of each class is 5 cm, there are five classe s. – There are many other ways in which we can organi se these measurements into a frequency distribution. For example, we can choose the lowest class limit to be 152 cm, and class widths of 6 cm. We then have the following table. Classes (in cm) Class frequencies 152 – 157 10
158 – 163 164 – 169 170 – 175
43 34 13
Table 7.1.8
Because the classes are wider, there are now only four classes.
You have been learning many concepts that are possibly new to you. Have you understood them? To check, try to do Activity 7.1.1.
7.1.1 A group of MAT011–K students obtained the percentages shown in the table for their first assignment.
308 Name Mark (%) Name Mark (%) Silas 34 Francina 57 Marian 38 Francois 85 Gill 45 Sean 63 Zita 68 Shamus 47 Sumaya 33 Frank 42 Dimitri 74 Dhaya 77 Amos 54 Eddie 23
James Jan Celeste
35 66 53
Isaac Abram Petrus
58 61 35
(a) Find the lowest mark, and the highest mark. (b) Suppose you represent the data by means of a grouped frequenc y distribution, with classes of width 10%, so that 20% is the the lower limit of the first class and the highest mark is in the last clas s. How many class es of marks will there be? (c) Use the information obtained in (a) and (b) to set up a grouped frequency distribution for the marks obtained by these students.
(a) The lowest mark is 23%; the highest mark is 85%.
Note that in practice the lowest class limit can be any percentage that is less than or eq ual to 23 %.
Sup-
pose we keep the class width 10%.
We ca n ch oose to
start at 0%; the first class will then have limits of 0%– 9% and a frequency of zero, since no students obtained marks in this ca tegory. In this case there will be nine classes. We can also choose to start at 22%; the first class will then have limits of 22% – 31%, and a frequency of 1.
There will ag ain be 7
classes.
(b) We start at 20% . We want cla sses of wi dth 10%. If we start at 20% , the class limits of the first class will be 20% – 29%, the second class will have limits 30% –39%, etc. The last class will have limits 80% – 89% and thus there will be 7 classes. (c) We count the marks in each class. Marks from 20% to 29%: Marks from 30% to 39%: Marks from 40% to 49%: Marks from 50% to 59%: Marks from 60% to 69%: Marks from 70% to 79%:
Tally: Tally: Tally: Tally: Tally: Tally:
| Frequency: 1 | | | | Frequency: 5 | | | Frequency: 3 | | | | Frequency: 4 | | | | Frequency: 4 || Frequency: 2
Marks from 80% to 89%:
Tally: |
Frequency: 1 Total: 20
We check to make sure that the total in the number column is the same as the total number of students we are considering. We now have the follow ing frequency table. Classes(%) 20–29 30–39 40–49 n 1 5 3 4 4
50–59 60–69 70–79 80–89 2 1
Table 7.1.9
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We have included, for interest only , an article on smoking from the South African Medical Journal that relies heavily on statistical sampling and various other statistical techniques. The article appears at the end of this topi c. An amusing quote, attributed to Fletcher Knebel, and given in the book Agnesi to Zeno, by Sanderson Smith, is “It has now been proved beyond a shadow of a doubt that smoking causes statistics.” Certainly smoking (or not) is a topical issue, and you may find the article interesti ng. Note that the purpose of the survey was to determine the extent to which people smoked, and the extent to which they were aware of the harmful effects of cigarette smoking. This article is thus an example of descriptive statistics, and not inferential statistics. 1 Section 7.1 presents an extremely simplified view of obtaining and organising data. On the first page of the article on smoking, in the section with the heading “Sampling” we have pointed out (by using italics in the quoted section below) some of the specialised statistical procedures used to ensure that the sample is representative (in terms of age, gender, race and so on) of the population as a whole. A sample of lung cancer sufferers was selected, in the following way: “A multistage cluster sampling method was used to select 2 238 households. A randomly selected adult from each household was interviewed in a standardised manner by trained interviewers in the preferred language of the respondent. ... The reliability of the interviewers’ survey techniques was ensured by repeating 10% of the interviews either telephonically or in person. The sociodemographic characteristics of the sample were compared with the 1992 census data and weighted accordingly.” It is clear that there is much more to sampling than we can discuss in this module. You may find it interesting to study a course in basic statistics at a later stage.
1
The article appears in Volume 86, No. 11, November 1996, pp. 1389–1393.
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7.1 1. State briefly what the following terms mean. tally, class, frequency, frequency distribution 2. An efficiency expert takes a sample of 100 “breakd own” times (in minutes) of the machines in a factory. (Breakdown time is the length of time during which a machine is out of action either awaiting repair or being repaired or replaced.) The times are given below. Represent this data by means of a grouped frequency distribution. Choose 5 minutes as the lower limit of the first class; choose class width 5 minutes, and include the longest ”breakdown” time in the last class. 2 21 21 38 59 36 22 40 25
23 38 24 27 32 42 13 37
19 36 11 12 24 19 24 16
27 49 22 23 44 47 25 52
43 23 45 27 16 29 22 17
24 20 21 10 13 31 23 28
20 18 9 7 27 10 19 24
45 22 29 34 23 48 20 18
33 17 21 22 33 15 34 23
22 35 41 33 27 21 39 53
35 27 20 28 31 21 28 23 25 12 44 17 30 24 32 40 29 14 30 15
2
This example is taken from Page, Berry and Hampson: Mathematics, a second start.
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7.2 USING GRAPHS TO REPRESENT DATA
7.2A PIE GRAPHS, HISTOGRAMS AND FREQUENCY POLYGONS
Do you remember why the circle does not represent a function?
Pictographs
We have already looked at a variety of different graphs, particularly in Topics 3, 4, 5 and 6 of this book. What these graph s have in commo n is that they each represent a particular equation, and, apart from circles, they are all graphs of functions. Statistical data are also often represented visually. One of the simplest types of visualisations we have is a pictogram, sometimes also called a pictograph. The name is easy to understand when we see that such graphs make use of pictures to represent information. If we want to use a pictogram to show the progress in the planting of indigenous trees, we could have the following exampl e.
7.2.1 Study the pictograph below, in which one
represents 1 000 trees.
Region
Tally
Gauteng
|||| |
Mpumalanga
||||
Limpopo
|||| |
Free State
|| |
Northern Cape
||||
Western Cape
|||| ||||
Table 7.2.1
From the pictograp h we see that 6 000 trees were planted in Limpopo, and 5 000 in Mpumalanga.
312 One of the limitations of pictographs is that it is usually difficult to represent parts of the whole. For example, if 5 300 trees are plante d in a certain region, how can we represent 300 trees using the given symbol for 1 000 trees? However, pictographs are useful for the representation of simple information, relatively free from language and numbers, and thus easy to understand. Although pictographs are useful, they are easy to misinterpret. Suppose a doughnut company uses the symbol of one doughnut to represent profits in rands, where every one centimetre of the thickness of the doughnut (i.e. the radius of the outer ring minus the radius of the inner ring) repr esents R100 000,00 profi t. If you are given the following pictogram, in which year would you say profits were highest?
1996
1997
1998 Figure 7.2.1
In 1996, the doughnut’s thickness is 0,6 cm, i.e. the profit is R60 000,00. In 1997, the doughnut’s thickness is 1,3 cm, i.e. the profit is R130 000,00. In 1998, the doughnut’s thickness is 1,2 cm, i.e. the profit is R120 000,00. If we do not carefully read the details such as the scale used to draw the graph, we may look at the picture and assume that the biggest profit was earned in 1998, since the biggest doughnut is used to represent the profit in that year. Pie graphs
Another type of statistical graph is called a pie graph, for the simple reason that it looks like a pie. Figure 7.2.2 shows a pie graph, where the slices, or wedges, of the pie represent different percentages. The intention is that, for example, half the pie represents 50%, and hence smaller slices represent smaller percentages. If we want to draw accurate pie graphs, we need to work out, as precisely as possible, the size of the wedges. The “pie” may be drawn to look like a three-dimensional pie, cut into various wedges. It may also be drawn in two dimensions, as a circle. We then refer to the “slices of the pie” as sectors.
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Definition 7.2.1 A pie graph shows percentages by means of areas of sectors of a circle, where the areas are proportional to the percentages that must be represented.
We commented earlier that the word “proportion” is sometimes used when “direct proportion” is meant.
For practical purposes it is not usually necessary that the wedges are drawn to scale. The infor mation in Figure 7.2.2 appea rs in the 1996/9 7 South Africa n Survey, published by the South African Institute of Race Relations.
Education levels of people over the age of 20:1995 Degrees No education
13%
3%
Diplomas
Matric Gr 1 − Std 5
7%
19%
24%
This graph is drawn in three dimensions. It was adapted and drawn in two dimensions in Topic 1 (Figure
Std 6 − Std 9
1.1.1(b)).
34%
Figure 7.2.2 Note
The graph has a label, or heading, to indicate what it represents.
In to draw the graph, the relevant datathen is obtained from the population asorder a whole, or from a suitable sample and manipulated in some way to reflect the whole population.
Each frequency is converted to a percentage, since pie graphs only show percentages, and not exact numbers or classes of numbers
If this is an accurate graph then the central angle of each sector is proportional to the percentage that the sector represents.
Figure 7.2.3 illustrates what we mean by “ central angle”.
314 The central angle is the name we give to the angle at the centre of the circle. B A
O
If we take the radius OA, and rotate it one complete circle, in an anti–clockwise direction, we travel through 360 degrees (we abbreviate this as 360◦ ).
ˆ The central angle A OB Figure 7.2.3
These questions relate to proportion, which is discussed in Topic 3 of Book 1. If yo u und erstood the concepts at that stage you should have no difficulty with the appli cation to angles here.
Since one complete revolution measures 360 degrees, the central angle for each slice of the pie will be some fraction of 360 degrees. Since 360 degrees represents 100%, what is the measurement of the central angle for a slice that represents 1%? What angular measurement will represent 25%? If the central angle of a sector in a pie graph measures 10 degrees, what percentage does the sector represent? (In case you trie d to find the answ ers: 1% can be represented by a sector with a central angle of 3 , 6◦ ; 25% can be represented by a sector with a central angle of 90 ◦ ; a sector with the central angle of 10 ◦ represents 25 9 %, i.e. approximately 2,78%.) A graph such as the one in Figure 7.2.2 is useful for making comparisons. We can see at a glance that the biggest slice of the pie represents people who have passed standards 6 to 9 (now called grades 8 to 11), and the smallest slice represents those with degrees.
7.2.2 Let us consider a group of 650 MAT011–K students. Suppose we have the information given in Table 7.2.2 and we want to represent it by means of an accurate pie graph.
Number of assignments submitted 1 2 3 4 5
Table 7.2.2
Number of students 35 140 258 153 64
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SOLUTION We
first convert the numbers of students to percentages
calculate the central angle of each sector
then use these angle measurem ents to sketch the sectors of the circle.
If we round the percentages to the first decimal place, we have 35 650 140 650 258 650 153 650 64 650
≈ 5, 4% ≈ 21, 5% ≈ 39, 7% ≈ 23, 5% ≈ 9, 9%.
Since one complete revolution measures 360 degrees, we have 35 × 360◦ 650 140 × 360◦ 650 258 × 360◦ 650 153 × 360◦ 650 64 × 360◦ 650 It will be difficult to measure 0,4, i.e. 25 , of a degree, with a normal protractor, so it is unlikely that graphs such as these are accurate.
≈ 19, 4◦ ≈ 77, 5◦ ≈ 142, 9◦ ≈ 84, 7◦ ≈ 35, 5◦ .
We now draw any circle. Draw a radius. At the centre, from this radius, measure an angle of 19 , 4◦ (using a protractor), and draw anot her radiu s. This sector thus represents 5,4% of the students. We continue in the same way so that, for example, we measure a central angle of 77 , 5◦ immediately next to the 19 , 4◦ angle, and draw another radius, so that we have a sector representing 21,5% of the students. Measure the remain ing angles (clearly it will not be necessary to measure the last angle), complete the sectors so that each one represents the required percentage, and write the percentages on each sector. Note that we need not have started with the 19 , 4◦ angle. We can start with any angle, and proceed in any direction aroun d the centre. The completed pie graph is given in Figure 7.2.4, on the next page.
316 5,4%
9,9% 1 5 2 23,5% 21,5% 4 3 39,7%
Number of different assignments submitted by MAT011–K students Figure 7.2.4
Try to apply the technique used in Example 7.2.2 in the next activity.
7.2.1 Consider the same group of students as in Example 7.2.2. Suppose we also have the following information. Academic performance
Cancelled the module before writing the exam Passedwithdistinction (75% or more) Passed the exam but without distinction (obtained 50% or more, but less than 75%) Failed the exam but qualified for a supplementary exam (obtained between 30% and 49%) Failed the exam, and did not qualify for a supplementary exam (obtained less than 30%)
Number of students (Total = 650) 38
3 438 123
48
Table 7.2.3
Draw a pie graph to show this information. Calculate the angles as though you were planning to draw an accurate graph, but your sketch need not be accurate.
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We convert the numbers to percentages, and round each percentage to the first decimal place. 38 650 3 650 438 650 123 650 48 650
≈ 5, 8% ≈ 0, 5% ≈ 67, 4% ≈ 18, 9% ≈ 7, 4%
We then calculate the required central angles correct to the first decimal place. 38 × 360◦ 650 3 × 360◦ 650 438 × 360◦ 650 123 × 360◦ 650 48 × 360◦ 650
≈ 21, 0◦ ≈ 1, 7◦ ≈ 242, 6◦ ≈ 68, 1◦ ≈ 26, 6◦
We now draw a circle, and represent these percentages as sectors of the circle, as far as possible estimating the size of each sector according to the central angles calculated above. Failed
Cancelled before writing Passed with distinction
Failed (qualified for supplementary)
7,4%
5,8% 0,5%
18,9%
Passed 67,4%
Academic performance of MAT011–K students Figure 7.2.5
318 Scatter graphs
Another way of representing data is by means of a scatter diagram or scatter graph or scatter plot . This kind of graph is use d when we wa nt to loo k at the data and for m an idea of what kind of equati on may best fit the data. The techniques for finding a suitable equation are beyond the scope of this module, and we thus only show an example of a scatter graph (in Figure 7.2.6), but we do not discuss such graphs further.
10 Number of students
8 6 4 2
0
10
20 30
40 50
60
70
80
90 100
Marks obtained by students in an exam (percentages rounded to the nearest ten percent)
Figure 7.2.6
Histograms
Another useful method of graphical representati on of data is the histogram.
Definition 7.2.2 A histogram represents each class of information by means of a rectangle whose width represents the class width and whose area is proportional to the frequency.
We do not necessarily use the same scale for the class
Note that area is proportional to length and breadth, since area = length × breadth. Since we use the same class width for each class, and a scale for class width as well as for frequency, we may also state the definition as follows.
width and the frequency.
A histogram represents each class of information by means of a rectangle whose width represents the class width and whose height is proportional to the frequency. We look at an example to understand what this definition means.
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7.2.3 Consider the data presented in Table 7.1.7 (table included here for convenience). Classes (in cm) Class frequencies 151–155 5 156–160 18 161–165 42 166–170 27 171–175 8
Measurements of heights are examples of continuous data.
The class width in this case is 5 cm. Heights are record ed to the nearest centimetre, i.e. we have the classes 151 cm – 155 cm, 156 cm – 160 cm, etc. However, if we want to draw a histogram to represent this data, the rectangles representi ng different classes would have gaps between them, for example a gap between the end of one class, say 160 cm, and the beginning of the next, 161 cm. We overcome this problem by extending class limits by one more decimal place . We call these the class boundaries. Extending the class limits in this way im plies that we make the lower class limit, for example 151 cm, smaller, and the upper class limit, for example 155 cm, bigger. The class boundary of a class lies half–way between its lower class limit and the upper class limit of the previous class. In this example the class boundaries are 150,5 cm, 155,5 cm, etc., and it is easy to draw a histogram when we know what boundari es are appropriate. The upper boundary value of one class is the lower boundary of the next class and hence the rectangles will not overlap. We repeat Table 7.1.7 again, including a column giving the class boundaries, i.e. 150,5 cm, 155,5 cm, etc. Classes (in cm) 151–155 156–160 161–165 166–170
Class boundaries (in cm) Frequencies 150,5–155,5 5 155,5–160,5 18 160,5–165,5 42 165,5–170,5 27
171–175
170,5–175,5
8
Note that measurements of heights are given to the nearest centimetre. There will thus be no recorded measurements such as 165,5 cm or 170,5 cm, so there will be no confusion about the interval in which a measurement lies.
320
Histogram 42
40 Number of students
30
27 18
20
8
10
5
We sometimes use a jagged line to show that the hori-
150,5
155,5
160,5
165,5
170,5
175,5
zontal (or vertical) axis has
Heights of students (in centimetres)
been shortened so that numbers from the srcin up to the point where the mea-
Figure 7.2.7
surements begin, have been omitted.
We can give an even more accurate graph if we use graph paper.
Sometimes the actual frequency is written on the top of each rectangle , as shown in Figure 7.2.7. Unless we do this, in a sketch such as this we have no way of being quite certain of the number of students whose heights are, for example, between 170,5 cm and 175,5 cm. Does the last recta ngle corr espond to 8 students, or 9? We can overcome this probl em to some extent if we use a grid. Figure 7.2.7 is repeated below, on a grid (see Figure 7.2.8), and the number of students in each class is a little easier to find. The last rectangle’s height seems to correspond to the number 8, i.e. there appear to be 8 students whose heights are between 170,5 cm and 175,5 cm.
50 Number of students
40 30 20 10
150,5
155,5
160,5
165,5
Heights of students (in centimetres)
Figure 7.2.8
170,5
175,5
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Note
In Figures 7.2.7 and 7.2.8 it is necessary to choose an appropriate scal e for both axes. We use two different scales. We use Horizontal axis: 1,5 cm represents a change in height of 5 cm Vertical axis: 1 cm represents 10 students.
The class width is 5 cm, represented by 1,5 cm. The area of each rectangle (and hence its height) is proportional to the frequency. The class boundary values are given to one decimal place more than the class limits.
In a histogram the classes do not always have to be represented on the horizontal axis. It is also acceptable to represent them on the vertical axis, as we see in the next example.
7.2.4 30 pupils in a class work on a project. Table 7.2.4 shows the distribution of the marks over the pupils in the class. In the table the marks have been divided into classes with class width of 10%. Since no one had a mark of less than 40% it is acceptable to begin with a class that has the limits 40% and 49%. Represent this information by means of a histogram. Classes
90% – 99% 80% – 89% 70% – 79%
Frequencies (Total = 30) 1 3 5
60% – 69% 10 50% – 59% 7 40% – 49% 4 Table 7.2.4
SOLUTION In order to draw a histogram we need the class boundaries. Since the class limits are given to the nearest unit we find the class boundaries by extending the limits by one decimal place, i.e. to tenths. We thus have the followin g table.
322 Class boundaries
89,5% – 99,5% 79,5% – 89,5% 69,5% – 79,5% 59,5% – 69,5% 49,5% – 59,5% 39,5% – 49,5%
Frequencies (Total = 30) 1 3 5 10 7 4
Table 7.2.5
We choose an appropriate scale. We have horizontal axis: 0,5 cm represents 1 pupil vertical axis: 1 cm represents 10%. Using this scale we thus draw rectangles with length 2 cm, 3,5 cm, 5 cm, 2,5 cm, 1,5 cm and 0,5 cm, respectively. Figure 7.2.9 shows how horizontally arranged rectangles represent this inform ation. The area (and hence the length) of each rectangle is proportional to the class frequency.
99,5
Histogram
89,5 Marks obtained 79,5 (percentages) 69,5 59,5 49,5 39,5
5 Number of pupils
10
Figure 7.2.9
Note
We see again that without using graph paper, or writing the actual numbers next to each rectangle, it is difficult to be precisely sure how many pupils fall into each class.
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Frequency polygon
Yet another important statistical graph is the frequency polygon. In order to draw frequency polygons we need to know the midpoint of each class.
Class midpoint
Either class limits or class boundaries are used to give us the class midpoint. We can obtain each class midpoint by adding the two class limits (or class boundaries) and dividing by two. In Example 7.2.3, the first class has the limits 151 cm and 155 cm and the boundaries 150,5 cm and 155,5 cm. If we add the two class limits (or the class boundaries) and divide the total by two, we obtain 153 cm,
We will also use the value of the class midpoint for calculations of specific statistical measurements. We study these in the next section.
Class boundaries are useful as we are able to obtain the class width by subtracting the smaller class boundary from the larger one.
which lies in the middle of the class denoted by 151 cm – 155 cm. In this way we obtain (for Figure 7.2.7) the class midpoints 153 cm, 158 cm, 163 cm, 168 cm and 173 cm. Figure 7.2.10 shows the relationship between the class limits, the class boundaries and the class midpoint.
150,5 151
Note that the frequency polygon consists of the line segments joining the dots that correspond to the midpoints of the classes; the histogram is shown here as well for compariso n, but it is not part of the frequency polygon.
155,5 153
Figure 7.2.10
155
If we plot points corresponding to each class midpoint and class frequency, and join these points, we obtain a frequency polygon. Figure 7.2.11 shows the frequency polygon obtained from the data used to draw the histogram in Figure 7.2.7. In Figure 7.2.11 we show the class midpoi nts on the horizontal axis, instead of the class boundaries.
42 40 Number of students
27
30 18
20 10
8 5 153
158
163
168 173
Heights of students in centimetres
Figure 7.2.11
How do we interpret freque ncy polygons? We draw Figure 7.2.11 again, this time without the histogram.
324
40
Frequency Polygon Number of students
30 20 10
153
158
163
168 173
Heights of students (in centimetres)
From the above figure we can deduce the following information.
There are 23 students whose height s are less than 160,5 cm.
There are 35 students whose height s are at least 165,5 cm.
More students are in the class with class boundaries 160,5 cm and 165,5 cm than in any other class.
Try to consolidate your understanding of histograms and frequency polygons by doing the next activity.
7.2.2 Remember that we think of mass as weight, although when you study physics you will see that there is a difference.
In a sample of 50 rugby players, the follo wing masses (in kilograms) were recorded. 72 93 105 92 97 85 94 96 83 78
86 79 86 9 6 101 86 96 98 104 88 89 89 79 107 104 87 92 82 91 9 6 111 79 82 84 87 101 90 88 93 79 90 9 5 108 92 86 77 97 81 89 9 3
Set up a frequency table with class widths of 10 kg. Choose classes so that the lower limit of the first clas s is 65 kg. On the same syste m of axes draw a histogram and frequency polygon to represent the information.
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The classes must each hav e a width of 10 kg. We are told to choose 65 kg as the lower limit of the first class, and the classes thus have the limits shown in the table below. (Note that we do not necessarily hav e to choose these classes. If we want to keep the class width 10 kg, we can also choose, for example, the limits 71 kg and 80 kg; 81 kg and 90 kg, and so on. We must ensu re that we choos e class limits so that no measurements are excluded. If the class limits chang e, then obviously the class boundaries and midpoints will change as well.)
Classes Class boundaries Class midpoints Frequency (kg) (kg) (kg) (Total = 50) 65– 74 64,5– 74,5 69,5 1 75– 84 74,5– 84,5 79,5 11 85– 94 84,5– 94,5 89,5 22 95–104 94,5–104,5 99,5 12 105 – 114 104,5 – 114,5 109,5 4
We draw the required histogram and frequency polygon in Figure 7.2.12.
25 Number
20
of players
15 10 5
64,5 74,5
84,5
94,5
104,4 114,5
Mass in kilograms
Figure 7.2.12
7.2.3 Consider a group of 30 Grade 3 childr en. They did two spelling tests, one at the beginning of term, and one at the end. Their marks (percentages) on the two tests were grouped into classes with class width of 15%. The graphs are shown in Figure 7.2.13 on the next page. The solid line represents the first test and the dashed line represents the second test.
326
Frequency Polygons
Although the class limits are not indicated on the graph, it should be clear that we have the following class limits: 0 – 14
10 9 Number 8 of 7 children 6 5 4 3 2 1 7
14,5
22
29,5
37 52
44,5
67
59,5
74,5
82
89,5
97
104,5
Percentages
15 – 29 30 – 44
Figure 7.2.13
45 – 59 60 – 74 75 – 89 90 – 104.
(a) What type of graphs are sho wn in Figure 7.2.1 3? (b) If 60% is a pass mark, ho w many children passed the firs t test? How many passed the second? (c) Did the childrens’ marks improve in the second test? How can you deduce your answer from the graphs? (d) No one coul d obtain mor e than 100% in the test. Why is it neces sary to have 104,5% as the last class boundary?
(a) Figure 7.2.13 shows two freque ncy polygons. (b) In the first test, 8 children passed . We determine this from the graph . The solid line shows that – 6 children obtained marks in the class with m idpoint 67% – 2 children obtained marks in the class with m idpoint 82% – 0 children obtained marks in the class with midp oint 97%. In the second test, 23 children passed. The dashed line shows that – 10 children obtained marks in the class wit h midpoint 67% – 7 children obtained marks in the class with m idpoint 82% – 6 children obtained marks in the class with midp oint 97%.
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(c) From (b) it is clear that more children passed the second test. However, even without counting up the numbers of children who passed the two tests, we can see that the distribution of the marks is different for the two tests. The graph for the second test has a peak further to the right (i.e . closer to 100% and further from 0%), and more marks on the right side of 60%, than the graph for the first test. (d) Although no one can obtain mor e than 100% , it is possible that a child could have obtained exactly 100%, so the last class must make provision for a mark of 100%.
Bar graphs
In many graphs, information is often presented in columns, in which the area of each column is not proportional to the frequency represented by the column, and thus the scale used for the width of each column is not important. In these graphs, only the height of each column represents the frequency. When we draw such columns, the widths of all the columns are the same, and the columns do not overlap. There are gaps, usually the same width, between the columns. Graphs such as these are called bar graphs or bar charts. Definition 7.2.3 A bar graph consists of vertical (or horizontal) columns of equal width. The height (or length) of each column is proportional to the frequency of the data represented by that column.
The article appears at the end of this topic.
The graphs in Figures 1 and 2 of the South African Medical Journal article are examples of such graphs.
Stem–and–leaf plot
The final type of visual representation we look at in this study unit is the stem– and–leaf plot. An example will illustrate the name, and why we use this method.
7.2.5 Suppose we have the following masses (measured in kilograms) of the bags of potatoes harvested by 52 school–boy volunteers one weekend. 27 36 42 31 36 36
35 48 42 35 31 40
42 41 46 37 29
35 38 41 43 37
36 40 26 33 37
51 46 42 39 45
43 37 34 33 46
30 46 38 54 44
51 42 39 40 31
37 38 36 31 46
328 We want to represent this data by means of a stem–and–leaf plot.
SOLUTION We proceed as follows.
Consider separately the units digit and the tens digit in each of the given numbers. For example, the number 36 has 3 tens and 6 units.
Draw a vertical line and down the left side of the line write all the tens digits that occur in the numbers in the table, from smallest to biggest. Thus we have: 2 3 4 5
Every time each tens digit occurs in the data, write down , on the right side of the vertical line, the corresponding units digit. These digits are written down in the order in which they occur, without any rearrange ment. We begin at the top left corner of the table , and read acr oss the row s. For example, the numbers andwrite 43 in2,the first hence nextwe tohave the number 4 we 42 first then 3, row andofsothe on.table, We and have the following completed diagram. Note that since there are 52 data items, there must be 52 digits on the right side of the vertical line.
Stem–and–leaf plots
2
7 6 9
3
5 5 6 0 7 6 8 7 8 4 8 9 6 1 5 7 3 9 3 16 17 71 6
4
2 3 8 1 0 6 6 2 2 2 6 1 2 3 0 5 6 4 6 0
5
1 1 4
Figure 7.2.14
Now you can see why it is called a stem–and–leaf plot: the arrangement looks a little like a leaf. One advantage of this arrangement of data is that it is easy to see straight away, without first grouping the data, that the majority of boys harvested between 30 and 39 kilogr ams of potatoes. If we count the numbers in the row on the right side of the line where 3 appears, we see that 26 boys harvested bags of potatoes weighing between 30 and 39 kilograms . Only 3 boys harveste d 50 kilograms or more.
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It is sometimes useful, for comparison purposes, to draw “back–to–back” stem– and–leaf plots. We can do this in a classroom situation, if we want to compare the marks of girls and boys in a class test. In such a case we put the tens digits in a column in the middle, and write down the units digits for the girls on one side of the column , and for the boys on the other side of the colum n. Figure 7.2.15 is an example of this. (Note that we give only the final plot, and not all the data used to create it.) One difficulty in this display is that we read the numbers on the left side of the middle column back to front. For example, in the first row , the first entry on the left represents 17%, and not 71%. 7127 3 3 7 263521 9135268 32 4 5 1 8 3 5 0 480927165613 3 0 4 5 3 3 0 1 7 7 95167792201 2 2 52 78 11 08 1 06 26 03 27 3 1 3 75201 0 4 8 4 4 7 2 9 3 4
Girls’ marks
Boys’ marks
Percentages obtained in a class test Figure 7.2.15
7.2B LINE GRAPHS Line graphs
At the beginning of this book we discussed situations in which it makes sense to join dots in the Cartesian plane that represent data item s. In the case of histograms, although we make use of a horizontal and a vertical axis, we have not been working in the Cartesian plane. In the case of frequency polygons, although the points that are joined represent points in the Cartesian plane, the other points on the lines are not relevant to the graph and do not represent specific points in the plane. We now consider another represent ation, called a line graph where we represent data by means of points in the Cartesian plane.
330 Definition 7.2.4 A line graph connects points in the plane by means of line segments. Each point corresponds to an item on the horizontal axis and on the vertical axis. Items along both axes are evenly spaced, according to an appropriate scale. The vertical height of each point is proportional to the quantity being represented.
Each specific point is represented by an order ed pair of real numbers. The point represents specific data, but the other points on the lines joining these specific points again hav e no meanin g in terms of the graph . Consider the following example.
7.2.6
You will study the meaning of averages in the next section. Note that the price of gold is always quoted in US dollars ($), and it is always given as dollars per ounce, even in
Consider the gold price over an eight week period at the beginning of 1998. (Note that internationally the gold price is quoted on a daily basis, and will be given in dollars and cents. For the purpose of this example we are looking at the average price over a week, rounded to the nearest dollar.) We have the following table. Week 1 2 3 4 5 6 7 8 Price per ounce 274 279 276 281 285 283 281 270 (in dollars)
countries which use the met-
Table 7.2.5
ric system of measurement. Do not confuse a line graph with a frequency polygon.
Time (in weeks) is a discrete variable; however, the price in dollars is theoretically a continuous variable. We show the change in the price by means of a series of line segme nts. A line such as this shows the trend in the gold price quite clearly, i.e. it shows whether the price is increasing or decreasing. Figure 7.2.16 on the next page is the graphical representation of the information in Table 7.2.5.
331
Line graph
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290 Price 285 in dollars 280 275 270
1234567
8 Weeks
Figure 7.2.16
Note
In line graphs
it is necessary to choose an appropriate scale for both axes
we join the dots to highlight the trend in the variable we are measuring
there are two varia bles, but neither is dependent on the other
items along the horizontal axis are often successiv e periods of time.
7.2.4 The table below gives Susan’s temperature (measured in degrees Celsius), taken every four hours, while she spent two days in hospital. Time 12:00 16:00 20:00 0:00 4:00 8:00 12:00 16:00 Temperature ( ◦ C) 39,7 38,5 38,0 38,0 37,4 37,4 37,5 37,5
(a) Draw a horizontal axis for time, usi ng a suitable scale. (b) Draw a vertical axis for temper ature, using a scale that will allo w you to show clearly the difference between temperatures such as 37,4 ◦ C and 37,5◦ C. (c) Label the axes.
332 (d) Plot the data against the axes. (e) Join the dots. (f) At what time was Susan’ s temperature the highest? (g) What was the trend of the temperature o ver the period that the tem perature was recorded?
The graph below answers questions (a) to (e).
39,8 39,6 39,4 Temperature
39,2
(in degrees Celsius)
39,0 38,8 38,6 38,4 38,2 38,0 37,8 37,6 37,4
12:00 16:00 20:00 0:00
04:00 08:00 12:00 16:00
Time in hours
Figure 7.2.17
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(f) The temperature was highest at 12:00 on the first day . (g) The temperature dropp ed quickl y to 38 ◦ C within the first 8 hours, then stabilised at that level for the next four hours. It dropped further over the next four–hour period to 37, 4◦ C, remained at that level for four hours, and moved up to 37 , 5◦ C for the next 8 hours. Interpreting a graph such as this is more meaningful if we know that normal body temperature is 37, 5◦ C. Medical staff treating Susan would know whether the temperature change s were due to medication; they would know what the expected trend in the temperature should be, and they would be able to react accordingly.
7.2 1. Draw a histogram to represent the inform ation in Example 7.1.3 (Table 7.1.4). Use a class width of 2 ◦ C. 2. We have the following information about Sports Valley Secondary School. There are 750 children at Sport V alley Secondary School. There are 420 girls. The boys play baske tball, cricke t and soccer . The girls play basketball and netb all. There are 30 girls and 10 boys who do not play any sport. Of those who do play sport, a third of the girls play netball and the rest play basketball. There are 500 basketball player s in the school. All the boys who play sport play soccer in the winter, and either basketball or cricket in the summer. Draw a pie graph to represent this situation. (Your graph should look accurate, but you need not measu re the angles. Round calcul ations to the nearest integer.) 3. Look at the bar graph on the next page. It represents the approximate areas, in millions of square kilome tres, of some well–kno wn land masses. Use the graph to answer the followin g questions. (a) (i) What is the size of Africa in square kilometres? (ii) How much bigger is Africa than Asia? (iii) Which is the smalle st land mass represe nted by this bar chart? (iv) Which two continents are toget her just bigger than Asia ? (b) Draw a pie graph to represe nt the same information. (Your graph should look accura te, but you need not measure the angles. Round calculations to the nearest integer.)
334
Africa
Asia
Europe
North America
South America
4
8
12
16 20
24
28
32
Area (in millions of square kilometres)
4. Look at Table 7.1.9. (a) Write down the table again, including appropr iate class boundarie s. (b) Draw a histogram to represent the information. 5. Look at Figure 7.2.15. (a) How many pupils wrote the test? (b) How many girls scored 50% or more? (c) How many boys scored betw een 45% and 50%, includ ing these marks? (d) How many times did any pupil sco re 57%?
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7.3 SOME STATISTICAL MEASUREMENTS
7.3A WHAT IS AVERAGE? What do you understand by the word “av erage”? Before reading on, ask a few different people what the y understand by this term . You may find that it is a much–abused word. One source of confusion is that we use the word average to mean different things. We speak of someone being “an average cricket player”, and someone whose marks are “average”, implying mediocrity rather than any specific measurement. Can you think of an example where almos t everybody is above average? Can you think of a measure according to which everybody is exactly average? 3 Most people have ten fingers. Hence we assume that 10 is the average numbe r of fingers people have. A few people have lost fingers or been born with less than ten. Very few people have been born with more than ten fingers. The number of people with more or less than ten fingers is small, but will still make it impossible to state that “everyone has ten fingers”. Thus the average number of fingers must be a little less than ten. Hence most people hav e an above–average number of fingers. Every living perso n has exactly one head. In this respec t we are all exac tly average. Mathematically, an average is a value which represents the data in some way. For example, suppose 11 people are asked to “choose any single digit number”, and they choose the following numbers:
5, 3, 3, 4, 8, 6, 2, 3, 1, 5, 4.
The number chos en most oft en is the num ber 3. It appe ars to be the “most popular” number, and in one sense we could regard it as an average. If we arrange these numbers from smallest to biggest, we have
1, 2, 3, 3, 3, 4, 4, 5, 5, 6, 8. 3
See One Equals Zero and Other Mathematical Surprises.
336 The number “in the middle” is the number 4, and it also gives us an idea of some kind of average. We also obtain 4 if we arrange the numbers in the reverse order, i.e. from biggest to smallest. If we add up all the numbers and divide by 11, we obtain 44 = 4. 11 The last case is the kind of average most of us first think about, but it can be misleading, as the next example shows.
7.3.1 An advertisement for a job in a small business tells you that “the average salary is R84 000 per year”. Does this lead you to believe that you would earn approximately that amount if you worked there?
SOLUTION Suppose the business has 10 employees. Respectively they earn approximately R12 000, R12 000, R60 000, R60 000, R60 000, R60 000, R60 000, R70 000, R200 000 and R250 000. Although the average (obtained by adding all the amounts and dividing the total by 10) is approximately R84 000, no one actually earns that amount; only one person earns an amount of R70 000, the closest figur e to R84 000. A better “average” for this data is R60 000, since this is the amount earned by the majority of employees.
Because the word “average” can be misleading, we prefer to be more specific. All three examples we have considered (i.e. “most popular” number, number “in the middle” and “sum of numbers divided by how many numbers there are”) are important statistical measurements. They have special names so that we can avoid ambiguity. These values that describe the characteristics of data often lie centrally within the data, and they are hence called measures of central tendency.
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7.3B ARITHMETIC MEAN, MEDIAN, MODE Arithmetic mean
We use the word “arithmetic” to distinguish this mean from others, such as the geometric mean and har-
We first define the arithmetic mean. Definition 7.3.1 The arithmetic mean (or average) of a set of data is the sum of all the numbers divided by the number of numbers.
monic mean, which we do not consider in this module.
We have a useful way of expressing “ the sum of all the numbers ”. We write n
∑ is the Greek capital letter
∑
S, and we call it sigma.
i=1
Remember that the dots indicate that we include, but do not necessarily write out, all the relevant numbers.
xi ; i ∈ Z
to denote the sum of all the numbers, represented by xi , from the first number (i.e. where i = 1) to the last number (i.e. if there are 10 numbers, then we have n = 10, and ∑10 i=1 xi = x 1 + x2 + x3 + . . . + x10 .) We use the symbol x to denote the arithmetic mean. We read it as “x bar”. Definition 7.3.1 means the following.
Arithmetic mean of ungrouped data
x =
n
1
∑
n
x i ; i ∈ Z.
(7.3.1)
i= 1
For example, if x 1 = 2, x2 = 5, x3 = −1, x4 = 3 and x 5 = 1 then We say this as: x bar is equal to 1 over n times sigma xi where i is an integer from 1 to n .
x =
1 5
5
∑
xi ; i ∈ Z
i=1
1 ( x1 + x2 + x3 + x4 + x5 ) 5 1 = (2 + 5 − 1 + 3 + 1) 5 1 = (10) 5
=
= 2.
If the mean is a decimal, we can decide (if we are not told) how many decimal places we will use in the answer. Usually we will use one decimal place more than the number of decimal places in which the data are expressed.
338
7.3.2 Consider the information given in Table 7.1.4. The arithmetic mean of the midday temperatures in Gauteng in February, 1998, is x=
1 28
28
∑
xi ; i ∈ Z.
i=1
Thus using (7.3.1) we find that the mean temperature is 1 (25, 4 + 26, 3 + . . . + 33, 6 + 32, 1) 28 1 = (811, 2) 28 ≈ 28, 97
x =
≈ 29. The mean temperature is thus approximately 29 ◦ C.
When data are presented in a frequency distrib ution we no longer know what the numbers were before they were grouped into intervals. We then take the class midpoint as the representative of the class, so that if there are, for example, 4 items in a class with midpoint 16, we add 4 × 16 to the terms from which x is calculated. We thus have n
Arithmetic mean of un-
∑ mi f i
x=
grouped data
i=1 n
; i∈Z
(7.3.2)
∑ fi
i=1
where m i represents the midpoint of the i th class, and f i represents the frequency of the i th class. Study the following example.
7.3.3 Look again at the data given in Exam ple 7.2.3. In the table belo w we hav e included an additional column, containing the class midpoints. Suppose we want to calculate the arithmetic mean for this set of data. Class (cm) 151–155 156 – 160 161 – 165 166 – 170 171–175
Class boundaries Class midpoints Frequencies (cm) (mi ) (cm) ( fi) 150,5–155,5 153 5 155,5 –160,5 158 18 160,5 –165,5 163 42 165,5 –170,5 168 27 170,5–175,5 173 8
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SOLUTION Using (7.3.2) we have x =
=
n ∑i=1 mi fi n ∑i=1 fi
5 ∑i=1 mi fi 5 ∑i=1 fi
1 1 2 2 3 3 4 4 5 5 = m f +m f +m f +m f +m f f1 + f 2 + f 3 + f 4 + f 5 (5 × 153) + (18 × 158) + (42 × 163) + (27 × 168) + (8 × 173) = 5 + 18 + 42 + 27 + 8 = 163, 75 cm
and hence the approximate mean height of the children is 163,8 cm.
For interest, the next activity is given as a further illustration of the way the concept of average can be misused. We include it here, since you now have an idea of what is meant by the arithmetic mean .
7.3.1 Suppose a motorist travels from Cape Town to Johannesburg at an average speed of 120 km/h. He returns at an average speed of 130 km/h. What is his average speed for the round trip, i.e. for the whole journey from Cape Town and back again?
Did you obtain 125 km/h for your answer, by finding the arithmetic mean of 120 km/h and 130 km/h? You may be surprised to find that this is wrong. Remember that d = s × t.
From this equation we have s =
d t
.
Now speed is calculated by taking into account two different variables, namely distance and time. We thus have to calculate the average speed for the round trip
340 by means of the equation average speed for the round trip =
total distance . total time
This means that the average speed for the round trip is not the same as the arithmetic mean of the speeds for the two stage s of the trip. Since we do not know the actual distance, let the distance from Cape Town to Johannesburg be x km. Thus the total distance is 2 x km. From the equation t =
d s
we have t (Cape Town to Johannesburg) =
x
120
and
x
t (Johannesburg to Cape Town) =
Now average speed for the round trip =
130
hours hours.
total distance total time
and thus we have 2x
average speed for the round trip =
x
120
x
km/h
+ 130
i.e. we have average speed =
2x 130x+120x 120×130
= 2x ×
15 600 250x
= 124, 8 km/h i.e. the average speed is 124,8 km/ h and not 125 km/h, as you may hav e expected.
Median
The second measure of central tendency we consider is the median. (The first was the arithmetic mean)
Definition 7.3.2 The median of a set of data is the number that occurs in the middle, once we have arranged the data in ascending (or descending) order.
341 The median thus divides the data so that there are exactly the same number of items to the left of and to the right of the median.
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It is easy to select the median if we have an odd number of measurements. If we have an even number of measurements, then there is no “middle” score, and we find the median by calculating the arithmetic mean of the two middle measurements. For example, the median of the numbers 1, 3, 3, 5, 7, 8, 9 is 5 (there are three numbers to the left of 5, and three numbers to the right of 5). The median of 1, 3, 3, 5, 7, 8 is
3+5 = 4. 2 When we have data such as we had in Example 7.3.1, we see that the median is a better measure of central tendency than the arithmetic mean. When we arrange the numbers in order, we have R12 000, R12 000, R60 000, R60 000, R60 000, R60 000, R60 000, R70 000, R200 000, R250 000. The median is thus R60 000 + R60 000 = R60 000 2 which is a better indication of what a potential employee might earn than the arithmetic mean. The median is less affected by extremes in the data than the arithmetic mean, i.e. in the calculation of the median the figures R12 000, R200 000 and R250 000 play a smaller role than they do in the calculation of the mean. It is also possible to calculate the median for grouped data . The formula for this is beyond the scope of this module .
Mode
The final measure of central tendency we consider is the mode.
Definition 7.3.3 The mode of a set of numbers is the number that occurs with the greatest frequen cy, i.e. it is the most common value.
342 For example, the mode of the numbers 1, 3, 3, 5, 7, 8, 9 is 3 (3 occurs twice, whereas the other numbers all occur once). Note
The mode does not always exist (for example when there are no values repeated more than once).
The mode need not be unique (for example when sever al values occur the same number of times).
There are times when the mode is a more appropriate measure of central tendency than the arithmetic mean or the median. Have a look at the next example.
7.3.4
Suppose youold, arewould considering joini ng a where gym class consisting 40 people. you are 25 years you join a class the modal age isofclose to 25 If years, or would you join a class where the arithmetic mean of the ages is close to 25 years?
SOLUTION As we showed earlier (in Example 7.3.1) it is quite possible that the arithmetic mean does not represe nt the majority. In that example, the deviation of most of the annual incomes from the mean is large, which shows that the mean is not a reliab le measure of what you can expect to earn. In the same way , it is possible to have a group of 40 people in which most of the ages are much older or much younger than 25 years, even though the arithmetic mean of the ages is 25 years. In this case you would probabl y prefer to join a class in which the ages of most of the members are close to 25 years, i.e. a class in which the modal age is approximately 25 years, regardless of what the mean age is.
Now try the following activity.
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7.3.2 The following masses (in kilograms) of babies were recorded at a clinic. 5,6; 6,3; 5,9; 5,6; 5,2; 7,4; 7,6; 5,6; 8,7; 4,9; 6,5; 6,7; 6,9; 5,9; 5,2; 7,0; 6,8; 5,6; 6,3; 5,6; 4,4; 4,9 (a) Calculate the median of the babies’ mas ses. (b) Calculate the mode of the bab ies’ masses.
(a) We arrange the masses in numerical order , from smallest to biggest. 4,4; 4,9; 4,9; 5,2; 5,2; 5,6; 5,6; 5,6; 5,6; 5,6; 5,9; 5,9; 6,3; 6,3; 6,5; 6,7; 6,8; 6,9; 7,0; 7,4; 7,6; 8,7. There are 22 measurements; we find the two masses in the middle and calculate their arithmetic mean: 5, 9 + 5, 9 = 5, 9. 2 Hence the median is 5,9 kg. (b) From the data we see that the mass 5,6 kg occurs mos t often; hence the mode is 5,6 kg.
Mean deviation (Measure of spread)
We may end up with an inaccurate result if we round the mean to this extent, but using 90,7 will make the calculations more complicated than necessary for the purpose of explaining the concept of mean deviation.
Apart from measures of central tendency, we also have measures of spread . One of these is the range of a set of data, and we have not discussed this concept. We now consider another important measure of spread, the mean deviation. Let us look again at the raw data giv en in Activity 7.2.2. We have the masses of 50 rugby players. If we calculate the arithmetic mean of these masses, we find that it is approximately 90,7 kg. (Do this yourself to check that you have understood how to calculate the arithmetic mean.) Suppose we round this to the neare st whole number, and take the mean as approximately 91 kg. If we want to know how much an individual rugby player’s mass differs from the arithmetic mean, we calculate the difference between his mass and the arithmetic mean. We call this the deviation of his mass from the mean, and we denote it by means of xi − x; i ∈ Z .
Thus, if we consider, for exampl e, a player whose mass is 72 kg, the deviation is 72 kg − 91 kg = −19 kg .
344 However, if we consider another player whose mass is 108 kg, then the deviation is 108 kg − 91 kg = 17 kg . Since some of the answers we get will be positive and others negative, you should see that the sum of all the deviations will be zero, i.e. n
∑ (xi − x) = 0. i= 1
The deviation is a measure of how close to, or how far from, the arithmetic mean an individual score is. The mean deviation is the arithmetic mean of all the individual deviations, calculated using only the numerical values of the deviations (i.e. ignoring the negative signs ). Mathematically we “ignor e the minus signs” by using a concept called absolute value, which we will not discuss further in this module. We write Mean deviation
Mean deviation
=
1
n
∑ | xi − x | n i=1
where we interpret | x i − x | to be the numerical value of x i − x. For example, if xi − x = 3, then | xi − x |= 3; if x i − x = −5, then | x i − x |= 5. The deviation gives us an idea of howthe thedata dataare aredispersed spread about thethe mean. The mean smaller the mean deviation, the closer about mean; a large mean deviation tells us that many data items are far from the mean.
7.3.3 Calculate the mean deviation for the data given in Activity 7.2.2, using x = 91 kg.
345
Mass (kg) 72 86 79 86 96 93 101 86 96 98 105 104 88 89 89 92 79 107 104 87 97
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Deviation (kg) Mass (kg) −19 85 −5 111 −12 79 −5 82 5 84 2 94 10 87 −5 101 5 90 7 88 14 96 13 93 −3 79 −2 90 −2 95 1 83 −12 108 16 92 13 86 −4 77 6 78
92 82 91 96
1 −9 97 0 89 5 93
81
Deviation (kg) −6 20 −12 −9 −7 3 −4 10 −1 −3 5 2 −12 −1 4 −8 17 1 −5 −14 −13
6
−10 −2
2
sum of all deviations ( ignoring the minus signs ) number of players 353 = kg 50
Mean deviation =
= 7, 06 kg We consider 7,06 kg “small” in relation to the masses of the players.
Since the mean deviati on is small, we may conclude that the data is spread fairly close to the arithmetic mean.
346
7.3 1. Refer to Activity 7.2.2. (a) Determine the median. (b) Find the mode. 2. Refer to Figure 7.2.15. (a) Calculate the arithm etic mean of the boys’ marks and of the girls ’ marks (to the nearest integer). (b) What is the median of the girls’ m arks? (c) What is the mode of the girls’ marks? (d) Calculate the mean deviation of both sets of marks (i.e. the girls’ marks and the boys’ marks) . (Give your answer to the nearest integer.) 3. Refer to Activity 7.2.4. (a) Calculate the arithmetic mean of Susan’s tempe ratures measured while she was in hospital. (b) What is the median for this data? 4. The monthly salar ies earned by eleven emplo yees of Company SQS are approximately (in rands): 5 100, 5 500, 5 900, 6 200, 6 500, 8 300, 8 400, 7 200, 6 800, 12 000 and 20 000. (a) Would the arithm etic mean , the media n or the mode give you the best idea of what salary you could expect if you were an employee of SQS? (b) Calculate the arithmetic mean (correct to the nearest ten). (c) What is the mode? (d) What is the median?
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7.4 PROBABILITY
7.4A A QUICK LOOK AT ELEMENTARY PROBABILITY THEORY You may know the meaning of these words: random, probable, possible, and chance. If you are not quite sure, look them up in a dictionary, then consider the discussion that follows.
According to one dictionary, we have: chance
the way things happen through no known cause luck arisk possibility, or likelihood
possible
capable of existing, or of happening, or of being done
probable
likely to exist, or to happen, or to be done
random
happening, existing, or being done, in no particular order
You can see that these words are quite similar in meaning, and we use them often, without necessarily really being sure what we mean. In our everyday language there are also other words and phrases which are ambiguous. Suppose several people were asked to assess how probable certain events were, on a scale of 1 to 10 (where 1 means almost no chance and 10 means almost certain), if the events were described by the following words or phrases: doubtful perhaps it could be nearly certain small chance high chance one can expect unlikely chances are not great most likely likely reasonable chance We would find that there would be a great variation in their interpretations.
4
It is useful to consider briefly the difference between unusual events, i.e. events that are outside our field of experience, and unlikely events, i.e. events with a low 4
See Smith: Agnesi to Zeno.
348 probability. We are influenced by the media to imagine that conspicuous or memorable events occur often, and are thus more probable than others which occur more frequently but are less newsworthy. We also need to distinguish between possibility and probability. It is possible that all students studying MAT011–K will obtain 100% in their exams, but it is not probable that this will happen! We use the word event to describe the particular result we are interested in, during an experiment or activity in which a variety of results, called outcomes, are possible.
In this example the event we are looking at is achieving one particular result, namely obtaining a 3 when we throw the die.
Probability theory deals with events that happen randomly, or by chance. Have you ever played a game involving the throwing of dice? (Incidentally, we speak about one die , and several dice.) In a game you may want to obtain a particular number, such as 3, when you throw a die. What is the likelihood that you will obtain a 3? Usually a die is in the shape of a cube, with 6 sides or faces. Each face has a different number of dots which represent the integers from 1 to 6. If we throw a die, there should be an equal chance, i.e. one chance out of six , for any one of these faces to finish upper most. Since we have a one in six chance of obtaining a 3 when we throw the die, we have a five in six chance of not obtaining a 3. Pascal was a French mathematician, who lived at about the same time as Descartes. During 1654 Pascal and another mathematician, Pierre Fermat, worked together on problems dealing with possible events and probability theory was born. Intuitively we recognise that if we throw a die exactly six times, we may not obtain the number we want, but if we throw the die enough times we expect that one out of every six throws should be the number we want. But what do we mean by “enough times”? If you have a die, try the following activity.
7.4.1 Throw a die 24 times, and record the numbers you obtain. How many times did you obtain 4? Calculate the fracti on you get when you divide the numbe r of times you obtained 4 by the total number of throws, i.e. 24.
YOUR ANSWER (Obviously each person’s answer will be different.) Fill in the blank spaces in Table 7.4.1 and in the statements that follow, according to the answers you obtain.
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We use the word “score” to mean the result we obtain.
Throw 1 3 5 7 9 11 13 15 17 19 21 23
Score
Throw 2 4 6 8 10 12 14 16 18 20 22 24
Score
Table 7.4.1
Number of times you score the number 4 = ....... . (We call this the frequency with which the number 4 was obtained.) Number of 4 s
= ...... .
24 (We call this the relative frequency of scoring 4, i.e. the number of times 4 was scored, divided by the total number of throws (24).)
350 Note that when we calculate Relative frequency
frequency of a particular event total number of events we always obtain a fraction between 0 and 1 (including 0 and 1), such as 16 . The more certain a give n event is, the closer the fraction approxim ates 1. When an event is impossible we recognise intuitively that we can expect the fraction to be 0.
7.4.2 If we can assign “probability values” between 0 and 1 (including 0 and 1) to various events, what value would you give to each of the following? (a) The sun will set in the west tomor row. (b) A coin will land heads up when it is spun. (c) If I put my hand into a packet con taining equal numbers of red and gre en sweets, I will take out a red sweet. (d) An apple falling out of an apple tree will stop in mid–air wit hout anyone or anything touching it. (e) I will score 3 on an y throw of a normal die.
(a)1
(b)
(c)
1 2
(e)
1 6
1 2
(See the discussion below.)
(d) 0
Before you began Activity 7.4.1 you may have expected to obtain the number four, four times. Because the chance of obtaining a four is 16 , if you throw the die twenty four times, you might expect to obtain a four 16 × 24 times, i.e. four times. Some of you may have obtained this resul t. However, when we carry out the experiment we see that we may obtain the number four more (or perhaps less) times than we expec ted. We see that the “one in six” chance will only becom e
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evident after a large number of throws. The more we throw the die, the closer we get to scoring a four on one sixth of the throws. Spinning a coin
We will have the same experience when we spin a coin. In this case there are only two outcom es: we call thes e “heads” or “tails”. We thus have a “one in two chance”, or a “fifty–fifty chance” of obtaining either heads or tails when we spin a coin. We realise that it is possible to spin the coin six time s, and obtain heads five times, and tails only once, but in the long run we should have an equal chance of obtaining either heads or tails. The fact that we can only determine the probability of an event after a significantly large number of trials highlights two points.
We should never have undue confidence in small samples.
If conclusions about a population as a whole are to be made on the basis of an experiment involving a small sample, it is essential that the sample is representative of the population and that correct statistical technique s are used in generalising the results obtained.
7.4.1 Suppose we throw two dice and add the two numbers that we obtain . For convenience we refer to the two different dice as a red die and a green die. The following table gives us all the possible totals we can obtain, and how these totals can be achieved. 5
1 2 Red 3 die 4 5 6
Green die 1 2 3 4 5 6 2 3 4 5 6 7 3 4 5 6 7 8 4 5 6 7 8 9 5 6 7 8 9 10 678 91011 7 8 9 10 11 12 Table 7.4.2
We find the possible totals from the table. For example, if we throw a 3 with the red die, and a 4 with the green die, then we have a total of seven. In the table we find this number by finding the place where the “3” row and the “4” column cross each other. The table shows that if you throw the two dice, there is for example only one way of obtaining a total of two, although there are, for example, three ways of obtaining a total of four. (We can only obtain two by throwing 1 twice; 5
See Smith: Agnesi to Zeno.
352 we can obtain four by throwing 1 and 3, or 2 and 2, or 3 and 1.) Note that the events do not all occur in the same number of ways : the event “obtaining a total of two” can only occur in one way; the event “obtaining a total of four” can occur in three different ways. From the table, determine which event is more likely: obtaining a total of seven or a total of eight?
SOLUTION There are six ways of obtaining a total of even, but only five ways of obtaining a total of eight, so it is more likely that we will obtain a total of seven.
Outcomes, sample spaces and events
We have seen that there are six possible outcomes when we throw a die: we can only score 1, 2, 3, 4, 5 or 6. In any statistical experiment, an outcome is one of several possible results. The sample space is the set of all possible outcomes of the experiment. Thus in this case, the sample space is { 1, 2, 3, 4, 5, 6}. If we are particularly interested in obtaining a 4, we call “obtaining a 4” an event. If we denote a sample space by S , and an event by E , then the number of outcomes in the sample space S is denoted by n(S), and the number of outcomes in the event E is denoted by n(E). We use this notation in the next example.
7.4.2 Drawing a coloured ball
I have a box containing 3 red balls, 4 blue balls and 6 green balls. There are thus 13 balls altogether. Suppose I want to determine the chance of taking out a green ball when I take out a ball without looking inside the box.
SOLUTION The sample space can be represented by the set S = { r1 , r2 , r 3 , b 1 , b 2 , b 3 , b 4 , g 1 , g 2 , g 3 , g 4 , g 5 , g 6 }
where r1 represents the first red ball, g3 represents the third green ball, and so on. There are thus 13 possib le outcom es, and the sample space consi sts of 13 elements. Thus n (S) = 13. If I am particularly interested in the event E which I describe as “taking out a red ball”, then the set E = {r1 , r2 , r3 } Revise the meaning of ⊂ in Topic 1 of Book 1.
consists of the three elements in the sample space that make up the event in this particular experiment. We have n (E ) = 3. Note that E ⊂ S .
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7.4.3 Remember that there are 13 hearts, 13 spades, 13 clubs and 13 diamonds in a pack of cards; we exclude the jokers in this case.
You have a pack of 52 cards. Consider the following two ev ents: obtaining hearts, or obtaining a diamond picture card, each time you draw a card from the pack .
(a) What is the sample space S for this experiment? Write down n (S). (b) Assume an even t H is described as draw ing a card that is a heart. How many elements are there in H ?
Remember that a Venn diagram involves drawing shapes to represent the sets you are co nsidering. See Topic 1 of Book 1.
(c) Assume that the event E is described as drawing a diamond picture card. (A picture card is any one of ace, king, queen or jack.) Write down the set E . How many elements are there in E ? (d) Draw a Venn diagram to repre sent S , H and E .
(a) S = {H1 ,..., H13 , D1 ,..., D13 , S1 ,..., S13 , C 1 ,..., C13 } where D i represents H
S
C
i represents a heart; i represents a spade and a diamond; club; and i is any integer from 1 to 13. We have n (S) = 52.
i
represents a
(b) Since H = {H1 , ... H13 } there are 13 elements in H , i.e. n (H ) = 13. (c) E = {a, k, q, j } (We have used a to denote an ace, k to denote a king, q to denote a queen and j to denote a jack (of diamonds).) There are 4 elements in E , i.e. n (E ) = 4. (d)
H S E
Disjoint sets have no elements in common.
We see that E and H are disjoint sets.
354 We explain “relative frequency” in the answer to Activity 7.4.1.
We denote the frequency of the event E by F (E ); the relative frequency after n trials is thus F (nE ) .
Up to now we have considered the relative frequency of specific events that are measurable a large number of times (even if we do not actually measure them that many times). Figure 7.4.3 shows that initially the relative frequency of the occurrence of an event E may vary considerably from the probable long–term result, but over time the relative frequency tends towards a particular number, which we denote by P (E ).
Relative frequency of E
y = P(E)
F(E) ____ n
10
100
1000
10000
100000
1000000
Number of trials (n)
Figure 7.4.3
In Figure 7.4.3 we note the following.
The number of attempts or trials indicated on the horizontal axis is not drawn to scale.
Each dot represents the relative frequency of obtaining the event particular number of trials (or attempts), n , i.e. each dot represents
As n becomes larger, so the number ber P (E ) and 0 ≤ P(E ) ≤ 1.
F (E ) n
E in a F (E )
.
n
gets closer to a particular num-
The horizontal line is defined by y = P(E ).
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Definition 7.4.1 Suppose we have an experiment in which there is a set S of equally likely possible outcomes and we are interested in the occurrence of an event E which is made up of specific outcomes. The probability of the event E is given by the formula P (E) =
n (E) n (S)
,
(7.4.1)
where n (E ) denotes the number of outcomes in the event E , and n(S) denotes the total number of possible outcome s.
In terms of this definition, we note that in Figure 7.4.3, as F (E ) n
using
gets closer to
n( E ) n( S )
n becomes larger,
. This definition of the probability P (E ) of an event E ,
n(E )
over a large number of trials, is called the empirical probability (also n(S) the experimental probability) of the event E . A more accurate formula for probability allows the number of trials to “tend to infinity”, which is a mathematical way of expressing that the number of trials is theoretically allowed to become indefinitely large. Calculations of P(E ) then involve the concept of limits, which we do not study in this module. When we incorporate the concept of limit we obtain a better definition of the probability P (E ). At this stage, we approximate P(E ) by means of the formula (7.4.1). We can put this another way. Suppose an event E can happen in n (E ) different ways, and that n(S) equally likely outcomes are possible. Then the probability of the occurrence of E (i.e. the chance of success) is found by calculating
n(E ) n(S)
.
There are various Probability Laws that follow from the definition of probability. We state them here with out proof, since you are not required to prove them. Law 1 Law 2 Law 3
If the probability of an event E occurring is P (E ), then the probability of E not occurring is 1 − P(E ). If A and B are two events that cannot occur at the same time, then the probability of either A or B occurring is P (A) + P(B). If A and B are two independent events (i.e. the events do not affect each other in any way) then the probability that both A and B occur is given by P (A) × P(B).
Although we do not prove these laws it is a good idea that you verify for yourself that they are valid. Try to do this in the next activity.
356
7.4.4
A pack of cards
(a) Take a pack of cards (without any joker s). What is the probabilit y of not getting a spade? (b) What is the probabil ity of getting either a heart or a diamond ?
A picture card is any ace, king, queen or jack.
(c) What is the probabil ity of getting a red picture card, i.e. wh at is the probability of obtaining a card that is both red and a picture card?
(a) There are 52 card s in the pack. If we do not want to get a spad e, we need to get either a heart, diamond or club. There are 13 of each type of cards, hence there are 39 cards that are not spades. Hence P ( not a spade ) =
Now
39 3 = . 52 4 13
3
1 − P (spade) = 1 − 52 = 4 . Hence P (not spade ) = 1 − P ( spade).
This illustrates Law 1. (b) There are 13 hearts and 13 diamonds in the pack, and henc e a total of 26 red cards. Thus P ( heart or diamond ) =
26 1 = . 52 2
We also have P(heart) + P(diamond) =
Thus
13 13 1 1 1 + = + = . 52 52 4 4 2
P(heart or diamond ) = P(heart) + P(diamond).
This illustrates Law 2. The different types of cards (i.e. hearts, diamonds, spades and clubs) are called “suites”.
(c) There are 16 picture cards in the pack (i.e. 4 in each su ite). Half the picture cards are red, and the other half are blac k. There are thus 8 red picture cards in the pack. Hence P(red picture card ) =
8 2 = . 52 13
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Also P(red card ) × P(picture card ) =
26 16 1 4 2 × = × = . 52 52 2 13 13
Thus P(red card and picture card ) = P(red card ) × P(picture card ).
This illustrates Law 3.
7.4.5 If you throw a die, calculate the following. (a) The probability of scoring a 5. (b) The probability of obtaining an even number. (c) The probability that you will not score a 4.
(a) Sample space S = {1, 2, 3, 4, 5, 6}. Thus n(S) = 6, i.e. there are 6 elements in S , since there are 6 possible outcomes when a die is thrown. We are considering the event “throwing a 5”. We have E = {5}, and n (E ) = 1 (since there is only 1 element in E ). Thus P (E ) =
number of ways of obtaining 5 n(E ) 1 = = . number of possible outcomes n( S ) 6
(b) Sample space S = {1, 2, 3, 4, 5, 6}. Thus n (S) = 6. The event “getting an even number” can be denoted by E = {2, 4, 6}, and hence n (E ) = 3. Thus P(E ) =
=
number of ways of obtaining an even number number of possible outcomes n(E ) n(S)
3 = 6 1 = . 2
358 (c) Sample space S = {1, 2, 3, 4, 5, 6}. Thus n (S) = 6. In order not to score 4, you need to score 1, 2, 3, 5 or 6. Thus the event “not scoring 4” can be denoted by E = {1, 2, 3, 5, 6}; hence n (E ) = 5. Thus P (E ) =
n( E ) n(S)
=
5 . 6
Consolidate your understanding of probability by considering the next example.
7.4.3
Spinning a coin
(a) Spin a coin 10 times and record the numb er of heads (H), and the number of tails (T), you obtain. What is the relative frequency of the event “coin lands heads uppermost”? What is the probability of this event? (b) What is the probability of obtain ing at least two head s if you spin a coin three times.
SOLUTION (a) Each person will obtain a different answ er for the first part of this question. Suppose one answer is the following.
Throw 1 2 3 4 5 Result H H H T T H
6 7891 TTT
0 T 4
The relative frequency of obtaining heads in this case is 10 . However, the sample space for this event is S = {H, T }, and n (S) = 2. The event is E = {H}, and hence n (E ) = 1. Thus P(E ) =
n(E ) n(S)
1 2
= .
This will be true regardless of the answer you obtained for the first part of this question.
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H H H T
T H T H
H T T
T H T
Spin:
1
st
2
nd
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(b) When you spin a coin three tim es there are several possible outcomes. In cases like these it is helpful to draw a tree diagram like that in the margin. We see that S = {(H,H,H), (H,H,T), (H,T,H), (H,T,T), (T,H,H), (T,H,T), (T,T,H), (T,T,T)}
and hence n (S) = 8. The event “obtain at least two heads” means “obtain two heads or three heads”. Thus E = {(H,H,H), (H,H,T), (H,T,H), (T,H,H)} and thus n (E ) = 4.
We thus have P(at least two heads ) =
rd
n(E ) n(S)
=
4 1 = . 8 2
Today probability theory is the foundation on which much of statistics is based. It plays an important role in many fields, for example medicine, genetics, and in determining the premiums that must be paid on insurance policies.
7.4 1. There are 5 blue pens, 3 red pens and 4 green pens in a closed box. Su ppose you want to work out the probability (according to Definition 7.4.1) of taking out a red pen without first looking at the pens. (a) What is the sample spa ce S in this case? Give your answer as a set, and write down n (S), the number of outcomes in S . (b) Write down the event E “taking out a red pen”, in terms of a set. Write down n (E ), the number of elements in E . (c) Calculate
n(E ) n(S)
.
(d) What does the answer in (c) represent? 2. Spin a coin 50 times. (a) Each time you spin the coin, record wh ether it lands with heads uppermost (H) or tails uppermost (T).
360 (b) Calculate the relative frequency of
heads (i.e. count the number of times you obtained H and write number of heads down ) 50 tails ( i.e. count the number of times you obtained T and write number of tails down ). 50
(c) Draw a graph plotting number of heads number of spins against 5 spins, 10 spins, 15 spins and so on (at intervals of 5 spins, up to 50 spins). This means, for example, that the first point on your graph will represent the fraction number of heads obtained after 5 spins . 5 (d) What is the differen ce between the first answer in (b) (i.e. the relati ve frequency of achieving heads) and the answer you would expect to get? 3. In a card game a player dra ws five red cards in a row from a pack of wel l– shuffled cards from which the jokers have been remo ved. Will the next card more likely be black, or more likely be red? 4. Suppose you are playing a card game in which all 52 cards (not the jokers) in the pack are dealt out. Calculate the probability (a) of being dealt any ace (b) of being dealt the ace of diamonds. 5.
(a) What is the probability of obtaining at least one head, if you spin a coin twice? (b) What is the proba bility of obtaining at most two tails, if you spin a coin three times?
6. A ball is taken at random (i.e. wit hout looking into the box) out of a box containing yellow balls, 4 white balls and 5 blue balls. What is the probability 6that (a) the ball will be yellow (b) the ball will not be white? 7. Refer to Example 7.4.1. Use one of the probability law s to determine (a) the probability of throwing 7 twice (b) the probability of throwing 8 and then 6.
361
MAT0511/003
We use examples to illustrate the relevant terminology.
• Raw data: represent either a sample or a total population. The following table gives discrete data. 24 28 21 20 26 29 25 18 21 24 27 26 28 27 22 26 25 21 21 21 24
• We use tally marks to determine the frequency (the number of times a particular number occurs) of each score. Score Tally 18 | 19 20 | 21 | | | | 22 |
23 24 25 26 27 28 29
||| || ||| || || |
Frequency 1 0 1 5 1
0 3 2 3 2 2 1
• Classes We can group the data into four classes, each with class width 3. Note that this is an arbitrary arrangement; we can just as easily group the data into six classes, with class width 2.
• Class boundaries and class midpoints Class boundaries are necessary when we draw histograms, and class midpoints are used to draw frequency polygons, and in various statistical calculations. We find the boundaries by extending the class limits by one decimal place so that the upper boundary of one class is the lower boundary of the next class. When the class midpoint is not obvious, it is calculated by adding the lower and upper class boundaries (or class limits), and dividing the answer by 2. We thus have the following table.
362 Classes
18,19,20 21,22,23 24,25,26 27,28,29
Class limits lower upper 18 20 21 23 24 26 27 29
Class widths
3 3 3 3
17 20 23 26
Class boundaries lower upper ,5 20,5 ,5 23,5 ,5 26,5 ,5 29,5
Class midpoints
19 22 25 28
• Frequency table or frequency distribution From the raw data we have the following frequency distribution. x 18 19 20 21 22 23 24 25 26 27 28 29 n 101 510 323 221
• Grouped frequency distribution Class 18–20 21–23 24–26 27–29
Frequency 2 6 8 5
• Pictographs The following pictograph illustrates the number of soccer players and the number of cricket players at four schools, where
represents 50 soccer players, and
represents 50 cricket players.
School A
School B
School C
School D
• Pie graphs 100 consumers make the following purchases: 50 people buy tea, 40 people buy coffee, 10 people buy cooldrink. We use the concept of central angle to detemine the sizes of the sectors of the circle:
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Since one revolution measures 360 ◦ , each item is represented by a sector of a circle in which the central angle is proportional to the percentage represented by the sector.
Product
Number
Tea
50
Coffee
40
Cooldrink
10
Fraction 50 100
Percentage
40 100 10 100
40%
Central angle 50 × 360◦ = 180 ◦ 100
50%
40 × 360◦ = 144 ◦ 100 10 × 360◦ = 36◦ 100
10%
This information gives us the following pie graph.
Tea 50%
10% Cooldrink
40% Coffee
• Histogram and Frequency Polygon We use the concept of class boundary. If we use the data given for the example at the beginning of this summary, we have the following histogram and frequency polygon. 10 8
8 Frequency
6
6
5
4 2
2
17,5
19
20,5
22
25
23,5
Class
28
26,5
29,5
364
In the histogram the area of each rectangle is proport ional to the frequency it represents.
In the frequency polygon, the height of each point corresponding to the class midpoint is proportional to the class frequency it represents.
• Stem–and–leaf plots Consider the raw data given in the following table. 26 28 34 33 39 31 22 41 18 21 27 35 16 43 45 49 36 37 30 19 18 40 41 45 We have the following plot. 1
8 6 9 8
2
6 8217
3
4391
4
1359015
5 0 6 7
• Line graphs Often used to show trends with respect to change over time.
5 Production costs
4
(in millions of Rands)
3 2 1
1990
1991
1992
1993
1994
1995
1996
Year
• Measures of central tendency
Arithmetic mean of raw data (calculated with or without a calculator) 1 n x = ∑ xi ; i ∈ Z n i=1
where n is the number of items. Refer to the raw data given at the beginning of this summary. For this data we have x = 24.
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Arithmetic mean of grouped data (calculated with or without a calculator) n mi × f i ∑ x = i=1n ; i∈Z ∑i=1 fi
where mi is the midpoint of the i th class f i is the frequency of the i th class n is the number of classes
The median of a set of numbers (arranged in ascending or descending order) is the number that – occurs in the middle (if we have an odd number of numbers) For example: 1, 3, 8, 8, 9, 9, 11: median = 8
↑ – is obtain ed by finding the arithm etic mean of the two middle numbers (if we have an even number of numbers) For example: 1, 3, 8, 8, 9 , 9, 11, 13
median =
8+9 = 8, 5 2
The mode of a set of numbers (if it exists) is the number that occurs most often. For example: 1, 3, 7, 8, 9, 9 , 11, 13
The number 9 occurs twice: there is one mode, 9. 1, 3, 8, 8 , 9, 9 , 11, 13
The numbers 8 and 9 occur twice: there are two modes, 8 and 9. 1, 3, 7, 8, 9 10 , 11, 12 No number occurs more than any other: there is no mode.
• Measures of spread
Mean deviation We calculate the arithmetic mean of the positive difference between the individual scores (xi ) and the arithmetic mean (x) of all the scores.
366
• Outcomes, sample spaces and events Consider a square piece of cardboard divided into four triangles, numbered 1, 2, 3 and 4. Assume a pointe d stick (e.g. a penci l) passes through the centre of the square.
When the pencil is placed on its point and spun, the carboard turns round several times, before coming to rest on one of the triangles.
Suppose we carry out an experiment to find out how many times the square lands on a particular number. The sample space S is the set of all possible outcomes of the experiment. The square can land on the numbers 1, 2, 3 or 4, hence S = {1, 2, 3, 4}, and n (S) = 4. An event E is a particular result. If we are interested in the number of times the square lands on 2, then E = {2} and n (E ) = 1 . Suppose we conduct an experimen t to find out how many times the square lands on a particular type of number, for example an odd number. The event is now E = {1, 3}, and n (E ) = 2. Outcomes and events do not all occur in the same number of ways. Suppose we consider two consecutive spins, and the event “obtain a total of 5 on two consecutive spins”.
367
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First spin 12 3 4 12 3 4 5 Second 23 4 5 6 spin 3 4 5 6 7 45 6 7 8
There are 16 possible outcomes (totals of the two scores) and 4 possible ways of achieving the event “obtain a total of 5”. Thus S = {(1, 1), ( 1, 2), (1, 3), (1, 4), (2, 1), (2, 2), ( 2, 3), (2, 4), (3, 1), (3, 2),
(3, 3), (3, 4), (4, 1), (4, 2), (4, 3), (4, 4)} i.e. n(S) = 16
and E = {(1, 4), (2, 3), (3, 2), (4, 1)}
i.e. n(E ) = 4.
• Empirical probability
The probability of an event E occurring in an experiment in which there is a set S of equally likely outcomes, is given by the formula P(E ) =
n(E ) n(S)
.
• Probability Laws
If P (E ) represents the probability of event E , then the probability of “not E ” is 1 − P(E ).
If A and B are two events that cannot occur at the same time, then the probability of obtaining either A or B is P (A) + P(B).
If A and B are two independent events, then the probability of obtaining both A and B is P (A) × P(B).
368
CHECKLIST Now check that you can do the following. SECTION 7.1
1. Use the terminology of statistics: data (discrete and continuous), population, sample, tally, class, class limit, class width, class boundary, frequency, frequency distribution, grouped frequency distribution. Examples 7.1.2, 7.1.3, 7.1.4, 7.1.5; Activity 7.1.1 2. Organise data in tables in a meaningf ul way. Examples 7.1.2, 7.1.3, 7.1.4, 7.1.5; Activity 7.1.1 SECTION 7.2
1. Use pie graphs to represent data. Definition 7.2.1; Example 7.2.2; Activity 7.2.1 2. Use histog rams and freque ncy polygo ns to represent data (this includ es application of the terminology class midpoint and class boundary). Definition 7.2.2; Example 7.2.3, 7.2.4; Activities 7.2.2, 7.2.3 3. Use stem–and–leaf plots to represent data. Example 7.2.5 4. Use line graphs to represent data. Definition 7.2.4; Example 7.2.6; Activity 7.2.4 SECTION 7.3
1. Determine the arithm etic mean of a set of data (gro uped or raw , with or without a calculator). Definition 7.3.1; Examples 7.3.2, 7.3.3; Activities 7.3.1, 7.3.2 2. Determine the median of a set of data. Definition 7.3.2; Example 7.3.4; Activity 7.3.3 3. Determine the mode(s) (if any) of a set of data . Definition 7.3.3; Example 7.3.4; Activity 7.3.3 4. Determine the mean deviation for a set of dat a. Activity 7.3.4
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SECTION 7.4
1. Apply the termino logy of probability : outcome, sample space, ev ent. Examples 7.4.1, 7.4.2, 7.4.3; Activity 7.4.3 2. Calculate the empirical probab ility of a given event . Definition 7.4.1; Probability Laws 1, 2 and 3; Example 7.4.3; Activities 7.4.2, 7.4.4, 7.4.5
370
ANSWERS TOPIC 1 Exercise 1.1 1.
Temperature o (in C)
(a)
24 22 20 18 16 14 12 10 8 6 4 2 0 Jan
Feb
Mar Apr May Jun
Jul
Aug Sept Oct Nov Dec
Month
(b) This reflects a place that has its lowest average daily temperatures between April and September, which are winter months in the southern hemisphere. We can conclude that this place is thus in the southern hemisphere. 2.
(a) The time between 10:00 and 9:00 is one hour. In the first hour they travelled 100 km. (b) They made no progress between 10:30 and 11:00 (the graph shows that they were stationary after they had travelled for 1 12 hours, and they only began travelling again 2 hours after their departure). Thus they spent 12 an hour changing the tyre. (c) 300 km (since the journey begins at 0 km and ends at 300 km). (d) They reached the 150 km stage 1 i.e. at 10:30.
1 2
hours after leaving Polokwane,
(e) The second stage during whic h they made no progress was from 3 hours after departure to 3 34 hours after departure. They spent 34 of an hour filling up with petrol and having something to drink. (f) They reached Pretoria 4 hours after leaving Polokwane, i.e. at 13:00.
371 3.
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(a) The dot at 0 represents the info rmation that at the be ginning (i.e. before the car begins to move) it has not travelled any distance. (b) 15 km (c) 35 km (d) faster (e) after 4 seconds the graph begins to look like a straig ht line; i.e. we can assume that speed is constant from this stage
Exercise 1.2
1.
(a) b (b) (0, a ) (c) x
≥b
(d) 0 < x < b (e) If P = ( p, q) satisfies the equation y = x+k m + l, then it lies on the graph; i.e. if we substitute x = p into the right hand side of y = x+k m + l and obtain q as the answer, then ( p, q) lies on the graph. (f) We need to solve x+k m + l > 0. 2.
k
We know y > 0 when x < b, thus x+m + l > 0 for all x such that x < b. (a) Yes. The equation is defined for all x 1. The x coordinate is 3 and we have 3 > 1. Also, ( 3, 7) satisfies the equation, since we obtain y = 7 when we substitute x = 3 into the equation.
−
≥−
(b) No, since the equation is only defined for x (c)
−1 < x ≤ 0
−
≥ −1, and −2 < −1.
(d) 1 3.
(a) (i) D (iv) B
(ii) E (v) C
(iii) A
(b) 5 2 + 02 = 25 = 16. Since the coord inates (5, 0) do not satisfy the equation of the circle, the point does not lie on the circle.
(c) If the coordinates of B are ( a, b) and if a2 + b2 = 16, then B lies on the circle. (d) 4 units (the radius) (e) 8 units the diameter of the circle (f) 4 units the radius of the circle
372 Exercise 1.3
1. Let the length of the side be x units. Then x2 + 42 = 10 2 , i.e. x2 = 84. Thus the length of the third side is 84 units. (We may write this as 2 21 units).
√
2.
(a) A sketch shows that
√
ABC is a right triangle, with right angle at C.
(b) Using the distance formula we find that b = AC = 3 units, a = BC = 2 units and c = AB = 13 units. If a 2 = b 2 + c2
√
or
b 2 = a 2 + c2 or
c 2 = a 2 + b2 then
ABC is a right triangle.
If ABC is a right triangle we expect that the longest side (in this case AB, i.e. c) will be the hypo tenuse. We thus nee d to find out whether c 2 = a 2 + b2 . Now LHS =
√
2
= 13. RHS = 2 + 3 = 4 + 9 = 13. 13
2
Hence LH S = RH S and thus
2
ABC is a right triangle.
3. If we sketch ABC we find that the longest side is AB . We thus need to determine whether AB2 = AC2 + BC2 .
√
By the distance formula we have AB = 20, AC = we substitute into the above equation we have
√10 and BC = √2. If
LHS = AB2 = 20 RHS = AC 2 + BC2 = 10 + 2 = 12 = 20.
Since LH S = RH S it follows that
ABC is not a right triangle.
373 4.
5.
√13, √ (b) d (P, Q) = 5, √ (c) d (P, Q) = 2 13,
M (x, y) = ( 2, 11 ) 2
(a) d (P, Q) =
2
− 1)
+ (y
2
(c) (x + 2) + (y (d) x2 + (y
2
− 2)
=9
3)2 = 12
− 1) −= 1 2
(a) C = ( 0, 0);
r=
−
(b) C = ( 1, 2); 7.
3 2
− , −2) M (x, y) = ( −1, 1) M (x, y) = (
(a) x2 + y2 = 2 (b) (x
6.
MAT0511/003
√3
r=6
(a) C = ( 2, 3)
√10 (c) ( x − 2) + (y − 3) = 10 √ 8. d (A, B) = 13 √13 = . Hence radius = M (x, y) = ( , −2) is the centre of the circle. Thus the equation of the circle (b) d (C , A) = 2
2
1 2
13 4
1 2
is
(x
1
)2 + (y + 2)2 =
−2
which we may write as 4(x
13 4
− 12 )
2
+ 4(y + 2)2 = 13.
TOPIC 2 Exercise 2.1 1. y = h (x) = 3x, x 2.
∈N
(a) Each value of x is associated with a unique value of y . y = f (x) = 13 ( 2x + 2)
−
(b) Each value of y is associated with a unique value of x . x = f (y) = 12 ( 3y + 2)
−
3.
(a) D f = N0 (b)
x 0 1 23 4 5 y 3 4 7 12 19 28
(c) The dots representing ( 0, 3), (1, 4), (2, 7), (3, 12), (4, 19), (5, 28) cannot be joined, since D f = R.
4.
(a)
R
374
−{1} (c) {x ∈ R : x ≥ 0}
(b)
R
(d)
R
1 2 1 2
{x ∈ R : x ≥ } (f) {x ∈ R : x > }
(e)
5.
(g)
R
(h) (i)
{x ∈ R : x ≤ −√3 or x ≥ √3} R
(a) 4 (b) 1 (c)
−3x
2
+1
(d) 3 x + 1 6.
(a) (b)
1 3 5 3
√x − 2x − 2 √x 3x √x +− 1 1−a 1+a √2 ≈ 1, 41 (i) 2
(c)
(d) (e) 7.
(a)
(ii)
(b) 8.
1 3
{x ∈ R :
− 1, and D f = Dg = R. f = g. We have f (x) = g(x) = x − 1, but D f = R −{−1} = D g = R.
Exercise 2.2 (a) 3 (b) 7
√
(c) 15 2 (d) 2.
1 2
−}
(a) f = g , since f (x) = g (x) = x 2 (b)
1.
x>
35 2
(a) 5 (b) 7 23 (c)
7 2
375 (d) 3.
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1 44
(a) x2 + 4x
−5 − 15 (c) (x + 5) (x − 2) (d) x − 2 (e) x + 6x − 12x − 70x + 75 (f) x + 6x − 10x − 50x + 125 (b) x2 + 2x
2
(g)
3
2
4
3
2
R
−{−5} √ (a) x + 3x − 2 √ (b) x − 3x + 2 √ (c) x(3x − 2) √ (h)
4.
4
(d)
R
x
3x
−2 2
−9x + 13x − 4 (f) 9 x − 11x + 4 (g) {x ∈ R : x ≥ 0} R (h) {x ∈ : x ≥ 0}−{ } √ (a) x + 3x − 2 √ (b) x − 3x − 2 √ (c) x 3x − 2 x (d) √ 3x − 2 (e) x − 3x + 2 (f) x + 3x − 2 (g) {x ∈ R : x ≥ } (h) {x ∈ R : x > } x − 2x + 2 (a) (e)
2
2 3
5.
2 2
2 3 2 3
4
6.
2
x2
1
− −
x2 (x2 2) (b) x2 1 (c) 1 (d) (x2
2
− 1) (x − 1) − 1 (x − 1) (x − 1) + 1 (x − 1) 2
(e)
4
2
2
(f)
2
“Simplifying” this does not really produce a simpler expression.
4
2
2
Once again we do not try to simplify this further.
376
−{−1, 1} (h) R −{−1, 1} (g)
R
TOPIC 3 Exercise 3.1 1.
y
0
x
_1 3
–1
2. y 2
0
3.
x 1
y
2
0
x 4
377 4.
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y
0
x
3
– _32
5.
3 2
6. No slope 7.
−2
8. 0 9.
y 3
10.
x
1
–1
y
(2,3)
0
x
378 11.
y
1
12.
x
0
–7
–2
(1 , – 2 )
(2,0)
5
y
2
0
13.
x
y
y = f ( x) = 1
1
x 0
y y = i ( x) = x (1,1) 0
x
379
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14.
y l2
l1
l5
l4
l3
1
x 1
Slope of l 3 is zero. 15.
(a) The slope of l 2 , i.e. (b) The slope of l3 , i.e. slope of l 2 .
−
5 , is the same as that of 2
2 , 5
l1.
is the negative reciprocal of
−
5 2
which is the
16. p = 4 17. q = 0 18.
y l3 (1,2)
l2 –1
– _12
(– 1 , – 2 )
19. l2 : 2y + 4x + 1 = 0
⇒l
2
: y=
x
l1
−2x −
1 2
1
l2 is obtained by shifting l 1 downwards by 3 2 units. 20. The line l1 represents a function since each value of x is associated with only one value of y . The line l 2 does not represent a function since more than one y –value is associated with the x –value of 2. In fact every value of y is associated with the x –value of 2.
380 21.
y 4 2
x –2
–1
1
2
3
1
2
3
1
2
3
–2 –4 –6
22.
y 4 2
x –2
–1 –2 –4 –6
23.
y 4 2
x –2
–1 –2 –4 –6
381
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24.
y 4 2
x –2
–1
1 –2 –4 –6
Exercise 3.2
− 4y − 1 = 0 2. 4x + 14y − 21 = 0 1. 16 x
3. 2x + 3y + 11 = 0
− y + 10 = 0 5. x + y − 3 = 0 4. 7x
6. 10 x + 9y + 13 = 0 7. y + 3 = 0
−2 = 0 9. x − 7 = 0 8. x
10. y + 10 = 0 11. y = 0 12. x = 0
−y+1 = 0 14. 2 x + 3y − 1 = 0 13. 3 x
15. x + 1 = 0 16. y = 0 17. x + 3y 18. 3 x
−3 = 0
− 2y + 5 = 0
2
3
382
−3 = 0 20. x − 2 = 0 21. x + 8y − 1 = 0 22. 8 x − y − 11 = 0 19. y
23. y = 23 x + 43 ; slope is
2 ; 3
y –intercept is
4 3
24. y = 25 x + 25 ; slope is
2; 5
y –intercept is
2 5
25. y =
−3x − 13; slope is −3; y–intercept is −13
26. y = 2x + 1; slope is 2; y –intercept is 1 4 3
15 4
5 4
− (b) y = (c) x = 28. y = f (x) = −x + 1; D f = R; R f = R 29. y = f (x) = x + 2; D f = {x ∈ R : x ≥ −3}; R f = {y ∈ R : y ≥ −2} 30. y = f (x) = −2; D f = {x ∈ R : −3 < x ≤ 8}; R f = {−2} 31. y = f (x) = −x; D f = {x ∈ R : −1 ≤ x < 2}; R f = {y ∈ R : −2 < y ≤ 1} 27.
(a) m2 =
4 3
32. Suppose l cuts the x–axis at a and the y–axis at b and a = 0 and b = 0. We then have the two points P(a, 0) and Q(0, b) on the line. Using the two–point form (3.2.3) we obtain
y
− 0 = b0 −− a0 (x − a)
i.e. we have
y= which we may rewrite as
b
− a ( x − a)
y=
b (a a
− x)
which becomes
ay + bx = ab.
If we divide both sides by ab ( ab = 0 since a = 0 and b = 0) we have
x y + = 1. a b
383
MAT0511/003
Exercise 3.3
− ,−
1. (2, 5) 2. ( 12 11
13 ) 11
2 3. ( 19 , ) 9 9
4. (1,
3 2)
− −3 6. y = −3x + 3 5. y = 2x
7. y = 1 8.
y l3
2
Q 1
P x 1
3
2
l2 l1
(a) P = ( 34 , 34 ); Q = (
−
3 5 , 4 4)
(b) l1 and l2 are parallel ( i.e. have the same slope) and are not coincident ( i.e. do not lie on top of each other). (c) 2 (d)
11 3
(e) x = 9.
9 4 1
(a) f x
−
x
( )= 3 + (b) P = ( 32 , 12 ) (c)
1
g x
( )=
x
1
−
14 3
−3 or x = 6 ≥ 3 (ii) x <1 (a) h(x) = 2x − 2 k(x) = 2 (b) y = −2 (d) x =
(e) (i) x
10.
(c) 6
(iii)
x
≤
3 2
384 (d) x =
−3
(e) (i) x >1 11.
x
(ii)
≤2
(a) f (x) = 6x + 24 (b) g(x) = 12x (c)
y 72
Cost in rands
48
f g
24
x 1 2 3 4 5 6 7 8 9 10 Number of cakes
(d) 4 (e) 9 12.
(a)
y 4 000 Cost in rands
income function: y = f ( x ) = 18x
3 000
cost function: y = g( x ) = 800 +10 x
2 000 1 000
x 50
100 150 200 250 Number of T-shirts
Break–even point = ( 100, 1 800 ) (b) Loss, R160 (c) 250 13.
−x (b) 2 x − y ≤ −2 (or y ≥ 2x + 2)
(a) y <
(c) x < 1
(d) y
≤ −2
385
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14. (a)
(b) y
y
x= –2
y = 2x
–2
x
0
(c)
x
0
(d) y
y
x
0
x
x + 2y = 3 x – 2y = 3
15.
y y=4 3
y = x – _2
x
0
y = x –1
2x = 3
16.
(a) direc tly proportional; x 3 ; 1 (b) directly proportional; x 2 ;
√3 4
(c) jointly proportional; h and x 2 ; 1 17.
(a) p = kq
√
(b) r = k s (c) p = kq 3 r 2 (d) p2 = ks 18.
(a) z = 5x (b) z = 55
√r 3
386 19. p = 3qr2 20. Let c be the cost in rands of printing n copies of a magazine that has p pages. Then c = kn p, where k is the constant of proportionality. (a) c = 240 000 where n = 4 000 and p = 120. Hence k = c = 12 np.
1 2
and we have
(b) The cost will be R230 000.
TOPIC 4 Exercise 4.1 1. Write y = a (x h)2 + k as y = az 2 + k, where z = x h. Then the axis of symmetry of the parabola defined by y = az 2 + k is the line defined by z = 0. Now z = 0 is the same as x h = 0, i.e. x = h . Thus the equation of the axis of symmetry of the parabola defined by y = a(x h)2 + k is the line defined by x = h.
−
−
−
2.
−
(a) 1 unit to the right and 2 units downwards. (b) 1 unit to the left and 2 units up wards. (c) 1 unit to the right and 2 unit s upwards. (d) 1 unit to the left and 2 units do wnwards.
−h units to the left and k units upwards. (a) (i) D f = R; R f = [0, ∞) (ii) y–intercept = 9; x–intercept = −3 (iii) x = −3 (iv) (−3, 0), the lowest point (b) (i) Dg = R; Rg = ( −∞ , 4] (ii) y–intercept = −8; x–intercepts are 2 − (e)
3.
(iii) x = 2
(iv) (2, 4), the highest point
√
2 3 and 3
√
2 + 2 33
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MAT0511/003
Exercise 4.2 1.(a)
(b)
y
( 2, 5)
y
5
( 2, 4)
−
4
−
−
( 1, 1)
x
−1
(c)
−1
x
(d)
y (1, 12 12 ) 12
y
3 1 12
−1
x
1 2
− (−1, −−1)
−4
1
6 x
(e)
y 1 3
− −2
5 6
2
x
1 ( 56 , 4 12 )
−
2. It does not cut the x –axis since ∆ = 3.
−7 < 0.
(a) down (b) x = 2 12 (c) maximum (d) two (e) p = ∆
−6
= 0.)
1 4
( Hint:
∆
= b2
− 4ac = 5
2
+ 4 p. Now find p such that
388
Exercise 4.3 1.
2
−(x + 2)
2
−x − 4 x + 2 (b) y = (x + 2)(x − 5) = x − x − 5 (c) y = −3x + 5x (d) y = − (x + 1) + 6 = − (x + 4)(x − 2) = − x − (e) y = − x(x − 4) = − x + x (a) y = (x − 2) = x − 3x + 3 (b) y = −2(x + 1) + 4 = −2x − 4x + 2 (c) y = − x − 1 (d) y = (x − 1)(x − 5) = x − x + 4 (e) y = x − 4x + 9 (a) y =
+6 =
1 2
1 2 2
3 2
2
2 3 1 5
2.
2 3
2
1 2 5
3 4
2 2 3
4 51 3x + 3
4 5
3 2 4
2
2
2
3 2 2
4 5 1 2 2
4 2 5
24 5
Exercise 4.4 1.
≤ −3 or x ≥ −2 (b) −1 < x < (c) x ≥ 3 or x < −2 (d) x ≤ − or x > 1
(a) x
3 2
1 3
2. The numbers are 10 and 10 and their prod uct is 100. 3. 242 m 2 4. 20 m; 2 seconds 5. Suppose the price of a ticket increases by x rands. Then the price of a ticket is ( 20 + x) rands. The estimated number of tickets sold will be ( 20 000 500x). The organisers will receive ( 20 + x)(20 000 500x) rands. Now
−
−
y = (20 + x)(20 000
500x)
= 400 000 + 10 000x
− − 500x . 2
(a) In order for the organisers to make the maximum amount of money
x=
−b = −10 000 = 10. 2a −1 000
Thus a ticket must cost R (20 + 10), i.e. R30.
−
×
(b) The number of spectators should be (20 000 500 10), i.e. 15 000. The amount of money that the organisers will make is R(15 000 30), i.e. R450 000.
×
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TOPIC 5 Exercise 5.1 1. (a)
y=
−x
3
(b)
y=
7 4x
(d)
y=
3 5x
(e)
y=
− mxn
(c)
y=
1
−x
2. Suppose f (x) = y = kx . The point (a, b) lies on the graph of f . Hence b = ak , i.e. ab = k . Since ab = k , it follows that a = bk . In this form a is the y–value and b is the x –value, and hence ( b, a) lies on the graph of f . 3.(a)
(b)
y
xy = 10
d
P
P
x d=
xy = 10
y
4xy + 9 = 0
d 0
d=
√20
x
9 2
4xy + 9 = 0
4.
y=
− 10x
y y=
− 3x x
y=
− 3x y=
10
−x 5. xy = ab (or y = 6. y =
ab x)
8 9
− x (or 9xy = −8)
7. We rewrite mxy + 2 = 0 as y = (why?) we have
−
2 m > 0.
Thus
− − == 2 mx , i.e.
2
y
2 m
k 2 x where k = m . Since k > 0 1 4 1 , i.e. m = 4 , i.e. m = 16. 2
−
−
−
390
Exercise 5.2 1.
(a) c = 22, r =
2.
(a) p =
3. 4.
22 t
k qr
(b) t = 11
(b) z =
kx 2 y3
(c) l =
√
km n pq
q is directly proportional to p and inversely proportional to r . (a) z is directly proportional to the square of x and inversely proportional to y . (b) Q is directly proportional R and inversely proportional to S and T . (c) y is jointly proportional to x and the square of z and inversely proportional to r .
5.
(a) R =
kl , k= d2
(b) 166 23
7 2 400
Ω
TOPIC 6 Exercise 6.1 1.
(i) (1, 0) ; (4, 3) (two solutions) (ii)
5 2,
3 4
− √ − −√ −√ (one solution)
(iii) no solution (remember to use the discriminant) 2.
(i)
1+ 7 , 2
1
7
2
;
1
2
7
,
−1+√7 2
(two solutions)
(ii) no solution (use the discriminant) (iii) (1, 2) (one solution) 3.
(a)
y y= 4 x
l1
4
x
4
l3 y= 4 x
l2
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MAT0511/003
(b) (i) 2 solutions
(ii) no solution
(iii) one solution
− −
(c) (i) (2, 2) ; ( 2, 2) (ii) no solution (iii) (2, 2) 4.
−9 (b) y = − x (c) M = ( 1, −8) (Solve y = x − 9 and y = − x simultaneously.)
(a) y = x
8
8
Thus the length of N M is 8 units.
(d) Length of QP = length of RP
=
8 3
− 6
units
− length of RQ
= 3 13 units (e) Length of OR = 5 units. Thus the length of RQ is 5.
8 5
units.
2
−x + x + 6 g (x) = x − 2x − 3 (b) P = − ,
(a) f ( x) =
2
3 9 2 4
2
(c) The length of MN is given by f ( x) g (x), i.e. by 2x + 3x + 9. Thus we solve 2x2 + 3x + 9 = 4. The x–coordinate of M and N is 1 or 52 . (d)
−
−
−
−
2
−2x + 3x+ 9 −2 x − + 10
Length of MN =
3 2 4
=
1 . 8
Thus the maximum length of M N is 10 18 units and this occurs when x = 34 . (e) (i) (ii)
−2 < x < − 1 2 x2
⇔ ⇔ ⇔ ⇔ 6.
(a) h : x 2 + y2 = 4
f : y = 12 x2 g:y=
−2 −x + 2
x2
− 2x − 3
− −
− 3x − 9 < 0
x2 + x + 6 <
x2
0
− 2 x − 3 < −x
2
+x+6
g (x) < f ( x) 3 < x <3 2
−
392
−
(b) P = ( 4, 6)
√
(c) Length of SN = 2 3 units. (d) The length of M T is given by g (x) by
−
1 2
2
(x + 1)
+ 4 12 .
1 2 2
− f (x), i.e. by − x − x + 4, i.e.
Thus the maximum length of MT is 4 12 units and this occurs when x = 1.
−
(e) The values of x for which
− 4
x2 + x
≥2
are the values of x for which
− 4
x2
≥ −x + 2
i.e. for which the upper semi–circle lies above or on the line. Hence 0 x 2.
≤ ≤
TOPIC 7 Exercise 7.1 1. Tally: the word we use for counting items of data. For each item we draw
|. Every4time ahorizontal vertical line: we lines: have a||||colle line through vertical . ction of 5 items we draw a Classes: Categories into which data are grouped. Frequency: the total number of data items in any given group or class. Frequency distribution: A table in which data are grouped into specific classes, showing the frequency of items in each class. 2. Class limits Tally marks 5–9 || 10–14 |||| ||| 15–19 |||| |||| || 20–24 |||| |||| || 25–29 |||| |||| || 30–34 |||| |||||
35–39 40–44 45–49 50–54 55–59
Frequency
2 8 12 29 15 11
|| |||| |||| |||| ||
8 7 5 2 1
|| || || || || || | |||| || |
This gives us the following frequency distribution. 5–9 10–14 15–19 Classes Frequency 2 8 12 29
20–24 25– 29 15 11 8
30–34 7
35–39 40–44 5 2 1
45–49
50–54
55–59
393
MAT0511/003
Exercise 7.2 1. We use a class width of 2 ◦ C. Lowest temperature: 24, 5◦ C; highest temperature : 33 , 6◦ C. Thus lowest class limit = 24, 5◦ C lowest class boundary = 24, 45◦ C number of classes = 5. Class limits ( ◦ C) Class boundaries ( ◦ C) 24,5 – 26,4 24,45 – 26,45 26,5 – 28,4 26,45 – 28,45 28,5 – 30,4 28,45 – 30,45 30,5 – 32,4 30,45 – 32,45 32,5 – 34,4 32,45 – 34,45
Frequency 8 5 6 5 4
This gives us the following histogram.
10 9 8 7
Frequency
6 5 4 3 2 1 24,45
26,45
28,45
30,45
32,45
34,45
Temperature
2. Children who play no sport: 30 girls, 10 boys, i.e. 40 children. Netball: 13 390, i.e. 130 girls (there are 420 play sport).
×
Soccer: 750
− 30 girls, i.e. 390 girls, who
− (420 + 10), i.e. 320 boys.
Basketball: 500 children. 2 of the girls who play sport play basketball, i.e. 260 girls play basketball. 3 Thus 500 260, i.e. 240, boys play basketball.
−
−
Cricket: There are 750 420 = 330 boys in the school. Thus there are 330 (240 + 10) cricket players, i.e. 80 cricket players.
−
We do not set up the pie chart based on the number of children in the school, but on the total number of participants in all the sports that are played. Thus several children are counted more than once, for example, a child who plays both basketball and soccer is counted in both categories. The sports and numbers of players represented are thus
394 No sport 40 Cricket 80 Netball 130 Soccer 320 Basketball 500 Total 1 070 Sport played
Fraction
Percentage
Degrees (central angle)
None
40 1 070
(to nearest integer) (to nearest integer) 4% 13
Cricket
80 1 070
7%
Netball
130 1 070
12%
44
Soccer
320 1 070
30%
108
Basketball
500 1 070
47%
168
27
We now have the following pie graph.
No sport Basketball
Cricket
4%
7% 47% Netball 30%
12%
Soccer
Sports played at Sport Valley Secondary School 3.
(a) (i) Approximately 30 million square kilometres. (ii) Africa is approximately 2,5 million square kilometres bigger than Asia. (iii) The smallest land mass is Europe. (iv) Europe and North America are together just bigger than Asia.
395
MAT0511/003
(b) Land mass
Approximate size Percentage of Degrees (in millions of total mass (to the square kilometres) (to the nearest nearest integer) integer)
Africa Asia
30 27,5
29 26
Europe NorthAmerica SouthAmerica Total
5,5 24 18 105
5 23 17
104 94 18 83 61
We now have the following pie graph.
South America 17%
Africa 29%
North America
23% 26% 5%
Asia
Europe
Approximate land masses of Africa, Asia, Europe, North and South America 4.
Class limits (%) Class boundaries (%) Frequency 20–29 19,5–29,5 1 30–39 29,5–39,5 5 40–49 39,5–49,5 3 50–59 49,5–59,5 4 60–69 59,5–69,5 4 70–79 69,5–79,5 2 80–89 79,5–89,5 1
396
5 Number of students
4 3 2 1
19,5 29,5 39,5 49,5 59,5 69,5 79,5 89,5
Percentages obtained in first assignment
5.
(a) 90 pupils wrote the test. (b) 29 girls scored 50% or more. (c) 7 boys scored between 45% and 50% (including these marks). (d) 57% was scored 4 times.
Exercise 7.3 1. We arrange the data (mas s in kg) in numerical order from smallest to biggest. 72, 77, 78, 79, 79, 79, 79, 81, 82, 82, 83, 84, 85, 86, 86, 86, 86, 87, 87, 88, 88, 89, 89, 89, 90, 90, 91, 92, 92, 92, 93, 93, 93, 94, 95, 96, 96, 96, 96, 97, 97, 98, 101, 101, 104, 104, 105, 107, 108, 111 (a) Median =
90 + 90 2
kg = 90 kg
(b) Mode: There are 3 modes: 79 kg, 86 kg, 96 kg 2.
(a) Boys’ marks: x Girls’ marks: x
≈ 51% (we have rounded 50,73% to 51%) ≈ 54% (we have rounded 54,47% to 54%)
(b) Girls’ marks (percentages) in numerical order (42 marks): 17, 23, 23, 27, 31, 39, 40, 41, 43, 44, 45, 45, 48, 50, 50, 51, 53, 53, 53, 54, 55, 57, 57, 59, 60, 60, 61, 61, 61, 62, 62, 62, 65, 67, 68, 68, 71, 73, 73, 80, 84, 92 Median: 55 + 57 112 = = 56 2 2 Hence the median is 56% (c) In the rows representing the girls’ marks (percentages), the number 3 appears 3 times in the 5 row; the number 1 appears 3 times in the 6
397
MAT0511/003
row; the number 2 also appears 3 times in the 6 row. Thus the marks occurring most frequently are 53%, 61% and 62%; hence there are three modes, namely 53%, 61% and 62%. (d) The arithmetic mean is approximately 52,47. For the purpose of this calculation we round the mean to x = 52%. Mark (%) 12 17 17 21 22 23 23 25 26 27 .. .
94 Total (ignoring Mean deviation = 3.
(a) x = (b)
1 8 (304, 0)
961 90
Deviation 52 = 52 = 52 = 52 = 52 = 52 = 52 = 52 = 52 = 52 = .. .
94
− 52 =
−) ≈ 10, 7 ≈ 11
− − − − − − − − −
− 40 35 − 35 − 31 − 30 − 29 − 29 − 27 − 26 − 25
42 961
= 38, 0◦ C
Median = 4.
12 17 17 21 22 23 23 25 26 27
38, 0◦ C + 37, 4◦ C = 37, 7◦ C 2
(a) Your most likely salary would be the mode, if there is one. (b)
x= Hence x
91 900 = 8 354, 545454 ... 11
≈ R8 350.
(c) In this case there is no mode, since no number occurs more frequently than any other. (d) We arrange the data in numerical order, from smallest to biggest. We have R5 100, R5 500, R5 900, R6 200, R6 500, R6 800, R7 200, R8 300, R8 400, R12 000, R20 000. The median is thus R6 800. Note that there are two data items that differ markedly from the rest, i.e. R12 000 and R20 000. These items influence the arithmetic mean, and hence in this case the median (R6 800) is a more likely salary than the mean (R8 350).
398
Exercise 7.4 1.
(a)
S =
{b , b , b , b , b , r , r , r , g , g , g , g } 1
2
3
4
1
5
2
3
th
1
th
where bi represents the i blue pen, ri the i green pen (i N). Thus n (S) = 12.
2
3
4
red pen and gi the ith
∈
(b)
E =
{r , r , r } 1
2
3
n(E ) = 3 (c)
3 1 n(E ) = = n(S) 12 4 (d) The probability of choosing a red pen is
1 . 4
2. Note: the answers obtained by each student will differ. Read through our solution, and try to determine whether or not your answers make sense. Our answer:
(a) H: 30 times T: 20 times (b) Relative frequency of obtaining heads: 30 50
=
3 5
= 0, 6
Relative frequency of obtaining tails: 20 50
=
2 5
= 0, 4
(c)
Relative frequency: No of heads No of spins
1 0,9 0,8 0,7 0,6 0,5 0,4 0,3 0,2 0,1 After 10 After 20 After 30 After 40 5 spins 15 spins 25 spins 35 spins
Number of spins
After 50 45 spins
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MAT0511/003
(d) Answer obtained: 35 Expected answer: 12 Since S = H, T , we have n(S) = 2; E = and thus P (E ) = 12 .
{
}
{ H}, hence n(E ) = 1,
3. Number of cards in the pack: 52 Number of red cards in the pack: 26 Number of black cards in the pack: 26 After 5 red cards have been drawn there are 47 cards in the pack, of which 21 are red, 26 are black. Then it follows that probability of drawing a red card = 21 47 probability of drawing a black card = 26 . 47 21 Since 26 > 47 there is a greater chance that the next card will be black 47 than that it will be red. 4.
(a) 52 cards in the pack, hence n (S) = 52. 4 aces in the pack, hence n (E ) = 4.
P(ace) = (b) If E = 5.
n(E ) 4 1 = = 52 13 n(S)
{ ace of diamonds } then n(E ) = 1 and hence P(E ) =
1 . 52
(a) Consider the situation when the coin is spun twice. From the tree diagram given in the margin on p. 368 it follows that the sample space is S = (H,H), (H,T), (T,H), (T,T) , i.e. we have n(S) = 4. If the event E consists of obtaining heads at least once , i.e. obtaining heads once or twice, then
{
{
}
}
E = (H,H), (H,T), (T,H) , and hence n (E ) = 3. n (E ) 3 = . n(S) 4 (b) Consider the situation when the coin is spun three times . From the tree diagram given in the margin on p. 368 it follows that the sample space is S = (H,H,H), (H,H,T), (H,T,H), (H,T,T), (T,H,H), (T,H,T), (T,T,H), (T,T,T), i.e. we have n (S) = 8. Thus: Probability of obtaining at least one head = P(E ) =
{
}
If the event E consists of obtaining at most two tails, i.e. of not obtaining tails or obtaining tails once or twice, then E = (H,H,H), (H,H,T), (H,T,H), (H,T,T), (T,H,H), (T,H,T), (T,T,H), and hence n (E ) = 7. n (E ) 7 Thus: Probability of obtaining at most two tails = P(E ) = = . n(S) 8
{
}
6. Let yellow balls be denoted by y1 , etc., white balls by w1 , etc., and blue balls by b 1 , etc. Then S = y1 , y2 , y3 , y4 , y5 , y6 , w1 , w2 , w3 , w4 , b1 , b2 , b3 , b4 , b5 and n (S) = 15.
{
}
400 (a) E =
{ y , y , y , y , y , y } and thus n(E ) = 6. 1
2
3
4
5
6
n(E ) 6 2 = = 15 5 n(S)
P(yellow ball) = P (not white) = 1
(b) Now
− P(white)
P(white) = hence 7.
P(not white) = 1
−
(See Law 1.)
4 15 4 15
=
11 . 15
(a) There are 6 possible ways of scoring 7, out of 36 possible totals when two dice are thrown. Let P(7) denote “probability of scoring 7 once”. 6 Now P (7) = 36 . Thus
× P(7) × ≈ 0, 03.
Probability of scoring 7 twice = P(7)
= =
6 36 1 36
6 36
(b) There are 5 ways of scoring 8. Thus probability of scoring 8 (denoted by P (8)) = There are 5 ways of scoring 6. Thus probability of scoring 6 (denoted by P (6)) = Thus probability of scoring 8 and then 6 is
P(8)
5 36
× P(6) = ×
5 36
=
25 1 296
5 . 36 5 36 .
≈ 0, 02.
401
MAT0511/003
REFERENCES Bohlmann C.A. and Singleton J.E.: Interactive Maths Book 2 : Polynomial and Rational Functions , University of South Africa, 1997. Freeman, R.: How to Learn Maths , National Extension College, 1994. Gonin, A.A., Archer, I.J.M. and Slabber G.P.L.: Modern Graded Mathematics
for Senior Certificate, (2nd edition), NASOU Limited. Jones, E. and Childers, R.: Contemporary College Physics, (3rd edition), WCB/McGraw-Hill, 1999. Movshovitz–Hadar N. and Webb J.: One Equals Zero and other Mathematical Surprises: Paradoxes, Fallacies and Mind Bogglers , Key Curriculum Press, 1998. Page S., Berry J. and Hampson H.: Mathematics: A second start (2nd edition), Prentice Hall, 1995. Sidiropoulos E., Jefferey A., Mackay S., Forgey H., Chipps C. and Corrigan T.: South Africa Survey 1996/97, South African Institute of Race Relations, 1997. Smith S. M.: Agnesi to Zeno , Key Curriculum Press, 1996. Sobel, M.A. and Lerner, N.: Precalculus Mathematics, (2nd edition), Prentice Hall, 1991. Stewart J., Redlin L. and Watson S.: Mathematics for Calculus (2nd edition), Brookes/Cole Publishing Company, 1993.
