A. RATIONAL NUMBERS: THE SET Q 1. Understanding Rational Numbers Definition
The ratio of an integer to a non-zero integer is called a rational number. The set of rational a numbers is denoted by Q. Q = { | a, b Z, b 0} b
2. The Set of Positive Rational Numbers If a rational number represents a point on the number line on the right side of zero, then it is called a positive rational number. a is a positive rational number if a and b are both positive integers or both negab tive integers. 2 –2 2 and For example, are positive rational numbers, and denoted by . 7 –7 7
In short,
Definition
The set of positive rational numbers is denoted by Q+. Q+ = {
a a | 0 and a, b , b 0} b b
3. The Set of Negative Rational Numbers If a rational number represents a point on the number line on the left side of zero, then it is called a negative rational number. a In short, is a negative rational number if a is a positive integer and b is a negative integer, b or if a is a negative integer and b is a positive integer. –5 5 are negative rational numbers. We can write negative rational and 4 –4 5 –5 5 numbers in three ways: – . 4 4 –4
For example,
Definition
The set of negative rational numbers is denoted by Q–. a a Q– = { | 0 and a, b , b 0} b b 10
Algebra 8
A. THE SET OF REAL NUMBERS 1. Understanding Real Numbers In algebra we use many different sets of numbers. For example, we use the natural numbers to express quantities of whole objects that we can count, such as the number of students in a class, or the number of books on a shelf. The set of natural numbers is denoted by N. N = {1, 2, 3, 4, 5, ...} -6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
Natural numbers
The set of whole numbers is the set of natural numbers together with zero. It is denoted by W. W = {0, 1, 2, 3, 4, 5, ...} -6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
Whole numbers
The set of integers is the set of natural numbers, together with zero and the negatives of the natural numbers. It is denoted by Z. Z = {..., –5, –4, –3, –2, –1, 0, 1, 2, 3, 4, 5, ...} -6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
Integers
We use integers to express temperatures below zero, distances above and below sea level, and increases and decreases in stock prices, etc. For example, we can write ten degrees Celsius below zero as –10°C. To express ratios between numbers, and parts of wholes, we use rational numbers. 8 2 3 0 17 For example, , , – , , and are rational numbers. 3 5 7 7 1 The set of rational numbers is the set of numbers that can be written as the quotient of two integers. It is denoted by Q. Q={
a | a, b and b 0} b 7 2 -7
-6
-5
-4
1 2 -3
-2
-1
2 3 0
5 4 1
2
13 2 3
4
5
6
7
Some rational numbers
Radicals
11
We can write every rational number as a repeating or terminating decimal. Conversely, we can write any repeating or terminating decimal as a rational number. 3 321 0.6, and 0.324 = 0.324242424... 5 990 –– 0.6 is a terminating decimal, and 0.324 is a repeating decimal.
For example,
There are some decimals which do not repeat or terminate. For example, the decimals
0.1012001230001234000 ... 3.141592653 ... =
R = R+ {0} R– R+ is the set of positive real numbers R– is the set of negative real numbers
2.71828 ...
=e
1.4142135 ...
= ñ2
do not terminate and do not repeat. Therefore, we cannot write these decimals as rational numbers. We say that they are irrational.
Definition
A number whose decimal form does not repeat or terminate is called an irrational number. The set of irrational numbers is denoted by Q or I. Definition
The union of the set of rational numbers and the set of irrational numbers forms the set of all decimals. This union is called the set of real numbers. The set of real numbers is denoted by R. R = Q Q For every real number there is a point on the number line. In other words, there is a one-to-one correspondence between the real numbers and the points on the number line.
Real Numbers
-2.35 -5
-4
-3
-2
-0.5
0.6
-1 0 1 2 Some rational numbers
e
p 3
4
5
The real numbers fill up the number line.
NWZQR Q R
12
We can summarize the relationship between the different sets of numbers that we have described in a diagram. As we know, the set of natural numbers is a subset of the set of whole numbers, the set of whole numbers is a subset of the set of integers, the set of integers is a subset of the set of rational numbers, and the set of rational numbers is a subset of the set of real numbers. This relationship is shown by the diagram on the left. Algebra 8
1. Understanding Square Roots Remember that we can write a a as a2. We call a2 the square of a, and multiplying a number by itself is called squaring the number. The inverse operation of squaring a number is called finding the square root of the number. Objectives
After studying this section you will be able to: 1. Understand the concepts of square root and radical number. 2. Use the properties of square roots to simplify expressions. 3. Find the product of square roots. 4. Rationalize the denominator of a fraction containing square roots. Definition
If a2 = b then a is the square root of b (a 0, b 0). We use the symbol ñ to denote the square root of a number. ñb is read as ‘the square root of b’. So if a2 = b then a = ñb (a b, b 0). Here are the square roots of all the perfect squares from 1 to 100.
ñ1 = 1
2 =4
ñ4 = 2
32 = 9
ñ9 = 3
2
ò16 = 4
2
ò25 = 5
12 = 1 2
4 = 16 5 = 25
ò36 = 6
7 = 49
ò49 = 7
82 = 64
ò64 = 8
ò81 = 9
ó100 = 10
62 = 36 2
2
9 = 81 2
10 = 100
2
The equation x = 9 can be stated as the question, ‘What number multiplied itself is 9?’ There are two such numbers, 3 and –3. Rule
If x R then
x if x 0. x2 | x | – x if x 0.
In other words, if x is a non-negative real number, then if x is a negative real number, then Radicals
x2 x, and
x2 – x. 13
For example,
32 3, ( 32 9 3), and (–3)2 –(–3) 3 ( (–3) 2 9 3).
We can conclude that the square root of any real number will always be greater than or equal to zero. ò–9 is undefined. Negative numbers have no square root because the square of any real number cannot be negative. ò–9 3, since 32 is 9, not (–9). ò–9 –3, since (–3)2 is 9, not (–9).
Note x = ñ9 and x2 = 9 have different meanings in the set of all real numbers. ñ9 = 32 = |3| = 3 If x2 = 9 then x = 3 or x = –3.
EXAMPLE
1
Solution
EXAMPLE
2
Solution
Evaluate each square root. a. ò81
b. ñ1
g. –ó100
h. –ó0.09 i. ò–4
d. ò64
b. ñ1 = 1
d. ò64 = 8
e.
g. –ó100 = –10
h. –ó0.09 = –0.3
j.
k.
(–4)2 16 4
e. ñ9
(–4)2
j.
a. ò81 = 9
k.
f. ó0.64
–4 2
c. ñ0 = 0
4 2 9 3
f. ó0.64 = 0.8 i. ò–4 is undefined
–42 –16 is undefined
Evaluate each square root. a. ó100
b. ó121
c. ó144
d. ó169
g. ó400
h. ó625
i.
j.
1225
e. ó225
f. ó361
10000
a. ó100 = 10
b. ó121 = 11
c. ó144 = 12
d. ó169 = 13
e. ó225 = 15
f. ó361 = 19
g. ó400 = 20
h. ó625 = 25
i.
j. 14
c. ñ0
1225 35
10000 100 Algebra 8
2. Properties of Square Roots Property
For any real number a and b, where a 0, and b 0, ñañb = óab. For example,
25 16 25 16 5 4 20, 3 27 3 27 81 9, 36 a2 36 a2 6 a ( a 0), and 5 5 5 5 25 5.
Note
b0
If a 0 then
a0
ña ña = óa a =
óa . 6 =
ña . ñ6
a2 a.
Mathematics is a universal language.
3
Simplify each of the following.
Solution
a. ñ2ñ8 = ó28 = ò16 = 4
b.
7 7 7 7 49 7
c. ò50ñ2 = ó502 = ó100 = 10
d.
25 1 25 1 25 5
EXAMPLE
a. ñ2 ñ8
b. ñ7 ñ7 c. ò50ñ2
d. ò25ñ1
e.
576 36 16 36 16 6 4 24
f.
10 90 10 90 900 30
e. ó576
f. ò10ò90
Property
For any real numbers a and b, where a 0, and b > 0, a b
If a > 0 then a a
=
a a
Radicals
= 1 = 1.
For example,
24 6
1 49
a . b
24 4 2, and 6 1 49
1 . 7 15
EXAMPLE
4
Simplify the expressions. a.
g.
Solution
25 9
b.
24a 3
h.
6a
50
c.
2 a5 b6 ab
1 64
d.
e. –
1 100
f.
x3 y3
a.
25 25 5 = = 9 3 9
b.
c.
16 16 4 49 49 7
d.
1 64
f.
625 625 25 144 144 12
e. – g.
h.
i.
1 1 1 – – 100 10 100 24a3
a5 b6 ab2
xy 3
3
x y
50 2
50 25 5 2
1 64
1 8
24 a3 4 a2 4 a2 2a 6a
6a
625 144
xy
i.
2
16 49
a5 b6 a4 b4 ( a2 b2 )2 a2 b2 ab2
xy x3 y3
1 x2 y2
1 2
2
x y
1 xy
Property
For any real number a and n Z, ( a )n an
(a 0).
Proof ( a )n a a a ...
a
n factors of ña
a a a ... a
an
n factors of a
For example, ( a )2 a2 a, ( 5)3 53 125, and ( 2 )8 2 8 256 16. 16
Algebra 8
EXAMPLE
5
Solution
Evaluate (ñ2)4 + (ñ5)4 – (ñ5)2 – (ñ2)6. ( 2 )2 ( 5)4 – ( 5)2 – ( 2 ) 6 2 4 5 4 – 5 2 – 2 6 (22 )2 (52 )2 – 52 – (23 )2 22 52 – 5 – 2 3 4 25 – 5 – 8 16
3. Working with Pure and Mixed Radicals Definition
A radical expression is an expression of the form index
n
a. radical sign
ña n
radicand
Square roots have index 2. However, we usually write square roots in their shorter form, ña: 2
a a
Definition
A mixed radical is a radical of the form x n a (x Q, x {–1, 0, 1})
For example, 3ñ2, 6ñ7, and 9ó115 are mixed radicals. ò55, ò99, and ò27 are not mixed radicals. We say that they are pure radicals. We can convert between mixed and pure radical numbers to simplify radical expressions. Property
For any real numbers a and b, where a 0 and b 0, a2 b a b and a b
For example,
a 2 b.
8 4 2 2 2 2 2 2 2 2 2, 27 9 3 32 3 32 3 3 3, 32 16 2 4 2 2 4 2 2 4 2, and 50 25 2 5 2 2 5 2 2 5 2.
Radicals
17
EXAMPLE
6
Simplify the expressions. a. ñ8 + 2ò32 – ò18 + ò72 – ò98 b. 2ò48 + 3ò27 – ó108 + ó243
Solution
a.
8 = 22 2 = 2 2 2 32 = 2 4 2 2 = 8 18 = 3 2 = 3 2 2
2
72 = 6 2 = 6 2 98 = 7 2 = 7 2 2
2
8 2 32 18 72 98 2 2 8 2 3 2 6 2 7 2 2 (2 8 3 6 7) 6 2
b. 2 48 3 27 – 108 243 2 4 2 3 3 3 2 3 –
6 2 3 9 2 3
8 3 9 3 – 6 3 9 3 (8 9 – 6 9) 3 20 3 EXAMPLE
7
Solution
Write the numbers as pure radicals. a. 2ñ2
b. 3ñ5
c. 5ñ3
a. 2 2 2 2 2 2 2 2 b. 3 5 32 5 9 5 c. 5 3 52 3
25 3
4 2
d. 10ò10
e. xñy
8
45 75
d. 10 10 10 2 10 100 10 1000 e. x y x2 y Property
For any non-zero real numbers a, b, c, and x, añx + bñx – cñx = (a + b – c)ñx .
Note ña + ñb óa+b For example, ñ9 + ò16 = 3 + 4 = 7, but ó9 + 16 = ò25 = 5. 18
Algebra 8
EXAMPLE
8
Perform the operations. a. ñ3 + ñ3
b. 2ñ5 + ñ5
c. 3ñ6 + 4ñ6
e. ò50 + ò98 + ó162 Solution
d. 10ñ5 – 3ñ5
f. 5ñx – ò9x + ó64x
a. ñ3 + ñ3 = (1 + 1)ñ3 = 2ñ3 b. 2ñ5 + ñ5 = (2 + 1)ñ5 = 3ñ5 c. 3ñ6 + 4ñ6 = (3 + 4)ñ6 = 7ñ6 d. 10ñ5 – 3ñ5 = (10 – 3)ñ5 = 7ñ5 e. ò50 + ò98 + ó162 =
ó252 + ó492 + ó812 = 5ñ2 + 7ñ2 + 9ñ2
= (5 + 7 + 9)ñ2 = 21ñ2 f. 5ñx – ò9x + ó64x = 5ñx – 3ñx + 8ñx = (5 – 3 + 8)ñx = 10ñx EXAMPLE
9
Solution
Compare the following numbers. a. ñ7 ... 3
b. 3ñ5 ... 2ò10
a.
b. 3 5...2 10
7...3
c. 2ñ7 ... 3ñ3
d. –2ñ3 ... –3ñ2
c. 2 7...3 3
d. –2 3... – 3 2
7... 9
32 5... 2 2 10
22 7... 32 3
– 22 3... – 32 2
7 9
45 40
28 27
– 12 – 18
2 7 3 3
–2 3 –3 2
7 3
3 5 2 10
Property
Let a, b, m, and n be four real numbers, satisfying a = m + n and b = m n. Then, 1.
2.
m n a2 b
m – n a–2 b
( m n)
Proof 1. In order to verify these expressions, suppose that t = òm + ñn. t2 = (ò m + ñn)2 = (ò m + ñn) (ò m + ñn) = (ò m ñn) + (òm ñn) + (ñn òm ) + (ñn ñn) (by the distributive property) = m + (ò m ñn) + (ñn ò m ) + n = m + n + 2ómn a
(by the commutative property)
b
t = a + 2ñb t = a 2 b 2
2. We can prove the second part in the same way. Try it yourself. Radicals
19
EXAMPLE
10
Solution
Simplify the expressions. Use the property to help you. a.
32 2
b.
52 6
e.
5 21
f.
2 3
a.
3 2 2 2 1 2 1 2+1
c.
c.
6 32
b.
52 6 3 2
21
3+2
d.
6–4 2
32
6 32 6 2 8 4 2 2 2 4+2
d.
42
6 – 4 2 6 – 2 2 2 6 – 2 2 2 2 6 – 2 8 4 – 2 2 – 2 4+2
42
e. We need a 2 in front of ò21 before we can use the property. Therefore, let us multi2
ply the expression by
2
. 7+3
5 21
2 2
5 21 3+1
2
f.
2
2 3
2 (5 21) 2
73
10 2 21 2
7 3 2
7 3 2 2
31
42 3 2
3 1 2
3 1 2 2
Check Yourself 1 1. Simplify the expressions. a. ñ2 ñ2 e.
32
b. ñ8 ò32 f.
12
2 2. Evaluate the following.
3
a. (ñ3)2 + (ñ4)4 – (ñ5)2 – (ñ2)4
c. ò3x ó12x g.
a b 3
a b
3
d. ñ2 ò18 h. –
1 49
b. (ña)4 + (ñb)2 – (ñc)6
3. Simplify the expressions. a. ò18
b. ò50
f. 2ñ2 + 3ñ2 – 4ñ2 20
c. ò48
d. ò20
g. ò50 – ò18 – ò32
e. 5ñ3 – 2ñ3 + ñ3 h. ó12x + ó27x – ó48x Algebra 8
4. Write each number as a pure radical. a. 5ñ3
b. 3ñ5
c. 4ñ2
d. 2ñ5
e. añb
5. Perform the operations. a. 6
10 15 2 3 2
b. 5ñ2 – ñ8
c. ò27 – ò48
d.
27 75 12 – 4 4 4
6. Compare the numbers. a. 3ñ5 and 2ò10
b.
1 and 2
1 3
c. –2ñ5 and –3ñ3
7. Write each expression in its simplest form. a.
3–2 2
b.
c.
62 8
g. ( 6 – 2 ) ( 8 2 12 )
7 – 2 10
d.
3 8 e.
h. ( 7 1) ( 8 – 28 )
i.
f.
9–4 5
7 – 48
3– 5 3 5
Answers 1. a. 2 b. 16 c. 6x d. 6 e. 4 f. 2 g. c. 4ñ3 d. 2ñ5 e. 4ñ3 f. ñ2
1 ab
h. –
1 7
2. a. 10 b. a2 + b – c3 3. a. 3ñ2 b. 5ñ2
g. –2ñ2 h. ò3 x 4. a. ò75 b. ò45 c. ò32 d. ò20 e. a2 b
1 1 c. –2ñ5 > –3ñ3 7. a. ñ2 – 1 2 3 b. 2 + ñ2 c. ñ5 – ñ2 d. ñ2 + 1 e. ñ5 – 2 f. 2 – ñ3 g. 4 h. 6 i. 2
5. a. 3ò30 b. 3ñ2 c. –ñ3 d. 3ñ3 6. a. 3ñ5 > 2ò10 b.
EXAMPLE
11
Simplify the following. a.
Solution
4 21 13 9
b.
6 6 72
1 16 16
c.
1 1
9 16
Start from the radical on the ‘inside’ of the expression and move outwards. a. Start with ñ9, on the inside, and work outwards. 4 21 13 9 4 21 13 3 4 21 16 4 21 4 4 25 4 5 9 3
b.
6 6 72
1 4 1 6 6 72 16 6 6 72 16 16 4 6 6 72
1 6 6 36 2
6 6 6 6 36 6 6 36 6
c. Radicals
1 1
9 25 5 9 3 1 1 16 16 4 4 2 21
EXAMPLE
12
a. Evaluate c.
Solution
a a a a ... 7. Find a.
b.
2 2 2 2 ...
x x x ... 5. Find x.
a. Let x 2 2 2 2 ... . x2 ( 2 2 2 2 ... )2
(take the square of both sides)
x2 2 2 2 2 2 ...
(remove a square root)
x
x 2x
( 2 2 2 ... x)
2
x x 2x x x
(simplify)
x 2. Therefore,
b.
2 2 2 2 ... 2.
a a a a ... 7 ( a a a a ... )2 7 2
( x x x ... )2 5 2
c.
x x x x ... 25 5
a a a a ... 49 7
a 7 49
x 5 25 x 20
a7
4. Multiplying Square Roots To multiply expressions containing square roots, we used the product property of square roots: ña ñb = óa b. We can also use the distributive property of multiplication over addition and subtraction to simplify the products of expressions that contain radicals. For example, 2ñ8 3ñ2 = 2 3 ñ8 ñ2 = 6ò16 = 64
Multiply the rational part by the rational part and the radical part by the radical part.
= 24 ñ2 (ñ3 + 2ñ2) = ñ2 ñ3 + ñ2 2 ñ2 = ñ6 + 2 ñ2 ñ2 = ñ6 + 2 2 = ñ6 + 4 22
Algebra 8
EXAMPLE
13
Solution
Perform the operations. a. ñ2(ñ5 + ñ3)
b. ñ3(3ñ3 + 2ñ2)
c. 2ñ5(ñ3 + ñ2 + 2ñ5 – ñ7)
a. ñ2(ñ5 + ñ3) = ñ2 ñ5 + ñ2 ñ3 = ó2 5 + ó2 3 = ò10 + ñ6 b. ñ3(3ñ3 + 2ñ2)= ñ3 3ñ3 + ñ3 2ñ2 = 3 ó3 3 + 2 ó3 2 = 3 3 + 2 ñ6 = 9 + 2ñ6 c. 2ñ5(ñ3 + ñ2 + 2ñ5 – ñ7)= 2ñ5 ñ3 + 2ñ5 ñ2 + 2ñ5 2ñ5 – 2ñ5 ñ7 = 2ò15 + 2ò10 + 4ò25 – 2ò35 = 2ò15 + 2ò10 + 20 – 2ò35
EXAMPLE
14
Solution
Multiply and simplify. a. (ñ2 + ñ3) (ñ2 + ñ3)
b. (5 + ñ5) (5 + ñ5)
a. (ñ2 + ñ3) (ñ2 + ñ3) = ñ2 ñ2 + ñ2 ñ3 + ñ3 ñ2 + ñ3 ñ3 = ñ4 + ñ6 + ñ6 + ñ9 = 2 + 2ñ6 + 3 = 5 + 2ñ6 b. (5 + ñ5) (5 + ñ5)= 52 + 2 5 ñ5 + (ñ5)2 = 25 + 10ñ5 + 5 = 30 + 10ñ5
EXAMPLE
15
Solution
Multiply and simplify. a. (ñ2 + 1) (ñ2 – 1)
b. (ñ5 + ñ3) (ñ5 – ñ3)
d. (ña + 1) (ña – 1)
e. (ña + ñb) (ña – ñb)
c. (1 – 2ñ2) (1 + 2ñ2)
a. (ñ2 + 1) (ñ2 – 1) = ñ2 ñ2 – ñ2 1+1 ñ2 – 1 1 = (ñ2)2 12 = 2 – 1 = 1 b. (ñ5 + ñ3) (ñ5 – ñ3) = (ñ5)2 – (ñ3)2 = 5 – 3 = 2 c. (1 – 2ñ2) (1 + 2ñ2) = 12 – (2ñ2)2 = 1 – 4 2 = 1 – 8 = –7 d. (ña + 1) (ña – 1) = (ña)2 – 12 = a – 1 (a 0) e. (ña + ñb) (ña – ñb) = (ña)2 – (ñb)2 = a – b (a, b 0)
EXAMPLE
16
Solution
Radicals
Multiply and simplify. a.
3 5 3– 5
b.
a.
3 5 3 – 5 (3 5) (3 – 5) 3 2 – ( 5) 2 9 – 5 4 2
b.
2 2 2 – 2 (2 2 ) (2 – 2 ) 2 2 – ( 2 ) 2 4 – 2 2
c.
a b a– b
2 2 2– 2
c.
a b a– b
a2 – b 23
EXAMPLE
17
Solution
Multiply and simplify. a. (ñ3 + ñ2) (ñ5 – 1)
b. (ñ5 + ñ3) (ñ7 + ñ2)
c. (2ñ3 + 1) (ñ5 + 1)
d. (3ñ2 – 2) (ñ5 – ñ3)
a. (ñ3 + ñ2) (ñ5 – 1)= (ñ3 ñ5) – (ñ3 1) + (ñ2 ñ5) – (ñ2 1) = ò15 – ñ3 + ò10 – ñ2 b. (ñ5 + ñ3) (ñ7 + ñ2) = (ñ5 ñ7) + (ñ5 ñ2) + (ñ3 ñ7) + (ñ3 ñ2) = ò35 + ò10 + ò21 + ñ6 c. (2ñ3 + 1) (ñ5 + 1) = (2ñ3 ñ5) + (2ñ3 1) + (1 ñ5) + 1 = 2ò15 + 2ñ3 + ñ5 + 1 d. (3ñ2 – 2) (ñ5 – ñ3)= (3ñ2 ñ5) – (3ñ2 ñ3) – (2ñ5 + 2ñ3) = 3ò10 – 3ñ6 – 2ñ5 + 2ñ3
5. Rationalizing Denominators 1
Look at the numbers
3
10
19
. They are all fractions, and each fraction 5 2 12 13 has an irrational number as the denominator. In math, it is easier to work with fractions that ,
,
, and
have a rational number as the denominator. Definition
Changing the denominator of a fraction from an irrational number to a rational number is called rationalizing the denominator of the fraction. Rationalizing the denominator does not change the value of the original fraction. To rationalize the denominator, we multiply the numerator and denominator of the fraction a by a suitable factor. For example, if the fraction is in the form , we multiply both the b numerator and the denominator by ñb. a
So,
b
a
b
b
b
ab
bb
a
ab . Note that b
b
and
ab have the same value: they are b
equivalent fractions. Look at some more examples: 3 2 3 3
3
2 2
3
2
2
3 5
24
3
3 5 2 2
2 3 3
2 2
3 2 2 2
3 3 3 3
3 2 2 2
6 4
6 , 2
3 3 3, and 3
3 52 2 2 2
3 10 3 10 . 22 4 Algebra 8
Definition
An expression with exactly two terms is called a binomial expression. Two binomial expressions whose first terms are equal and last terms are opposite are called conjugates, i.e. a + b and a – b are conjugates. If a 0 and b 0, then the binomials xña + yñb and xña – yñb are conjugates. We can use conjugates to rationalize denominators that contain radical expressions. 1 For example, let us rationalize . ñ3 – ñ2 is the conjugate of ñ3 + ñ2. 3 2 Therefore, we multiply the numerator and the denominator by ñ3 – ñ2 to rationalize the denominator. 1 3 2
3– 2 3– 2
1 ( 3 – 2) ( 3 2 ) ( 3 – 2 )
3– 2 2
( 3) – ( 2 )
2
3– 2 3– 2 3– 2 3–2 1
Remark
(a + b)(a – b) = a2 – b2 (ña + ñb)(ña – ñb) = a – b where a 0 and b 0. EXAMPLE
18
Rationalize the denominators. a.
Solution
a.
b.
5 3–2 2 5 3–2 2
3– 2 2 2 –1
6 2 1– 3
Radicals
3 2 –2 52 5
3–2 2
2 2 –1
32 2 3 2 2
6 2
d.
1– 3
5 (3 2 2 ) (3 – 2 2 )(3 2 2 )
3 2 –2 52 5
5 3 5 2 2 3 2 – (2 2 ) 2
3 5 2 10 3 5 2 10 3 5 2 10 9–8 1
( 3 – 2 ) (2 2 1) 3 2 2 3 1 – 2 2 2 – 2 1 (2 2 – 1) (2 2 1) (2 2 ) 2 – 12
d.
5
c.
c.
3– 2
b.
2 6 3–2 2 2 – 2 8 –1
2 6 3–4– 2 7
( 6 2 ) (1 3) 6 6 3 2 1 2 3 (1 – 3) (1 3) 12 – ( 3) 2 6 18 2 6 6 3 2 2 6 2 6 4 2 – 6 –2 2 1– 3 –2 –2
(3 2 – 2) (5 – 2 5) 3 2 5 – 3 2 2 5 – 2 5 – 2 2 5 (5 2 5) (5 – 2 5) 5 2 – (2 5) 2
15 2 – 6 10 – 10 – 4 5 15 2 – 6 10 – 10 – 4 5 25 – 20 5 25
EXAMPLE
19
Rationalize the denominators to find the sum. 3
3 2 2 3
Solution
3 2 2
2 3–2 2 3 3–2 2 2 3 2 2 3 – 2 2 3 2 2 3–2 2 3–2 2 3 2 2 2
( 3 ( 3 – 2 2 )) ( 2 ( 3 2 2 )) ( 3 2 2) ( 3 – 2 2) ( 3 3) – ( 3 2 2 ) ( 2 3) ( 2 2 2 ) ( 3)2 – (2 2 )2 3–2 6 6 4 7– 6 6 –7 3–8 –5 5
Check Yourself 2 1. Rationalize the denominators and simplify. a.
g.
3 7
5
b.
1
3
a b
h.
2 10
c. 2
d. –5
2 x y
i.
a3
1 2
e.
5
3 5
3 5
f.
3
2 6
a 3 b4
j.
x3 y3
3
a b2
2. Rationalize the denominators and simplify. a.
g.
1
2
b.
2 –1 2 5–2
3– 2 2 5– 7
h.
2 10 – 6
c.
i.
7– 5
2
3
d.
5 1 a
6 –2
a b
j.
a b
a– b
5 1
e.
f.
5 –1
3– 2 2 3– 2
3
k.
3 3–2 7
3. Rationalize the denominators and simplify. a. d.
1 5 2 2 2 1
–
1
b.
5– 2 2 2 –1
3
e.
2 2
2 2– 2 3
2 2 2
–
2
3 1 1– 3
1
c.
3 3
1 3– 3
3 3 3
Answers 1. a.
15 21 b. 3 7
b. ñ6 – 2 c. i. 26
30 4
c. ñ2 d. –ò15 e. ò15 f.
5 –1 3 6 6 d. 2 2
e.
3 5 2
f.
g.
4– 6 10
10 20
h.
b a
i.
2 xy
j. ab 2. a. ñ2 + 1
g. 5ñ2 – ò10 + 3 ñ5 – 3 h.
6 –16 3 2 a – ab a b 2 ab 2 5 j. k. –9ñ3 – 6ñ7 3. a. b. ñ2 c. d. 6 4 3 a–b a–b
35 3 2
e. 17 3 – 3 6 Algebra 8
EXERCISES
1 .1
1. Evaluate the square roots. a. ò36 16x2
d.
5. Perform the operations.
b. ó100 e.
c. –ó121
25 y2
f.
121 a4
a. 3ñ3 + 2ñ3
b. 6ñ5 + ñ5
c. –5ñx + 5ñx
d. ñ6 – 3ñ6
e. 3ò18 + 2ò72
f. ò80 – ó125 + ò45
g. ò75 + ó108 – ò48 + ò27
2. Simplify the expressions. a. ñ3 ñ3
b. ñ5 ñ5
c. ñ3 ò12
d. ñ3 ò27
e. ò2x ò8x
f.
g.
h. 3 2 x 4 18 xy2
3a 5a
6 xy 24xy
h.
9x3 16 x3 – 4 x 25 x
i.
2.25 – 2.89 1.44
j.
0.9 – 2.7 1.7 0.4
3. Simplify the expressions. 50
a.
d.
2
72 x3 2 2x
b.
e.
72 8 32 x3 y2 24 x
c.
f.
108
6. Write each expression in its simplest form.
27
3xy3
a.
32 2
b.
5–2 6
c.
8 2 12
d.
11 96
e.
8–2 7
f.
5 24
g.
7–4 3
h.
3 8 3– 8
i.
4 15 – 4 – 15
f.
a 2 8a – a
k.
2 2 4 12
3
12 x y
4. Write each number as a mixed radical. a. ñ8
b. ò72
c. ó243
d.
e. ó125
f.
Radicals
1000
x3 y2
27
7. Simplify the expressions. a.
32 21 – 23 4
b.
7 2 – 6 9 3 9
c.
d.
8 16
10. Rationalize the denominators. a.
3 3
d. 1 2 4 3
1 9 16 96
g. j.
13 6 6 9
3 2
b.
6
1
e.
3 1 3 –1 3
4 3 3 2
2 2
k.
2 3
f.
1– 2
h.
c. –
3– 5
i. l.
1 11 –2 3 2
10 2 7 –5 10 2 21 7 3
8. Find x in each equation. b. 3 3 3 3 ... x
a.
2 2 2 2 ... x
c.
3x 3x 3x ... 9
9. Find the products. a. ñ5 (ñ2 + ñ3) b. ñ7 (1 + ñ7) c. –ñ2 (ñ3 – ñ8 + 1)
11. Perform the operations. a. b. c. d.
d. ñ2 (ñ8 + ò32)
1 3
1
2
3 3 1
2
3 –1
5 2 3 – 11
4 2 3 2 2
–
2 2 3 11
2 3
–
3–2 2
e. ñ6 (2ñ3 + 3ñ2) f. (3 + ñ5) (3 – ñ5) g. (2ñ2 – 3) (2ñ2 + 3) h. (2ñ3+2) (2ñ3 – 2) i. (ò12 + ñ8) (ñ3 – ñ2) j. (–ò12 + 2ñ2) (ñ2 + ñ3) k. l. m. 28
2 1
2 –1
12. Perform the operations. a.
5–2
5 –1
1 2
b.
2 11 72
2 33 2 3–3 5 2 3 2 3 16 – 9 3
5 2 –
c.
1 2 1
1 3 2
1 4 3
...
1 100 99 Algebra 8
Objectives
After studying this section you will be able to: 1. Understand the concepts of nth root and rational exponent. 2. Write numbers in radical or rational exponent form. 3. Understand the properties of expressions with rational exponents. 4. Use the properties of rational exponents to solve problems.
A. RATIONAL EXPONENTS 1. nth Roots In section 8.2 we studied exponential equations. For example, 2n 2n = 2 is an exponential equation. Let us solve it. 2n 2n = 22n = 2
(an am = an+m)
22n = 21 2n = 1, n =
1
1
1 1 for n in the original equation we will get 2 2 2 2 2, . If we substitute 2 2 1
but we know that ñ2 ñ2 = 2. Therefore, 2 2 2 2 2. Remark
Let x R – {–1, 0, 1}. If xm = xn then m = n. Definition
For any natural number n and a, b R. 1
n If an = b then a = b n b. a is called the nth root of b. It is denoted by ñb. n
In the expression ña, a is called the radicand and n is called the index. n
Remember that we do not usually write the index for square roots: ña = ña.
2
Look at some examples of different roots: 52 = 25 and 5 = ò25
‘the square root of 25 is 5’,
2 = 8 and 2 = ñ8
‘the cube root of 8 is 2’,
3 = 27 and 3 = ò27
‘the cube root of 27 is 3’, and
2 = 16 and 2 = ò16
‘the fourth root of 16 is 2’.
3 3 4
Radicals
3
3 4
29
2. Rational Exponents Definition
If m and n are positive integers (n > 1), and b is a non-negative real number, n
m is called a rational exponent. n 2
1
m
bm b n .
2
For example, the numbers 8 3 , 4 2 , and 2 2 have rational exponents. EXAMPLE
20
Write the expressions in radical form.
Solution
b. 32 5
1
1
Solution
f.
1
( a 3 )2
2
c. 5 3 3 52 3 25
b. 32 5 5 32
3
21
e. ( x4 ) 4
1
a. 3 2 2 3 3
1
1
d. x 4
c. 5 3
d. x 4 4 x3 EXAMPLE
3
2
1
1
a. 3 2
1
4
e. ( x4 )4 x4
f.
1
3
( a 3 )2
a
Write the expressions with radical exponents. 3
a.
5
35
a.
5
35 3 5 31 3
d.
3
a2 a 3
b.
c.
64
5
7
3
d.
x14
a2
3
3
b.
3
64 43 4 3 41 4
e.
5
ab2 ( ab 2 )5 a 5 b 5
2
1
5
e.
1
ab2 7
c.
14
x14 x 7 x2
2
Rule
If b is any real number and n is a positive integer (n > 1): 1. If b > 0 then ñb is a positive real number. n
2. If b = 0 then ñb is zero. n
3. If b < 0 and n is odd, then ñb is a negative real number. n
4. If b < 0 and n is even, then ñb is not a real number. n
EXAMPLE
22
Solution
30
Simplify the following. b.
4
a.
3
8
a.
3
8 23 2 3 2
3
3
c.
81
3
(8 > 0) 3 3
–27
4
d. b.
4
e.
–16 4
4
81 34 3 4 3
c.
3
–27 3 (–3)3 (–3) –3
(–27 < 0 and 3 is odd)
d.
4
–16 is not a real number
(–16 < 0 and 4 is even)
e.
5
0 =0
5
0
(81 > 0)
Algebra 8
Check Yourself 3 1. Write the expressions in radical form. 1
1
2
3
a. 2 2
b. a 3
c. a 3
d. x 5
1
1
e. ( x2 y3 )6
1
1
1
g. ( a2 b2 ) 2
( x2 )3
f.
h. (( x y) x2 ) 3
2. Write the expressions with rational exponents. 3
a.
4
b.
2
4
5
c.
a2
3
d.
x y2
e.
625 e.
6
6
4
x4 f.
a8
g.
–512
g.
3
x2 y4 h.
12
x6 y4
3. Simplify the expressions. a. ò16
3
b.
3
27 c.
–64
d.
4
9
64 f.
x2 y4 h. 4 16 a4 b8
Answers 1. a. ñ2 b. 3 a c. 3 a2 d. 5 x3 e. 6 x2 y3 1
1
2
b. 2 2
c. a 5
d. ( xy2 )3
f. 6 x g. a2 b2 1
2
h. 3 ( x y)x2
1
e. x 3 f. a2 g. ( x2 y4 )3 h. ( x3 y2 )6
1
2. a. 2 3
3. a. 4 b. 3 c. –4 d. 5 e. 2
f. –2 g. xy2 h. 2ab2
B. PROPERTIES OF RADICALS Property
For all real numbers a and b, where a > 0 and b > 0, and for any integer m and n, where m > 1 and n > 1: 1.
2.
an a
if n is odd.
5.
n
a
n
an | a |
if n is even.
6.
m
a b
n
a b n a b
m
a
n
3.
n m
n
n
n
a
b n
n
7.
a b
4. a b
n
8. an b
and
n
an b a
n
n
am
n
b
m n
mn
mn
a
mn
an bm
mn
an
mn
mn
bm
m n
an bm
an bm
a
b
Look at the examples of each property. 1.
Radicals
3
b.
5
32 5 (2)5 –2
4
d.
4
(3)4 | 3| 3
a.
3
27 33 3
c.
4
16 2 4 | 2| 2
31
2.
a. ñ5 ñ4 = ó54 = ò20 5
c. 3.
8
8 4 2 2
2 4
4.
64 x2 4
4x
4
4
b. x x
5.
4
3
3
3
81 3 27 3 3
x4 x
4
3
24
x5 3
6
4 22 3
6.
81
23 2
x3 y6 z2 3 ( x y2 )3 z2 x y2 z2
6
a.
3
3
64 x2 4 4 4 4 16 x 4 2 4 x 2 4 x 2 x 4x
3
3
3
2 4 324 8
b.
a. 2 3 3 2 3 3 3 8 3
c.
3
3
x5 y5 xy
a. c.
b.
5 3 3
23
34
2
7.
a.
8.
a.
3
5
c.
4
x8
4
3
34
53
24 33
3 2
32
3
b.
22 2
2 3
12
32
6
5 3 32
6
125 9
b.
6
x8
8
3
3 2
52 25
32
a2
6
1125 3
2 4 12 16 27 33
5 5
4 2
6
5
a b
b.
4 3
32
b3
12
6
4 3 2
a2 b3
12
24
12
x8 x
Check Yourself 4 1. Simplify the expressions. a.
4
b.
16
4
(–2)4
c.
5
(–3)5
d.
7
–128
c.
5
x 2 x3
5
d.
3
x 3 x2 y2
4
x5
e.
3
27x3
2. Perform the operations. a. ñ3 ò12
b.
3
3
3 9
3. Simplify the expressions. 3
2
a.
b.
8
625 3
5
c.
4
d.
x
3 3
3
d.
4
x5 y6
d.
4
32 16
4. Simplify the expressions. a.
3
b.
40
3
81
c.
4
32
c.
3
a2 a3
5. Perform the operations. a. 32
3
2 4
b.
3
4
x x3
6
3
Algebra 8
6. Simplify the expressions. 3
a.
3
4
4
b.
3
4
6
c.
2
5
a3
d.
a2
6
x4
8
x3
7. Write each expression in its simplest form. 3
a.
b.
3
5 4
x
c.
3 3 3
x27
d.
4
x16
3
e.
3
3 3
Answers 1. a. 2 b. 2 c. –3 d. –2 e. 3x 2. a. 6 b. 3 c. x d. x 3 y2 4. a. 2 3 5 b. 3 3 3 c. 2 4 2 6. a. 12
256 27
b.
6
9 2
c. 20 a7
d. xy 4 xy2 d. 24 x7
5. a. 2 6 2 7. a. 6 3
3. a.
1 2
b. 5 c. x d. 3 9
b. x 12 x c. a 6 a
d. 4
12
27
b. 20 x c. x d. x2 e. 9 81
C. RADICAL EQUATIONS Definition
An equation that has a variable in a radicand is called a radical equation. For example, the equations ñx = 25,
x 1 3, and
2 x – 1 3 x 5 are radical equations.
Let us look at some examples of how to solve radical equations. EXAMPLE
23
Solution
Solve
x 1 3.
( x 1)2 32 (take the square of both sides)
Check:
x 1 9
?
8 13 ?
9 3
x8
33
Therefore, 8 is a solution. EXAMPLE
24
Solution
Solve
3x 1 5.
( 3x 1)2 52 3x 1 25 3x 24 x8
Radicals
Check:
?
3 8 15 ?
25 5 55
Therefore, 8 is a solution. 33
EXAMPLE
25
Solution
2 x 3 3 8.
Solve
2x 3 3 8
Check:
?
2 11 3 3 8
2x 3 8 – 3
?
25 3 8
2x 3 5
?
( 2 x 3)2 5 2
5 3 8
2 x 3 25
88
2 x 22
Therefore, 11 is a solution.
x 11 EXAMPLE
26
Solution
Solve
x2 12 x 6.
( x2 12 )2 ( x 6)2
Check:
?
(–2)2 12 – 2 6 ?
x2 12 x2 12 x 36
4 12 4
12 x –24
?
16 4
x –2
44
Therefore, –2 is a solution. EXAMPLE
27
Solution
Solve
3
3x 2 5.
3
( 3x 2 )3 5 3
Check:
3x 2 125
3
?
3 41 2 5 3
3x 123
?
123 2 5 3
x 41
?
125 5 5 5
Therefore, 41 is a solution. EXAMPLE
28
Solution
Solve
4x 1 – 5x – 1 0.
4x 1 – 5x – 1 0 ( 4 x 1)2 ( 5 x – 1) 2 4x 1 5 x – 1 x2
Check:
?
4 2 – 1 – 5 2 – 1 0 ?
9 – 9 0 ?
3 – 3 0 0 0
Therefore, 2 is a solution. 34
Algebra 8
EXAMPLE
29
3x 1 3x 6 5.
Solve
3x 1 5 – 3 x 6
Solution
(take the square of both sides)
( 3x 1)2 (5 – 3 x 6 ) 2
3x 1 25 – 2 5 3 x 6 3 x 6
(simplify)
10 3x 6 30 ( 3 x 6 )2 3 2 3x 6 9 3x 3 x1
Check:
?
3 1 1 – 3 1 6 5 ?
4 9 5 ?
2 35 5 5
Therefore, 1 is a solution.
Check Yourself 5 1. Solve each equation and check your answer. a. ñx = 5
b. ñx + 1 = 3
d. ñx = –3
e.
2x – 1 7
f.
3 x 1 –1 8
g.
3x – 5 4 6
h.
5x 1 –30 2
i.
2x 6 – x
j.
4x – 3 – 2 x – 2 0
k.
x3 x–2 5
l.
x2 1 2 – x
n.
x 2
o.
m.
3
x 1 3
p.
4
x 2x – 1
3
c. óx+1 = 6
3
3
3x 2 5x
4
Answer 1. a. 25 b. 4 c. 35 d. e. 25 f. 8 g. 3 h. Radicals
1 17 i. 2 j. k. 6 l. 3 m. 26 n. 2ñ2 o. 1 p. 1 2 5 4 35
EXERCISES
1 .2
1. Write the expressions in exponential form and simplify if possible. a. ò21 3 5
e.
7
b. f.
2
2
7
c.
5
( 3)6
g.
3
3
c. ò7x = x
1 3
3 –3 c. ( ) 2 2
2
1
b. b 3
d.
3
4x 6
5x – 4 6
f.
3
7x 6 5
g.
5x – 1 3 7
h.
2x – 3 – 2 4
i.
x2 5 x 1
j.
x2 9 5
7x – 6 4
l.
2x – 5 x – 4
x x 16 – 2
n.
6 x
e.
d. x
3 2
ab c
3. Simplify the expressions. 3
3
b.
4 5
d.
3
5 5
e.
5
g.
3
–27
h.
3
j.
3
3
2 4 3
m.
p.
k.
2
3
250
3
4
q.
c.
1016
243
f.
(–5)4
i.
a3 a2
4
3 27 21
6
7
n.
3 9 3
a2
4 3
4
2
3 3
2
o.
4
2
r.
3 3
s.
c. e.
3
4
4 3
1 9
8x – 2 6
1 4x – 3
1
1
2– x
–
2 x
2
6
11 – 3 4 – 27
d.
2
17 6 8
f.
3
9
2 x1
6. Simplify the expressions. 3
2 3 6
108
33x – 6
5
b.
6
3
x 1 2 x x 1– 2 x 8 3
q.
2
5
3
3 2
p.
162
l.
2
o. ñ6x – x= 5
3
4. Perform the operations. a.
3
m.
64x6
4
3
1
e. ( a )6
k.
a.
3x – 14 8
b.
3
2. Write the expressions in radical form. a. a 2
2x 1 3
a. 2
d. a x
5
5. Solve the equations.
7 3 10 2 9 27
( 2 – 6 )( 2 3 )
a. 23 24
b. 53 52
c. 43 42 4
d. 26 2–4 22
e. 3x 3–2x 33x+1
f. 2x 3x 5x
g. (x – y)2 (y – x)3 (x – y)–4 h. 298 + 298 + 298 + 298
36
Algebra 8
7. Simplify the expressions. a.
25 23
b.
x 1
d.
10. Solve the equations for x.
53 54
c. –3
a b x 1
e.
5
(–2) (–2)–4
f.
a. 4x+1 = 23x – 4
ax –1 ax – 2 5 2 104
4
c. 9x –2 (
b. 3x – 4 = 81
1 3– x ) 27
d. 8x+1 = 42x – 3
e. (2x – 3)3 = (x + 1)3 f. (2x – 4)6 = x6 g. 5 23x – 1 – 23x+1 = 256
8. Perform the operations. a. (–2)5 (–2)3
b. –32 (–3)3 (–3)
c. (–4)2 (–22)3 (–2–2)4
d.
(–2 3 )–3 (–2 2 )6 1 (– )3 2
e. (–a)7 (–a4) (–a)–2
h.
12 x 12 x 12 x 81 4x 4x 4x
i.
3 4x – 3 4 x –1 –
j.
3 93 x –1 ((–3)2 ) –4 81 (243) x
22 x 128 4
11. 3a = 25 and 3b = 5. Find
a . b
12. 3x = 4. Find 92x + 3x+1.
f. (–a–3)–2 (–a2)–3 (–a–1)–1
13. 2x = 3y = a. Write (12)xy in terms of a, x, and y. 14. 2a – 3 and 4b + 7 are integers with 32a – 3 = 54b + 7. Find a + b.
9. Simplify the expressions. 2
15. (x + 1)x
a. 2x – 1 + 2x – 1 + 2x – 1 + 2x – 1
– 16
= 1. Find the sum of the possible
values of x.
b. ax + 2 ax – 3ax c. 3x+1 + 3x – 1 + 3x+2 – 3x – 2
4
16. 44 = 16x. Find x.
10 x 10 x 10 x 10 x d. 5x 5x 5x 5x
e.
(2 –7 3–5 )2 36 –7
f.
313 311 – 39 312 310
g. (–1)101 (–1)125 (–1)100 (–1)99 (–1)49 h.
3
–4 0
((–39) (–2) ) (–1) (–3)125 6 –127 2126
Radicals
17. |x + y – 3| + |x – y – 1| = 0. Find x. 18. Simplify |x – 4| + 2|3 – x| if 3 < x < 4.
2003
37
19. Find the sum of the possible values of x that satifsfy the equation |(|2x + 4|) – 2| = 4.
22 . x 72 72 72 ... and y 42 – 42 – 42 – ... . Find x + y.
20. Find the product of the possible values of x that satisfy the equation |2x + 3| = |5 – 3x|.
23 . 2a 2a 2a ... 12 – 12 – 12 – ... 3. Find a.
21. Simplify the expressions. 3
a.
4
(–3)2 – –8 16
24 .Simplify
72 48
b.
2.89 6.25 – 1.96 2
d.
5 –1
2
1
4
2 4
1
4 6
...
1 252 254
1 254 256
27. How many different natural number solutions
12 2 16 2
5
g.
7– 2
h.
does the equation ñx + ñy = ó300 have? 5
2 7
28. ( 2 1)7 57122 57121. Find (ñ2 – 1)7.
4 12 4 – 12
(Hint: (
1 2 –1
) 2 1 )
( 6 20 ) ( 5 – 1) 2
x y x – y 2
k.
38
4
3
f.
m.
4
18 3 29 – 16
e.
l.
4
25 . x2 – x2 – x2 –... 2. Find x. 26 .Simplify
5 1 3
j.
3
27 11 – 2 30 30 30 ... .
54
c.
i.
4
2
xy if x < y < 0 x–y
2.7 5.4 7.1 2 0.2 4
15 – 6
4
15 6 – 3
4 7 4– 7
29.
1
15 1 9 16 9 24 – 16 . Find x . 8 x x x
30. Evaluate
31.
3
(50 51 52 53) 1.
3
x 2 – x – 2 4. Find x2.
(Hint: (a – b)3 = a3 – 3ab(a – b) – b3) Algebra 8
.
CHAPTER REVIEW TEST
1A
1. Simplify (|(–(–2)3) – 2|) – [3 |–|(–3)2 – 3||]. A) –8
B) –12
C) 8
D) 28
6. |x – 2| = |x + 3|. Find the value of x which satisfies the equation. A) –
2. Calculate |1 – ñ2| + |1 + ñ2|. A) 0
B) ñ2
C) 2
1 4
D) 2ñ2
3. How many elements are there in the solution set of |x – 2| + 1 = 0? B) 1
C) 2
D) 3
B) 2x – 2
C) 2
C) 1
D)
D) 4
5. ||x + 1| – 5| = 3. Find the sum of the possible
following numbers is negative? B) a–2
1
C) –a–3
D) –(–a)3
–1
8. Simplify (– )3 . 2 A) –
1 8
B)
1 8
C) –8
D) 8
A) 23x
10. Simplify
B) 24x
C) 29x – 1
D) 28x+1
4m 4m . 2m 2m 2m 2m
values of x. A) –3 Chapter Review Test 1A
3 2
9. Simplify 4x+1 8x – 1 16x.
4. –1 < x < 3. Calculate |x + 1| + |x – 3|. A) 2x
1 2
7. a is a positive real number. Which one of the A) a–1
A) 0
B) –
2 m 1
B) –2
C) 2
D) 4
A) 22m
B) 2m +1
C) 2m – 1
D) 2 m 2 39
11.
16. Evaluate (3ñ2 – ò12) (ò18 + 2ñ3).
3x 1 3x 2 3x 3 27. Find x. 39
A) 1
12. 2 a – b 2 –
B) 3
1 2 2 a b 1
A) 0
C) 13
D) 27
17. Evaluate
0. Find a.
B) –1
C) –3
D) 3
13. a = 2x, b = 3x, and c = 5x. Express 90x in terms of a, b, and c. A) (abc)
2
15. Evaluate A) –1 40
A) –ñ3
18. Evaluate A) 1
2
B) a bc
2
2
C) ab c
2
B) 8ñ2
D) 26ñ2
2.25 – 2.89 1.44.
B)
1 2
C) 1
D) 6
4 – 12.
B) ñ3
C) ñ3 – 1
D) ñ3 + 1
2 2 4 12 .
B) ñ3
C) ñ3 – 1
D) ñ3 + 1
D) a b c
19. Evaluate
C) ò38
C) ñ6
B) 180
4 2
14. Evaluate 3ñ8 – 2ñ2 + ò32. A) ñ6
A) 36
A)
3 2
20. Evaluate D)
2 3
A) 7
27 m3 9 m2 . 12 m3 4 m2
B) ñ2
5 7– 2
B) 2ñ7
9 4
C)
5 7 2
D) 3m+1
.
C) 5ñ7
D) 5 Algebra 8
CHAPTER REVIEW TEST 1. Evaluate A) –
(–1)55 (–172 ) – (–1) 64 . (–121 ) – (–1)82 – (–1)15
3 2
2. Evaluate
1B
B) –3
6. Evaluate
C) 1
D) –1
2 –2 4–1 . 2 –2 – 3–1
1 B) 3
A) 3
A)
569 – 568 . 568 567
1 6
5 6
B)
C)
5 3
D)
10 3
7. 3x = 5, 2y = 3, and 5z = 0.125. Find x y z. A) 3 C) 2
–1
B) 10
C) –3
D) 30
C) 3
D) 4
D) –6
8. 3x+1 + 2 3x – 1 = 33. Find x. 3
3.
x(2 ) 16. Find |x|. ( x2 )3
A) 2
B) 4
A) 1 C) 8
B) 2
D) 16
3
2
9. a = (52)3, b = 5(2 ), and c = 5(3 ). Which statement is true?
4. What is half of 220? A) 210
B) 221
5. Find the simplest form of A) 3x
B)
Chapter Review Test 1B
1 3x
C) 219
D)
1 210
3x 3x 3x . 9x 9x 9x
C) 32x
A) a > b > c
B) a > c > b
C) b > c > a
D) c > b > a
10. k = 1254 642. How many digits are there in the number k? D) 3x+1
A) 10
B) 11
C) 12
D) 13 41
11.
32 x 243x 9x 3. Find x. 81x1
A)
2 9
B) 2
41 A) 16
12 48 – 27 75 – 2 3
A) 2ñ3
15.
4
B) –1
4
C) 1
3
B) 2
19. Evaluate C) 3
D) 4
A) ñ8 6
3
11 24
B)
4 7
20. C)
11 7
D)
1 6
5
8 x 1
3 7
42 x 3
A) 9
.
B) –2
C) –ñ2
11 2 30 – 8 2 15 3–2 2
D) 4
.
B) ñ6 – ñ3 D)
3 5 – 3– 5 2
A) ñ5
D) 3ñ3
13 13 – 64 .
2 –2
C) ñ3
18. Evaluate
.
2 2
1
3 15
A)
2 D) 3
x x2 x xn . Find n.
A) 42
3 C) 2
1
A) 2ñ2
D) 10
17. Evaluate
41 B) 4
13. Simplify
A) 1
C) 9
9 1 1 . 16
12. Evaluate
14. Evaluate
16. Evaluate
B) 2ñ5
3
3 2
.
C) 2
D) 1
3
2 4 2 4 ... .
B) 2
C) 8
D) ñ8
C) 114
D) 120
3
2. Find x.
B) 105
Algebra 8
Objectives
After studying this section you will be able to: 1. Define statistics as a branch of mathematics and state the activities it involves. 2. Describe some different methods of collecting data. 3. Present and interpret data by using graphs. 4. Describe and find four measures of central tendency: mean, median, mode, and range.
A. BASIC CONCEPTS 1. What is Statistics? Statistics is the science of collecting, organizing, summarizing and analyzing data, and drawing conclusions from this data. In every field, from the humanities to the physical sciences, research information and the ways in which it is collected and measured can be inaccurate. Statistics is the discipline that evaluates the reliability of numerical information, called data. We use statistics to describe what is happening, and to make projections concerning what will happen in the future. Statistics show the results of our experience. Many different people such as economists, engineers, geographers, biologists, physicists, meteorologists and managers use statistics in their work.
Definition
statistics Statistics is a branch of mathematics which deals with the collection, analysis, interpretation, and representation of masses of numerical data. The word statistics comes from the Latin word statisticus, meaning ‘of the state’. The steps of statistical analysis involve collecting information, evaluating it, and drawing conclusions. For example, the information might be about:
44
what teenagers prefer to eat for breakfast; the population of a city over a certain period; the quality of drinking water in different countries of the world; the number of items produced in a factory. Algebra 8
The study of statistics can be divided into two main areas: descriptive statistics and inferential statistics. Descriptive statistics involves collecting, organizing, summarizing, and presenting data. Inferential statistics involves drawing conclusions or predicting results based on the data collected.
2. Collecting Data We can collect data in many different ways.
a. Questionnaires A questionnaire is a list of questions about a given topic. It is usually printed on a piece of paper so that the answers can be recorded. For example, suppose you want to find out about the television viewing habits of teachers. You could prepare a list of questions such as:
Do you watch television every day? Do you watch television: in the morning? in the evening?
What is your favourite television program? etc. Some questions will have a yes or no answer. Other questions might ask a person to choose an answer from a list, or to give a free answer. When you are writing a questionnaire, keep the following points in mind:
1. A questionnaire should not be too long. 2. It should contain all the questions needed to cover the subject you are studying. 3. The questions should be easy to understand. 4. Most questions should only require a ‘Yes/No’ answer, a tick in a box or a circle round a choice.
In the example of a study about teachers’ television viewing habits, we only need to ask the questions to teachers. Teachers form the population for our study. A more precise population could be all the teachers in your country, or all the teachers in your school.
Statistics and Graphics
45
b. Sampling A sample is a group of subjects selected from a population. Suppose the population for our study about television is all the teachers in a particular city. Obviously it will be very Xdifficult to interview every teacher in the city individually. Instead we could choose a smaller group of teachers to interview, for example, five teachers from each school. These teachers will be the sample for our study. We could say that the habits of the teachers in this sample are probably the same as the habits of all the teachers in the city.
population
sample
A sample is a subset of a population.
The process of choosing a sample from a population is called sampling. The process of choosing a sample from a population is called sampling. When we sample a population, we need to make sure that the sample is an accurate one. For example, if we are choosing five teachers from each school to represent all the teachers in a city, we will need to make sure that the sample includes teachers of different ages in different parts of the city. When we have chosen an accurate sample for our study, we can collect the data we need and apply statistical methods to make statements about the whole population.
c. Surveys One of the most common method of collecting data is the use of surveys. Surveys can be carried out using a variety of methods. Three of the most common methods are the telephone survey, the mailed questionnaire, and the personal interview.
3. Summarizing Data In order to describe a situation, draw conclusions, or make predictions about events, a researcher must organize the data in a meaningful way. One convenient way of organizing the data is by using a frequency distribution table. A frequency distribution table consists of two rows or columns. One row or column shows the data values (x) and the other shows the frequency of each value (f). The frequency of a value is the number of times it occurs in the data set. For example, imagine that 25 students took a math test and received the following marks.
46
8
7
9
3
5
10
8
10
6
8
7
7
6
5
9
4
5
9
6
4
9
3
8
8
6 Algebra 8
The following table shows the frequency distribution of these marks. It is a frequency distribution table. mark
(x)
1
2
3
4
5
6
7
8
9
10
frequency
(f)
0
0
2
2
3
4
3
5
4
2
Note The sample size is the number of elements in a sample. It is denoted by n. We can see from the table that the frequency of 7 is 3 and the frequency of 8 is 5. The sum of the frequencies is equal to the total number of marks (25). The number of students took test is called the sample size (n). In this example the sample size is 25. The sum of the frequencies and the sample size are the same. EXAMPLE
1
Twenty-five students were given a blood test to determine their blood type. The data set was as follows: A
B
AB
B
AB
A
O
O
AB
A
B
O
O
O
B
AB
A
O
B
O
O
B
AB
B
O
Construct a frequency distribution table of the data and find the percentage of each blood type. Solution
There are four blood types: A, B, O, and AB. These types will be used as the classes for the distribution. The frequency distribution table is: class
frequency
percent
A
4
16 %
B
7
28 %
O
9
36 %
AB
5
20 %
Total 25
Total 100
We can use the following formula to find the percentage of values in each class: f % 100% where n f = frequency, and n = total number of values (25). For example, in the class for type A blood, the percentage is 4 100% 16%. 25
Statistics and Graphics
47
B. PRESENTING AND INTERPRETING DATA A graph is a diagram that relates two or more different types of information.
When we have collected, recorded and summarized our data, we have to present it in a form that people can easily understand. Graphs are an easy way of displaying data. There are three kinds of graph: a line graph, a bar graph, and a circle graph (also called a pie chart).
1. Bar Graph The most common type of graph is the bar graph (also called a histogram). A bar graph uses rectangular bars to represent data. The length of each bar in the graph shows the frequency or size of a cooresponding data value.
48
Mark
Maths
9
Physics
7
Chemistry
7
Biology
8
Computer
10
History
5
Music
6
Music
History
Computer
10 9 8 7 6 5 4 3 2 1 0 Mathematics
Then we can draw bars to show the marks for each subject.
Subject
Biology
We begin by drawing a vertical scale to show the marks and a horizontal scale to show the subjects.
Chemistry
Solution
The following table shows the marks that a student received at the end of the year in different school subjects. Draw a vertical bar graph for the data in table.
Physics
2
Marks
EXAMPLE
Lessons
Algebra 8
2. Line Graph We can make a line graph (also called a broken-line graph) by drawing line segments to join the tops of the bars in a bar graph.
Music
History
Computer
Biology
Chemistry
Physics
Music
History
10 9 8 7 6 5 4 3 2 1 0 Mathematics
Lessons
Computer
Biology
Chemistry
Physics
Marks
10 9 8 7 6 5 4 3 2 1 0 Mathematics
Marks
For example, look at the line graph of the data from Example 5.2.
Lessons
To draw the line graph, we mark the middle point of the top of each bar and join up the points with straight lines.
EXAMPLE
3
Statistics and Graphics
The following table shows the number of cars produced by a Turkish car company between 1992 and 2000. Draw a bar graph and a line graph of the data in this table.
Car Production Year
Production
1992
110 659
1993
133 006
1994
99 326
1995
74 862
1996
65 007
1997
91 326
1998
88 506
1999
125 026
2000
140 159
49
160 140 120 100 80 60 40 20 1997
1998
1999
2000
1997
1998
1999
2000
1996
1995
1994
1993
0 1992
Number of cars (10 000)
Year 160 140 120 100 80 60 40 20 1996
1995
1994
1993
0 1992
First we need to choose the axes. Let us put the years along the horizontal axis and the production along the vertical axis of the graph. It will be difficult to show large numbers such as 133 006 on the vertical axis. Instead, we can choose a different unit for the vertical axis, for example: one unit on the axis means 10 000 cars. We write this information when we label the axis.
Number of cars (10 000)
Solution
Year
6
3
6%
6.4% 5.3%
5.2%
5 4
6%
3.4%
2 1
Germany
0 Canada
a. Which country spent the largest percentage of its GDP on education?
7%
7
Australia
domestic product. Look at the graph and answer the questions.
Share of education expenditures as a percentage of GDP in selected countries *
Norway
The graph below shows the amount of money that seven different countries spend on education in 2003, as a percentage of each country’s gross
8
United Kingdom
The gross domestic product (GDP) of a country is the total value of new goods and services that the country produces in a given year.
USA
4
Turkey
EXAMPLE
* Source: Education at a Glance 2003 - OECD indicators
b. Which country spent the smallest percentage of its GDP on education? c. Find the percentage difference between the countries which spent the largest and smallest percentage of their GDP on education. d. Which countries spent the same percentage of their GDP on education? 50
Algebra 8
Solution
a. The USA spent the largest percentage (7% of its GDP). b. Turkey spent the smallest percentage (3.4% of its GDP). c. 7 – 3.4 = 3.6% d. Norway and Australia spent the same percentage: both countries gave 6% of their GDP.
Practice Problems 1. The bar graph below compares different causes of death in the United States for the year 1999. Look at the graph and answer the questions. Comparative causes of annual deaths in the United States (1999)*
Cause * Source: World Health Organization
a. What was the most common cause of death? b. What was the least common cause of death? c. What is the ratio of the number of deaths caused by smoking to the number of deaths caused by alcohol? d. How many deaths are shown in the graph? e. On avarage, how many people died per day from each canse in 1999? (Hint: There were 365 days in 1999.)
Answers 1. a. Smoking b. Drug Induced c. Statistics and Graphics
16 3
d. 632000 e. 1732
51
EXERCISES
2 .1
1. The set of quiz scores in a class is as follows. 8
5
6 10 4
7
2
7 6
3
5
9
2
4
6
6 8
4 10 8
6
5
1
7
Construct a frequency distribution table for this data.
2. A student’s expenses can be categorized as shown in the table. Expenses Food Rent
4. The following table shows the amount of sea fish caught in Turkey in 2003. Fish Anchovy
Quantity (1000 tons) 416
Horse Mackerel
295
Scad
16
Gray mullet
12
Blue fish
11
Pilchard
11
Whiting
12
Hake
8
Other
32
Source: Turkey’s Statistical Yearbook 2004
Present this information in a circle graph.
Percent of total income. 30% 27%
Entertainment
13%
Clothing
10%
Books
15%
Other
5%
Present this information in a bar graph.
5. The following bar graph shows the hazelnut production in Turkey from 1999 to 2003. Use the graph to answer the questions. 700 000
Hazelnut production in Turkey (tons)
600 000 500 000
100 000
2003
200 000 2002
Number of class members 8
2001
Sport Football
300 000
2000
chosen by each of forty students in a class.
400 000
1999
3. The following table shows the favorite sport
Basketball
5
Volleyball
7
a. Estimate the total production for all five years.
Swiming
12
Wrestling
3
b. Which year had the highest production?
Karate
2
Judo
4
Present this information in a circle graph. 52
c. Find the combined production for 2002 and 2003. d. Draw a broken line graph of the data. Algebra 8
Definition
An equation that can be written in the form a0
ax2 + bx + c = 0, is called a quadratic equation. I’m sick of being an unknown
In the equation, a, b, and c are real number coefficients and x is a variable. A quadratic equation written in the form ax2 + bx + c = 0 is said to be in standard form. Sometimes, a quadratic equation is also called a second degree equation. For example,
1 x+3=0 2 are all quadratic equations. By the definition of a quadratic equation, a cannot be zero. However b or c or both may be zero. For instance, 2x2 = 0 and x2 – 9 = 0 3x2 + 5x = 0,
2x2 – x – 1 = 0 and ñ2x2 –
x2 + 3x – 5 = 0,
are also quadratic equations. We can see that quadratic equations are formed by second-degree polynomials. Polynomials of a different degree do not form quadratic equations. Let us look at the coefficients a, b, and c of some quadratic equations. Equation
a
b
c
3x2 + 5x – 9 = 0
3
5
–9
1 – x + 3x2 = 0
3
–1
1
ñ2x2 + 5x = 0
ñ2
5
0
–1
0
1 2
1 – x2 = 0
–1
0
1
(ñ3 + 1)x2 = 0
ñ3 + 1
0
0
x2 x 1 – + =0 2 3 4
1 2
– x2 +
EXAMPLE
54
1
1 =0 2
–
1 3
1 4
Determine whether the following equations are quadratic or not. 1 2 x – 2 x +5 = 0 2
a. x2 + 1 = 0
b.
d. x2 – 2x–1 + 3 = 0
e. (x – 1)(x + 2) = 0
c. 2x2 – 3x = 5 f. (x – 2)x2 = 0 Algebra 8
Solution
To the best of our knowledge, the origin of the term ‘quadratic’ is Latin. It is derived from quadratus which is the past participle of quadrare which means ‘to make square’. From this it is clear that part of the word is connected to the Latin word for ‘four’: it refers to squaring, and a square is a regular four-sided figure.
a, b, c, and e are quadratic equations. Equation d is not quadratic, since the power of x is –1, which does not meet the requirements for a quadratic. Equation f is a third degree equation, so it is not quadratic. To solve a quadratic equation we must find the values of the unknown x which make the left-hand and right-hand sides equal. Such values are called the solutions or roots of the quadratic equation. A number of techniques are available to help us obtain a solution to any quadratic equation.
A. SOLVING EQUATIONS OF THE FORM ax2 = 0 We have seen that the solutions (or roots) of a quadratic equation are the values of x that make the two sides of the equation equal. Let us find the roots of the simple quadratic ax2 = 0. ax2 = 0 (a 0) x2 = 0 xx=0 x = 0 or x = 0 x1 = x2 = 0 We can see that this equation has two equal roots. When the roots of a quadratic equation are the same, we say that the equation has a double root.
If A B = 0, then A = 0 or B = 0.
EXAMPLE
2
Solution
3 Solve the equation x2 0. 2 3 x2 0 2 x2 0 x1 x2 0
Quadratic Equations
55
B. SOLVING EQUATIONS OF THE FORM ax2 + bx = 0 Let us look at the solution of this more complex quadratic. ax2 + bx = 0 x(ax + b) = 0 x = 0 or ax + b = 0, so b x1 = 0 and x2 = – . a This kind of quadratic equation has two roots and one of them is always zero. EXAMPLE
3
Solution
Solve the equations. a. x2 + x = 0
b. –4x2 + 5x = 0
c. ñ3x2 – 2x = 0
a. x2 + x = 0
b. –4x2 + 5x = 0
c. ñ3x2 – 2x = 0
x(x + 1) = 0
x(–4x + 5) = 0
x(ñ3x – 2) = 0
x = 0 or x + 1 = 0
x = 0 or –4x + 5 = 0
x = 0 or ñ3x – 2 = 0
x1 = 0 or x2 = –1
x1 = 0 or x2 =
5 4
x1 = 0 or x2 =
2 3
=
2 3 3
C. SOLVING EQUATIONS OF THE FORM ax2 + c = 0 Look at the calculation. ax2 + c = 0,
c0
2
ax c x2
c a
x
c a
Here the sign of
c is important. a
c > 0, the equation has no real solution, because we cannot find the square root of a a negative number.
If
c < 0, the equation has two real solutions. These roots are symmetric, i.e. they are the a same numeral with opposite signs.
If
Note All positive real numbers have two square roots. One root is the positive square root and the other root is the negative square root, i.e. if a2 = b and a is a positive real number, then a = ñb. 56
Algebra 8
EXAMPLE
4
Solution
Solve the equations. a. 3x2 – 27 = 0
b. 2x2 + 6 = 0
c. 7 – 4x2 = 2
a. 3x2 – 27 = 0
b. 2x2 + 6 = 0
c. 7 – 4x2 = 2
3x2 = 27
2x2 = –6
x2 = 9
x2 = –3
x = 3
4x2 = 5 x2 =
no real solution (x2 cannot be negative)
x=
Check Yourself 1
5 4
5 5 = 4 2
Solve the equations. 1. –3x2 = 0
2. 5x2 – 20x = 0
3. 7x2 + 35 = 0
4. 2x2 – 8 = 0
Answers 1. 0 2. 0, 4 3. no real solution 4. 2 FIND THE MISTAKE! Let a and b be two arbitrary numbers such that a b. Then (a – b)2 = a2 – 2ab + b2 = b2 – 2ab + a2 (a – b)2 = (b – a)2 a–b=b–a 2a = 2b a = b. Can you find the mistake in this working?
D. SOLVING EQUATIONS OF THE FORM ax2 + bx + c = 0 There are three basic methods for solving a quadratic equation of the form ax2 + bx + c = 0: 1. factoring, 2. completing the square, and 3. using the quadratic formula.
1. Factoring If we can write ax2 + bx + c = 0 as the product of two linear factors, then we can easily solve the equation. To solve a quadratic equation by factoring, follow the steps. 1. Write the equation in standard form, ax2 + bx + c = 0, a 0. 2. Factor the left side of the equation. Quadratic Equations
57
3. Apply the zero product property, that is, set each factor equal to zero. 4. Solve each equation to obtain the roots. EXAMPLE
5
Solve by factoring. a. x2 + 3x + 2 = 0
Solution
b. 6x2 – 19x – 7 = 0
a. x2 + 3x + 2 = 0
c. 2x2 = x + 3
b. 6x2 – 19x – 7 = 0 (2x – 7)(3x + 1) = 0 2x – 7 = 0 or 3x + 1 = 0 7 1 x1 = , x2 = – 2 3
(x + 2)(x + 1) = 0 x + 2 = 0 or x + 1 = 0 x1 = –2, x2 = –1 c. 2x2 = x + 3 2x2 – x – 3 = 0 (x + 1)(2x – 3) = 0 x + 1 = 0 or 2x – 3 = 0 3 x1 = –1, x2 = 2
Note When you are solving an equation, do not divide both sides by an expression containing the variable for which you are solving. You may be dividing by zero. For example, to solve x2 – 2x = 0, do not divide both sides by x, because x may be zero and you will also lose one of the solutions.
Check Yourself 2 Solve the following equations. 1. 3x2 = 5x + 2 Answers 2 1 1. , 2 2. 2 , 5 3
2. (5x – 1)(x + 2) = x + 2
3. 4x(x + 1) = 3
3 1 3. , 2 2
2. Completing the Square The idea behind this method is to adjust the left side of the equation ax2 + bx + c = 0 so that it becomes a perfect square, that is, the square of a first-degree polynomial. Expressions in the form x2 + 2xy + y2 and x2 – 2xy + y2 are perfect square polynomials. Numbers whose square roots are integers or quotients of integers are perfect squares.
58
For example, x2 + 6x + 9 and x2 – 4x + 4 are perfect squares, because x2 + 6x + 9 = (x + 3)2 and x2 – 4x + 4 = (x – 2)2. To write a quadratic equation as a perfect square, follow the steps, Algebra 8
1. Make sure a = 1 in the quadratic. If it isn’t 1, divide each term by a.
x+2 x
1
1
2. Rewrite the equation so that the constant term is alone on one side of the equation. 3. Take half of the coefficient of the x term and square it.
x
x+2
4. Add this number to both sides of the equation.
1
5. Factor the left-hand side into a perfect square.
1
2
6. Solve for x by using the square root property. x2 + 4x
Let us look at some examples of completing the square.
(x + 2)2 = x2 + 4x + 4
Start _____________
Add _____________
Result ___________________________
x2 + 4x
4
x2 + 4x + 4 = (x + 2)2
x2 + 12x
36
x2 + 12x + 36 = (x + 6)2
x2 – 6x
9
x2 – 6x + 9 = (x – 3)2
x2 + x
1 4
x2 + x +
2
4
1 1 = (x + )2 2 4
Note The expression x2 + 2bx is equivalent to (x + b)2 – b2. EXAMPLE
6
Solution
Solve by completing the square. a. x2 + 6x – 7 = 0
b. 2x2 – 4x + 1 = 0
a. x2 + 6x – 7 = 0
b. 2 x2 – 4x+ 1= 0 1 x2 – 2 x+ = 0 2 1 x2 – 2 x = – 2 1 x2 – 2 x+ 1= – + 1 2 1 (x – 1)2 = 2 1 x – 1= 2
x2 + 6x = 7 x2 + 6x + 9 = 7 + 9 (x + 3)2 = 16 x + 3 = ò16 x1 = –7,
x2 = 1
x1 =1 – Quadratic Equations
2 , 2
x2 =1+
2 2 59
Check Yourself 3 Solve the equations. 1. x2 + 8x – 3 = 0
2. 2x2 – 5x – 3 = 0
3. 2x2 – 2 = 4x
Answers 1. –4 ò19
2. 3,
1 2
3. 1 ñ2
3. The Quadratic Formula The final method will work on any quadratic equation. Therefore, we can use it when the other easier methods fail or are not easy to apply. Look at the derivation of the quadratic formula. ax2 + bx+ c = 0
(a 0, a, b, c )
b c x2 + x+ = 0 a a
(divide both sides by a)
b c x2 + x = – a a
(try to get a perfect square)
b b c b x2 + x+( )2 = – +( )2 2a 2a a a
(add (
(x+
b 2 ) to both sides) 2a
b 2 4ac+ b2 ) = 2a 4a2
x+
b b 2 4ac b 2 4ac = = 2a 2a 4a2
x=
b2 4ac b b 2 4ac –b = , 4a2 2a 2a
x1
b+ b 2 4ac b – b 2 4ac , x2 2a 2a
QUADRATIC FORMULA If ax2 + bx + c = 0, a 0, then x =
EXAMPLE
60
7
– b b 2 – 4 ac 2a
Solve x2 + 2x – 8 = 0 using the quadratic formula. Algebra 8
Solution Make sure that you write a quadratic equation in standard form before you identify the values a, b, and c.
EXAMPLE
8
Solution
First we identify the coefficients a, b, and c. For this equation a = 1, b = 2, and c = –8. Let us substitute the values of a, b, and c into the quadratic formula. x= x1 =
9
Solution
2 6 4 = = 2, 2 2
x2 =
2 6 –8 = = –4 2 2
Solve 3x2 + 2x – 4 = 0. For this equation a = 3, b = 2, and c = –4. x=
EXAMPLE
2 b b 2 4ac 2 2 4 1 (–8) 2 36 = = 2a 2 1 2
2 b b2 4ac 2 2 4 3 (–4) 2 52 = = 2a 23 6
x1 =
2 + 52 2 + 2 13 1+ 13 = = 6 6 3
x2 =
2 52 2 2 13 1 13 = = 6 6 3
Find the real solutions of the equation 9 +
3 2 = 0 , x 0. x x2
3 2 = 0 , is not a quadratic equation. However, x x2 we can make it quadratic by multiplying each side by x2, since x 0. The result is
In its present form, the equation 9 +
9x2 + 3x – 2 = 0, x 0. Now, a = 9, b = 3, and c = –2. b2 4ac = 32 4 9 (–2) 81 x1 =
3 81 12 2 = = 29 18 3
x2 =
3 + 81 6 1 = = 18 18 3
Check Yourself 4 Solve the equations. 2. x2
1. 4x2 + 3x – 1 = 0
5 = 3 x 2
3. 2x2 – 4x = 5
Answers 1. 1, Quadratic Equations
1 4
2.
3 19 2
3.
2 14 2 61
4. Discriminant of a Quadratic Equation discriminant of a quadratic equation
Definition
The quantity b2 – 4ac is called the discriminant of a quadratic equation. The discriminant tells us whether the equation has real solutions, and also tells us how many roots of an equation exist. The discriminant is denoted by (delta). x=
b , 2a
= b2 4ac
For a quadratic equation ax2 + bx + c = 0, the value of determines the number of real roots. 1. If > 0, there are two distinct real roots. 2. If = 0, there is one real root (a double root). 3. If < 0, there is no real root. Use the discriminant to check the number of roots before you solve a quadratic equation. EXAMPLE
10
Solution
Solve x2 + 6x + 7 = 0. First, find . = b2 – 4ac = 36 – 4 7 1 = 8 So is positive. x1 =
6 + 8 = 3 + 2, 2
x2 =
6 8 = 3 2 2
We can see that > 0 and there are two real roots. EXAMPLE
11
Solution
Solve x2 – 4x + 4 = 0. First, check . = b2 – 4ac = 16 – 4 4 1 = 0 So = 0 and x1 =
4 0 , 2
x2 =
4+ 0 . 2
Hence x1 = x2 = 2. We can see that = 0 and there is only one real root (a double root). EXAMPLE
12
Solution
Solve x2 – 2x + 5 = 0. = b2 – 4ac = 4 – 4 1 5 = –16 is negative, so there is no real root.
62
Algebra 8
EXAMPLE
13
Solution
For which values of k does the equation 3x2 – 4x + k = 0 have no real solution? If there is no real solution, the discriminant must be negative. Therefore, = (–4)2 4 3 k = 16 12 k < 0 . 16 12 k < 0, 16 < 12 k, 16 < k, 12
4 k> . 3
So the equation has no real solution for k >
EXAMPLE
14
Solution
4 . 3
For what values of m does the equation x2 + 3mx – 5m – 1 = 0 have a double root? A quadratic equation has a double root if its discriminant is 0. Therefore, = (3m)2 – 4 1 (–5m – 1) = 0. 9m2 + 20m + 4 = 0 (9m + 2)(m + 2) = 0 9m + 2 = 0 or m + 2 = 0. So m1 =
EXAMPLE
15
Solution
2 and m2 = 2 . 9
The equation mx2 + (2m + 1)x + m – 1 = 0 has two real roots. Find m. If the quadratic equation has two real roots, then its discriminant is positive. = b2 4ac = (2m+ 1)2 4m(m 1) = 4m2 + 4m+ 1 4m 2 + 4m = 8 m+ 1 = 8 m+ 1 > 0 8 m > 1 m>
–1 8
Therefore, the equation has two real roots if m > Quadratic Equations
–1 . 8 63
EXAMPLE
16
Solution
Prove that (a2 + b2)x2 + 2(a + b)x + 2 = 0 has no real root if a and b are unequal.
= 4(a + b)2 – 4(a2 + b2) 2 = 4(a2 + 2ab + b2) – 8(a2 + b2) = 4a2 + 8ab + 4b2 – 8a2 – 8b2 = –4a2 + 8ab – 4b2 = –4(a – b)2 But (a – b)2 is always non-negative, and it is also non-zero, since a b. So < 0. Thus the equation (a2 + b2)x2 + 2(a + b)x + 2 = 0 has no real roots if a and b are unequal.
EXAMPLE
17
Solution
From each corner of a square piece of sheet metal, a man removes a square of side 3 cm. He turns up the edges to form an open box. If the box holds 48 cm3, what are the dimensions of the piece of sheet metal? Let x be the length of a side of the square. 3
x6
3 3
3
x6
x6
3
3 3
x6
3
Since the volume of the box is 48 cm3, we have 3(x – 6)2 = 48 (x – 6)2 = 16 x – 6 = 4 x1 = 10, x2 = 2. We discard the solution x = 2 since length cannot be negative. So the sheet of metal is 10 cm by 10 cm. 64
Algebra 8
EXAMPLE
18
Solution
A ladder 5 m long is leaning against a house. The distance from the bottom of the ladder to the house is 1 m less than the distance from the top of the ladder to the ground. How far is the bottom of the ladder from the house?
Let us use x to represent the distance from the top of the ladder to the ground. The ladder then forms the right triangle shown in the diagram. By using the Pythagorean Theorem, we get the equation x2 + (x – 1)2 = 52 x2 + x2 – 2x + 1 = 25 2x2 – 2x – 24 = 0
5
x
x2 – x – 12 = 0 (x + 3)(x – 4) = 0 x = –3 or x = 4.
x1
Since the length cannot be negative, x = 4. So the distance from the bottom of the ladder to the house is 3 m. EXAMPLE
19
Solution
A motorboat heads upstream a distance of 48 km on a river whose current is running at 3 km per hour (km/h). Then the motorboat returns. The trip upstream and back takes 12 hours. Assuming that the motorboat maintained a constant speed relative to the water, what was its speed? We use x to represent the constant speed of the motorboat relative to the water. Then the true speed going upstream is x – 3 km/h, and the true speed going downstream is x + 3 km/h. Since Distance = Velocity Time, we can write Time = Distance / Velocity. Therefore, the boat takes travel upstream and downstream.
48 hours to x–3
48 km
48 hours to travel x+ 3 x 3 km/h
Since the total time is 12 hours, 48 48 + = 12; x 3. x – 3 x+3
x + 3 km/h
Multiply both sides by (x – 3)(x + 3): Quadratic Equations
65
48(x – 3) + 48(x + 3) = 12(x – 3)(x + 3) x2 – 8x – 9 = 0 (x – 9)(x + 1) = 0 x = 9 or x = –1. Since the speed cannot be negative, the speed of the boat is 9 km/h.
Check Yourself 5 1. Evaluate the discriminant of the equation 2x2 – x – 3 = 0, and describe the roots. 2. The sum of two numbers is 10 and sum of their squares is 68. Find the numbers. 3. For which value(s) of m does the equation mx2 + mx – 1 = 0 have a double root? Answers 1. = 25, two real roots
2. 2, 8
3. –4
M AT H F U N A mathematician, a physicist, and an engineer were traveling through Scotland when they saw a black sheep through the window of the train. ‘Aha,’ said the engineer, ‘I see that Scottish sheep are black.’ ‘Hmm,’ said the physicist, ‘You mean that some Scottish sheep are black. We haven’t seen all the sheep yet.’ ‘No,’ said the mathematician, ‘All we know is that there is at least one sheep in Scotland, and that at least one side of that one sheep is black! We haven’t seen the other side of the sheep yet.’
66
Algebra 8
T HE G OLDEN R ATIO The Golden Ratio appears again and again in art, architecture, music and nature. Its origins go back to the days of the ancient Greeks, who thought that a rectangle with sides in the Golden Ratio, called a golden rectangle, exhibited the most aesthetically pleasing proportion. The use of the Golden Ratio has been of interest to artists and architects since before the building of the Parthenon in Greece in the fifth century B.C. The rectangle drawn around the Parthenon with its upper triangular structure intact, as shown in Figure 1, is a golden rectangle. 1
1
x1
1
P
x Figure 1
Figure 2
Figure 2 shows the dimensions of a golden rectangle. The ratio of length to width in the rectangle is x to 1. A square P with sides of 1 unit has been marked, leaving a smaller rectangle Q. For the smaller rectangle Q, the ratio of length to width is 1 to x – 1. In order for the larger rectangle to be a golden rectangle, the two ratios need to be equal, creating a proportion: x 1 = 1 x–1 When we apply cross multiplication to this proportion, we obtain a quadratic equation: x 1 x(x –1) = 1 ; x2 – x = 1 ; x2 – x – 1 = 0. 1 x 1 Since we cannot factor this equation, we apply the quadratic formula. The solutions are x1
1 5 1 5 and x2 . 2 2
Because x represents the length of a rectangle, the negative solution is discarded 1 5 1.618034 . This is the Golden Ratio. 2 The Golden Ratio occurs in nature as well as in art. For example, in sunflowers,
and the positive solution is
the ratio of the number of clockwise spirals to the number of counterclockwise spirals approximates the Golden Ratio.
Q
EXERCISES
3 .1
A. Solving Equations of the Form ax2 = 0 1. Solve the equations. a. ñ2x2 = 0
b. (ñ3 – 2)x2 = 0
1 c. x2 = 0 5
2
d. 0.07x = 0
B. Solving Equations of the Form ax2 + bx = 0 2. Solve the equations. 2
D.Solving Equations of the Form ax2 + bx + c = 0 9. Solve by factoring. a. x2 – x = 0
b. –3x2 + x = 0
c. x2 – 49 = 0
d. x2 – 25 = 0
e. x2 – x – 2 = 0
f. x2 + 3x + 2 = 0
g. x2 – 2x + 1 = 0
h. 6x2 + x – 15 = 0
i. 10x2 – 19x + 6 = 0
j. 12x2 – 5x – 2 = 0
10. Solve by factoring. 2
a. 2x + 5x = 0
b. –7x + 3x = 0
c. 3x2 – 8x = 0
d.
2 x 6 x2 = 0 3
a. x2 + (a + 1)x + a = 0 b. (x + 1)2 – 2(x + 1)(x – 3) + (x – 3)2 = 0 c. 2ax2 + (5b – 2a)x – 5b = 0
11. Solve by completing the square.
3. Solve (x + 2)2 = 2(x + 2). 4. Solve (2x – 1)(x + 3) = –3. 5. Solve (x + 1) – 3(x + 1) = 0.
a. x2 – 4x – 1 = 0
b. x2 + 4x = 3
c. x2 – 6x – 13 = 0
d. 3x2 – 2x + 4 = 0
e. 4x2 + 8x + 15 = 0
f. 2x2 + 7x + 11 = 0
2
g.
C. Solving Equations of the Form ax2 + c = 0 6. Solve the equations. a. x2 – 16 = 0
b. 7x2 + 3 = 0
c. 9x2 – 25 = 0
d. 0.6x2 – 15 = 0
e. –5x2 + 4 = 0
f.
2 x2 =0 25 8
7. Solve the equations. a. (x – 2)2 – 9 = 0 2
c. (x + 1) = 5
2
d. (1 – x) – 1 = 8
8. Solve (4x + 1)(x – 1) = (x – 1)(x + 1) – 3(x – 3). 68
h. 3x2 + x =
1 2
12. Solve by using the quadratic formula. a. x2 – 4x + 2 = 0
b. 2x2 + x – 1 = 0
c. 4x2 + 12x + 9 = 0
d. 3x2 – 5x + 1 = 0
e. 5x2 + 4x + 7 = 0
f. 3x2 – 7 = 2 – 2x
g. x2 + x + 1 = 0 h. 25x2 + 40x + 16 = 0 i. (2x – 3)2 = 11x – 19 j.
b. (x + 3)2 + 7 = 0
1 2 1 x 3x + = 0 2 2
3 2 1 1 x x =0 4 4 2
2 k. 4x 1 = x(10 x 9) 3
l.
x2 1 =11( x +1) 2 Algebra 8
13. Solve by using the quadratic formula. a. 4 b.
x2 – (4k + 2)x + 7k + 2 = 0 form a perfect square?
1 2 =0 x x2
1 1 2 + = x 1 x 1 3 2
2 2 c. (x 3) (x 2) = 1 x 16 4 2
d.
17. For which values of k does the equation
18. The longest side of a right triangle is 6 cm less than twice the length of the medium side. The shortest side is 6 cm. Find the length of the two other sides.
1 1 5 + = x 1 x 4 4
14. Use the discriminant to determine whether each quadratic equation has two real solutions, a double root, or no real solution, without solving the equation. a. 2x2 + 3x + 1 = 0
19. A wire that is 32 cm long is cut into two pieces,
and each piece is bent to form a square. The total area enclosed by the two squares is 34 cm2. Find the length of each piece of wire.
b. x2 + 5x – 6 = 0
c. 4x2 + 12x + 9 = 0 d. 25x2 – 20x + 4 = 0 e. x2 + 4x + 7 = 0
f. 2x2 – x + 2 = 0
Mixed Problems 20. a, b, c, and d are four consecutive even natural 15. Consider the equation ax2 + 3x + 10 = 0. For
which values of a does the equation have
numbers. The sum of a and c is one fifth of the product of b and d. Find a, b, c, and d.
a. two distinct real roots? b. one double root? c. no solution?
21. Two squares have sides (x + 6) cm and (2x + 1)
cm respectively. The sum of their areas is 697 cm2. Find the areas of the squares.
16. For which values of m does the equation 2x2 + 2x + m + 4 = 0 have a. two real solutions? b. one solution? c. no real solutions? Quadratic Equations
22. A year ago, a father was eight times as old as his
son. Now his age is the square of his son’s age. How old are they now? 69
We have seen that the roots of an equation depend on its coefficients. Therefore, there exist certain relations between the coefficients and the roots of an equation. In this section we will consider the relations between the roots and the coefficients a, b and c of a quadratic equation. We know that the roots of the quadratic equation ax2 + bx + c = 0, a 0 are x1 = François Viète (or Vieta) 1540-1603, French mathematician. Vieta was a founder of modern algebra, who introduced the use of letters as algebraic symbols and correlated algebra with geometry and trigonometry. Vieta presented methods for solving equations of second, third and fourth degree. He knew the connection between the positive roots of equations and the coefficients of the different powers of the unknown quantity. The word ‘coefficient’ is actually due to Vieta. When Vieta applied numerical methods to solve equations, he used methods which were similar to those used by earlier Arabic mathematicians.
b b + and x2 = . 2a 2a
Let us use these formulas above to find the sum of the roots of a quadratic equation. x1 + x2 =
b b+ + 2a 2a
=
b + ( b+ ) 2a
=
2b b = 2a a
Therefore,
x1 + x2 = -
b . a
We can use the same expressions for x1 and x2 to find the product of the roots of a quadratic equation. b b+ ( b ) ( b+ ) x1 x2 = = 4a2 2a 2a =
b2 (b2 4ac ) 4ac c = 2= 4a2 4a a
Therefore,
x1 x2 =
c a
.
These relations were discovered by François Vieta, a French mathematician, and so they are together called Vieta’s theorem. 70
Algebra 8
Vieta’s theorem
Theorem
Let x1 and x2 be the roots of the quadratic equation of the form ax2 + bx + c = 0, a 0. Then b x1 + x2 = , a
c x1 x2 = . a
By using Vieta’s theorem we can now find the sum and product of the roots of a quadratic equation without calculating the roots.
Note By using Vieta’s theorem we can also see the following. 1.
x + x2 – b / a b 1 1 (sum of the reciprocals of the roots) + = 1 = = x1 x2 x1x2 c/a c 2
c b2 2ac b 2. x12 + x22 = (x1 + x2 )2 2x1x2 = 2 = (sum of the squares of the roots) a a2 a
3. x13 + x23 = (x1 + x2 )3 3x1x2 (x1 + x2 ) = EXAMPLE
20
Solution
Derive a formula for the difference of the roots of a quadratic equation in standard form. Let us consider the two differences. First, x1 x2 =
b b+ b + b 2 = = = . 2a 2a 2a 2a a
Similarly, x2 x1 =
. a
So we can say that x1 x2 = EXAMPLE
21
Solution
3abc b 3 (sum of the cubes of the roots) a3
. a
Find the sum and product of the roots of the given equations, without solving the equations. a. 2x2 + 6x + 5 = 0
b. x2 – 3x – 5 = 0
a. a = 2, b = 6 and c = 5
b. a = 1, b = –3 and c = –5
b 6 x1 + x2 = = = 3 a 2 x1 x2 =
c 5 = a 2
b 3 x1 + x2 = = = 3 a 1 x1 x2 =
c –5 = = –5 a 1
Note The quadratic equation 2x2 + 6x + 5 = 0 has no real root. However, by Vieta’s theorem, the 5 sum of the roots is –3 and the product of the roots is , which are real numbers. Can you 2 say why? Quadratic Equations
71
Check Yourself 6 1. Find the sum and the product of the roots of the following equations, using Vieta’s theorem. 7 a. 3x2 + 5x – 1 = 0 b. x2 – 4x + = 0 c. –x2 + 7x – 1 = 0 2 2. For the previous equations, find a. 1 + 1 . b. x12 + x22. c. x13 + x23. x1 x2 Answers 5 1 7 1. a. , – b. 4, c. 7, 1 2. a. 5 b. 9 c. 322 3 3 2 EXAMPLE
22
Solution
x1 and x2 are the non-negative roots of the equation 3x2 + 2mx + 1 = 0. Given x1 = 3x2, find x1, x2, and m. x1 = 3x2 x1 x2 =
c 1 = ; a 3
1 3x2 x2 = ; 3
x1 = 3x2 then x1 = 3 x1 + x2 =
2m ; 3
1 x2 2 = ; 9
x2 =
1 since x2 > 0; 3
x2 =
1 3
1 =1 3
1 2m 1+ = ; 3 3
4 2m = 3 3
m = 2 EXAMPLE
23
Solution
x1 and x2 are the roots of the equation x2 – 3x + 1 = 0. Find the value of
Since x1 + x2 = –
x1 x2
+
x2 x1
.
(–3) 1 = 3 > 0 and x1 x2 = =1> 0, both x1 and x2 are positive. 1 1
x1 x + 2 =k x2 x1 k2 =
x12 2 x1x2 x2 2 + + x2 x1 x1x2
k2 =
x13 + x2 3 + 2 x1x2 x1x2
9 + 27 1 k = + 2 1 = 20 1 2
k1 = 2 5 , k2 = 2 5 k= 2 5 72
(why?) Algebra 8
EXAMPLE
24
Solution
x2 + x – 6m = 0 and x2 – 2mx + 3 = 0 have a common root. Find m. Let x1 and x2 be the roots of x2 + x – 6m = 0 and x1 and x3 be the roots of x2 – 2mx + 3 = 0, then x1 + x2 = – 1 (1) x1 + x3 = 2m
and (2)
x1x2 = – 6m . x1x3 = 3
From (1) we get x2 – x3 = –1 – 2m. From (2) we get
x2 = –2m ; x2 = –2mx3. x3
From (1) and (2), 2 mx3 x3 = 1 2m x3 (2 m+ 1) = 1+ 2m x3 =
1+ 2 m = 1. 1+ 2 m
If we substitute 1 for x in the second equation, we have 4 – 2m = 0, so m = 2.
Note We can apply Vieta's theorem to a cubic polynomial equation. Let x1, x2 and x3 be the roots of the equation ax3 + bx2 + cx + d = 0. Then, b 1. x1 + x2 + x3 = . a
c 2. x1x2 + x1x3 + x2 x3 = . a
d 3. x1x2 x3 = . a
Check Yourself 7 1. One of the roots of the equation 2x2 – mx + 8 = 0 is 3 more than the other root. Find m. 2. x1 and x2 are the roots of the equation x2 + x – m + 1 = 0. x1 – x2 = 5 is given. Find m, x1, and x2. Answers 1. 10 2. 7, 2, –3
Mathematics is one component of any plan for liberal education. Mother of all the sciences, it is a builder of the imagination, a weaver of patterns of sheer thought, an intuitive dreamer, a poet. The study of mathematics cannot be replaced by any other activity. Quadratic Equations
73
EXERCISES
3 .2
1. Find the sum and the product of the roots of each
6. One of the roots of the equation
equation, without solving it.
3 6x2 + 13x + n2 + 2n – 2 = 0 is – . Find n and 2 the other root of the equation.
a. 10x2 – 11x – 12 = 0 b.
7 2 9 5 x + x – =0 8 7 3
c. (x – 2)(3x – 4) = 13 d. x + 7 = (2x – 1)(3x – 2) 2
7. Let x1 and x2 be the roots of the equation
2
e. (4x + 3) = (3x + 1)
4x2 + 5x = 0. Find x12x2 + x1x22.
f. x2 = ñ2(3x – ñ2x)
2. The sum of the roots of the equation
8. Consider the equation
(5k + 2)x2 + 7kx – 8k = 0 is 3. Find the product
(m + 2n)x2 – (m + 2n)x + m – n = 0.
of the roots.
If the arithmetic and the geometric means of the roots of the equation are equal, find the relation between m and n.
3. The product of the roots of the equation (4m2 – 1)x2 + (2m + 1)x + 2m – 1 = 0 is 5. Find m.
9. Let x1 and x2 be the roots of the equation x2 – 8x + 5k = 0. x1 = 2x2 – 1 is given. Find k.
4. Find the sum of the squares of the roots of the following quadratic equations. a. 2x2 + 5x + 1 = 0 b. –x2 + 7x + 2 = 0 c. 3(x + 2)(x – 1) = 4(x – 2) – 1
10. Let x1 and x2 be the roots of the equation
x2 + (m + 1)x + m + 2 = 0. 2x1 + 3x2 = –13 is given. Find m.
d. 4x(3 – 4x) = x – 1
5. One of the roots of the equation 4 20x2 + (2m2 + 2m + 1)x + 20 = 0 is – . Find 5 m and the other root of the equation. 74
11. Let x1 and x2 be the roots of the equation (k + 1)x2 – kx + k – 4 = 0.
6 6 + = 30 is given. x1 x2
Find k. Algebra 8
12. Consider the equation x2 – (3k + 1)x + 8 = 0. 2 1
2 2
x + x = 20 is given. Find k.
13. Consider the equation x2 + (m + 1)x – m = 0. 2 1
2 2
x + x = 13 is given. Find m.
18. Let p and q be the roots of the equation
2x2 – 5x + p2 + q2 = 0. Find the discriminant of the equation.
19. Find x12 +
1 2
x1
, if x1 is the root of the equation
2
1 1 x + 6 x + +9 = 0. x x
14. Let x1 and x2 be the roots of the equation kx2 + (k – 1)x – 2 + k = 0. 2 2 3 + = is given. Find k. x1 + 3 x2 + 3 2
20. Consider the equation x3 + cx + 1 = 0.
x1 =
1 1 is given. Find the roots of the + x2 x3
equation and c.
15. Consider the equation x2 + (k – 2)x + k – 6 = 0. If the equation has two negative roots, find all the possible values of k.
16. Find two positive consecutive numbers such that the sum of their squares is 85.
21. The roots of the equation x2 – 5x + p = 0 are also
the roots of the equation x3 + qx + 30 = 0. Find p + q.
22. The sum of the two roots of the equation 2x3 – x2 – 7x – 3 = 0 is 1. Find the roots of the equation.
23. From each corner of a square piece of sheet 17. Find the product of the roots of the equation
1 1 1 1 , a + b 0. + = a b a+b+ x x
Quadratic Equations
metal, a man removes a square of side 2 cm. He turns up the edges to form an open box which holds 24 cm3. What are the dimensions of the piece of sheet metal? 75
Let x1 and x2 be the roots of the equation ax2 + bx + c = 0. b c We can write the equation ax2 + bx + c = 0 as x2 + x + = 0. a a b x1 + x2 = a , and We know that c x1 x2 = a .
So we can write the equation again as x2 – (x1 + x2)x + (x1 x1) = 0. In other words, if we know the sum and product of the roots of a quadratic equation then we can write the equation as x2 – Sx + P = 0 where S = x1 + x2 and P = x1 x2. This means that we can derive (find) a quadratic equation if we know its roots.
Note If the roots of a quadratic equation are x1 and x2, then (x – x1) (x – x2) = 0. EXAMPLE
25
Solution
Find a quadratic equation whose roots are –1 and 5. x1 = –1 and x2 = 5 S = x1 + x2 = –1 + 5 = 4 P = x1 x2 = (–1) 5 = –5 Hence, the equation is x2 – Sx + P = 0 x2 – 4x – 5 = 0.
EXAMPLE
26
Solution
Find a quadratic equation whose only root is x1 = x2 =
1 . 3
1 3
1 1 2 S = x1 + x2 = + = 3 3 3 1 P = x1 x2 = . 9 Hence, the equation is x2 76
2 1 x+ =0, or 9 x2 6 x+1 =0. 3 9 Algebra 8
EXAMPLE
27
Solution
Find the equation whose roots are 2 + ñ3 and 2 – ñ3. x1 = 2 + ñ3, x2 = 2 – ñ3 S = x1 + x2 = 4 P = x1 x2 = (2 + ñ3) (2 – ñ3) = 1 Hence, the equation is x2 – 4x + 1 = 0.
EXAMPLE
28
Solution
Find the equation whose roots are 1 more than the roots of x2 – 3x – 4 = 0. Let the roots of x2 – 3x – 4 = 0 be x1 and x2. Let the roots of the equation we are looking for be x3 and x4. x1 + x2 = 3,
x1 x2 = –4
x3 = x1 + 1,
x4 = x2 + 1
S = x3 + x4 = (x1 + 1) + (x2 + 1) = x1 + x2 + 2 = 3 + 2 = 5 P = x3 x4 = (x1 + 1) (x2 + 1) = x1 x2 + x1 + x2 + 1 = (–4) + (3) + 1 = 0 Hence the equation is x2 – 5x = 0. EXAMPLE
29
Solution
The equation x2 + (k + 4)x + k – 4 = 0 has symmetric roots. Find them. x1 = t, x2 = –t (symmetric roots) x1 + x2 = t +( t) =
b a
( k+ 4) 1
0 = k 4 k = 4
x1 x2 =
c a
t ( t) =
k 4 1
t2 = 8 t2 = 8
t = 2ñ2 Therefore, x1 = 2ñ2, x2 = –2ñ2.
Check Yourself 8 1. Find a quadratic equation whose roots are –2 and 3. 2. The roots of the equation x2 – x – 2 = 0 are x1 and x2. Find the equation whose roots are x3 and x4, where x3 = 2x1 + 1 and x4 = 2x2 + 1. Answers 1. x2 – x – 6 = 0 2. x2 – 4x – 5 = 0 Mathematics has beauty but not everyone sees it. Quadratic Equations
77
EXERCISES
3 .3
1. Find the equation with the given roots.
6. The roots of the equation x2 – kx – 3k + 1 = 0 are two more than the roots of the equation x2 – (k – 4)x – 4k = 0. Find k.
1 3 , 2 2
a. –1, 1
b.
c. 0, 4
d. 2 + ñ2, 2 – ñ2
e. ñ3 – ñ2, ñ3 + ñ2
f.
1 1 , p2 q 2
7. Find three consecutive integers a, b, c, such that a2 + b2 + c2 = 110.
8. Find the number that is 3 more than twice its 2. Find the quadratic equation whose sum of the
square root.
roots is –2 and product of the roots is 5.
9. A swimming pool can be filled by two pipes x1 and x2. Find the equation whose roots are x3 and x4, such that x3 = 2x1 and x4 = 2x2.
together in six hours. If the larger pipe alone takes five hours less than the smaller pipe alone to fill the pool, find the time in which each pipe alone would fill the pool.
4. Find the equation whose roots are 2 less than the
10. Alex can do a job in one hour less than Jane. If
3. The roots of the equation x2 – 4x – 3 = 0 are
roots of the equation 2x2 – 6x + 58 = 0.
5. The roots of each given equation are x1 and x2.
6 5 hours. How long would it take each person working
Alex and Jane work together the job takes
alone?
Write a new equation with roots x3 and x4. a. x2 – 2x – 3 = 0
b. x2 + 5x + 4 = 0
x3 = x1 – 1
x3 = 2x1 + 1
x4 = x2 – 1
x4 = 2x2 + 1
c. 6x2 – 9x – 6 = 0 x3 = x1 – x2 x4 = x2 – x1
11. A man completed a job for $156. It took him seven hours longer than he expected and so he earned $14 an hour less than he anticipated. How long did the man expect the job to take?
d. x2 2 x+ 1 = 0 x3 = x4 =
x12 x2 2 2
x x1
12. The sum of the numerator and denominator of a certain positive fraction is 11. If 1 is added to both the numerator and the denominator, the fraction 3 is increased by . Find the fraction. 56
e. x2 – 2mx + 3m – 2 = 0 f. mx2 – 2mx – 1 = 0
78
x3 = x1 – 2x2
x3 = 2x1 – x2
x4 = x2 – 2x1
x4 = 2x2 – x1
13. The area of an isosceles right triangle is 81 m2. Find the perimeter of the triangle. Algebra 8
14. The distance between two cities A and B is
19. Two painters working together can paint the front
140 km. A car driving from A to B left at the same time as a car driving from B to A. The cars met after one hour, then the first car reached city B 35 minutes later than the second car reached city A. Find the speed of each car.
of a house in 16 hours. One of the painters alone can finish this job in 24 hours less time than the other painter alone. How much time does each worker need to do this job alone?
20. A motorcyclist traveled at a constant speed for 60 km. If he had gone 10 km/h faster, he would have shortened his traveling time by one hour. Find the speed of the motorcyclist.
15. An aeroplane traveled a distance of 400 km at an average speed of x km/h. Write down an expression for the time taken. On the return journey, the speed increased by 40 km/h. Write down an expression for the time for the return journey. If the return journey took 30 minutes less than the outward journey, write down an equation in x and solve it.
21. The area of a triangle is 36 m2. The length of the base is twice the height. Find the length of the base and the height of the triangle.
22. A jeweler wishes to mix an alloy of 25% silver with another alloy of 40% silver. How much of each should he use to produce 60 kg of an alloy which is 30% silver?
16. A car drove from one city to another and returned
by a different route. The outward journey was 48 km and the return journey was 8 km shorter. The speed of the car on the return journey increased by 4 km/h. The return journey took one hour less time. Find the speed of the car on the outward journey.
23. One of two pipes can fill a pool 24 hours faster
than the other one. The slower pipe filled the pool for eight hours, then the other pipe was opened. The pipes filled together for twenty hours and 2 of the pool. Find the time that each pipe 3 requires to fill the pool alone.
filled
17. A group of women plan to share equally in the $14,000 cost of a boat. At the last minute three of the women decide not to pay. This raises the share for each of the remaining women by $1500. How many women were in the original group?
18. A car drove from a city A to a city B. The distance between the two cities is 350 km. After 200 km, the speed of the car decreased by 20 km/h. The total trip took 5 hours. Find the speed of the car in the first part of the journey. Quadratic Equations
24. A worker can clean a pool in four hours less time than it takes another worker. If the men work 8 together the job takes hours. How long would 3 it take each man working alone?
25. The length of a rectangle is 2 m more than its width and the area is 48 m2. Find the length and width of the rectangle. 79
A. WRITING EQUATIONS IN QUADRATIC FORM standard form of an equation
Definition
An equation is in standard form if the only term on the right-hand side of the equation is zero. For example, the equations 6x2 + 2x – 3 = 0 and x4 – 5 = 0 are both in standard form. The equation 6x2 + 2x = 3 and x4 = 5 are not in standard form. Certain equations that are not quadratic can be expressed in quadratic form using substitutions. These equations can be recognized because when they are written in standard form, the exponent of the variable in one term is half the exponent of variable in the other term. For example, we can write standard form equations such as x4 + 17x2 + 72 = 0 2x8 + 4x4 = 0 x – ñx – 12 = 0 as quadratic equations, because the exponent of the first variable is twice the exponent of the second variable. Look at the steps to write an equation as a quadratic. 1. Let t be a variable term with the half exponent. 2. Substitute t in all the terms with the variable. 3. Solve for t. 4. Back substitute for the original variable. EXAMPLE
30
Solution 1
Solve x4 – 13x2 + 36 = 0. The equation x4 – 13x2 + 36 = 0 is not a quadratic equation but we can write it as (x2)2 – 13x2 + 36 = 0. For this reason, it is a quadratic in x2. Let x2 = t. First we solve for t, then solve the resulting equations for x. (x2)2 – 13x2 + 36 = 0 x2 = t, so t2 – 13t + 36 = 0. By factoring, (t – 4)(t – 9) = 0 t = 4 or t = 9. Since t = x2 x2 = 4 x = 2
80
or
x2 = 9 x = 3
. Algebra 8
Solution 2
Alternatively, we can directly factorize the equation. x4 – 13x2 + 36 = 0 (x2 – 4)(x2 – 9) = 0 x2 = 4 x = 2
or
x2 = 9 x = 3
In both solutions, the roots of the given equation are –3, –2, 2, 3. EXAMPLE
31
Solution
Solve (5x – 1)2 + 4(5x – 1) – 5 = 0. For the equation (5x – 1)2 + 4(5x – 1) – 5 = 0, we let t = 5x – 1 so that t2 = (5x – 1)2. Then the original equation becomes t2 + 4t – 5 = 0. First solve for t: (t + 5)(t – 1) = 0, so t = –5 or t = 1. Now, solve for x. 5x – 1 = –5 5x = –4 x1 =
or
–4 5
5x – 1 = 1 5x = 2 x2 =
2 5
Hence, the roots of the equation are EXAMPLE
32
Solution
–4 2 and . 5 5
Solve (a2 – a)2 – 2(a2 – a) = 0. Let t = a2 – a. The equation becomes t2 – 2t = 0 t(t – 2) = 0 t = 0 or t = 2. Now, solve for a. 1. a2 – a = 0 a(a – 1) = 0 a = 0 or a = 1 2. a2 – a = 2 a2 – a – 2 = 0 (a – 2)(a + 1) = 0 a = 2 or a = –1 Hence, the roots of the equation are –1, 0, 1, 2.
Quadratic Equations
81
EXAMPLE
33
Solution
4
1 2 Solve x + 5 = 2 x2 + 2 . x x First x2 0, so x 0. Let x
1 = t. x
How can we write the right-hand side in terms of t? Let’s take the square of t: 2
1 1 t2 = x = x2 + 2 2, so x x 1 x2 + 2 = t 2 + 2. Now, let us multiply both sides of the equation by 2: x 2 2 x2 + 2 = 2t2 + 4. x Then we have the equation t4 +5 = 2t2 + 4 t4 2t 2 +1= 0 (t2 1)2 0 t2 1= 0 t =1 or t = 1. 1 Since t = x , x 1 1 x =1 or x = 1. x x Let us multiply the equations by x: x2 x 1= 0 or x2 + x 1= 0. By using the quadratic formula, x1 =
1 5 1+ 5 1+ 5 1 5 , x2 = , x3 = , x4 = . 2 2 2 2
Check Yourself 9 Solve the equations. 1. (x + 15)2 – 3(x + 15) – 18 = 0. 2. (x + 5)4 – 6(x + 5)2 – 7 = 0. 3. x – 9ñx + 14 = 0. Answers 1. –18, –9 2. –5 ñ7 3. 4, 49 82
Algebra 8
B. EQUATIONS INVOLVING PRODUCTS AND QUOTIENTS If the product of two or more numbers is zero, then at least one of the factors must be zero. If the quotient of a division is zero, then the dividend must be zero.
Note 1. P Q = 0 if and only if P = 0 or Q = 0, where P = P(x) and Q = Q(x). 2.
EXAMPLE
34
Solution
P = 0 if and only if P = 0 and Q 0. Q
Solve (x2 – 1)(x2 – 2x – 8) = 0. Try to factorize each part if possible: (x – 1)(x + 1)(x – 4)(x + 2) = 0. If the product is zero then at least one of the factors is zero. x–1=0;
x=1
x+1=0;
x = –1
x–4=0;
x=4
x+2=0;
x = –2
Thus, the roots of the equation are –2, –1, 1, 4.
EXAMPLE
35
Solution
Solve (x2 – 4x + 10)(x2 – 5x + 2) = 0. We cannot factorize the parts. Let us try to solve x2 – 4x + 10 = 0 and x2 – 5x + 2 = 0 as two different quadratic equations. x2 – 4x + 10 = 0; = 16 – 4 1 10 = –24 < 0. Therefore, this equation has no real root. x2 – 5x + 2 = 0 ;
= 25 – 4 1 2 = 17
This time, by the quadratic formula, the roots are x1 = Quadratic Equations
5 17 5+ 17 , x2 = . 2 2 83
EXAMPLE
36
Solution
Solve x3 – x2 – 4x + 4 = 0. First try to factorize the expression. x3 – x2 – 4x + 4 = x2(x – 1) – 4(x – 1) = (x – 1)(x2 – 4) = (x – 1)(x – 2)(x + 2). Now, the question becomes: solve (x – 1)(x – 2)(x + 2) = 0. So the solution is x–1=0;
x1 = 1
x–2=0;
x2 = 2
x+2=0;
EXAMPLE
37
Solution
Solve
x3 = –2.
x2 5x 6 = 0. x+ 2
The denominator of a fraction cannot be equal to zero, so x + 2 0, x –2. We need to solve x2 – 5x – 6 = 0. x2 – 5x – 6 = (x – 6)(x + 1) (x – 6)(x + 1) = 0 x1 = –1 or x2 = 6 Since these roots are not equal to –2, x1 = –1 and x2 = 6 are both solutions to the equation.
EXAMPLE
38
Solution
Solve
x2 + 7 x 8 = 0. x2 3x+ 2
First solve x2 – 3x + 2 0: x 2, x 1. x2 + 7x – 8 = (x – 1) (x + 8) = 0 So x1 = 1 and x2 = –8 are the roots of the numerator. We know that x 1, so the only solution is –8.
Note It is very important to check the roots of the numerator to see whether they make the denominator zero or not. We can do this either by substituting the roots of the numerator in the denominator, or by finding the roots of the denominator directly and checking whether they are common or not. 84
Algebra 8
EXAMPLE
39
Solution
Solve
x 7 x+ 3 = 1. x 4 x+ 4
x – 4 0, and x + 4 0, so x 4, and x –4 (since the denominators cannot be zero). x 7 x+ 3 1 = 0. x 4 x+ 4
Now we have
Let us make the denominators common: (x 7)( x+ 4) ( x+ 3)( x 4) ( x2 16) = 0. (x+ 4)(x 1) x2 2 x =0 (x 4)(x+ 4)
–x2 – 2x = 0 ; –x(x + 2) = 0 ; x1 = 0 and x2 = – 2.
EXAMPLE
40
Solution
Solve
4 x+ 1 x2 5 = . x+ 1 x 1 x2 1
4 x+ 1 x2 5 is not a quadratic equation. However, = x+ 1 x 1 x2 1 we can make it quadratic by multiplying each side by x2 – 1 since (x2 – 1 0 ; x 1). The result is
In its present form, the equation
4(x – 1) – (x + 1)2 = x2 – 5 4x – 4 – x2 – 2x – 1 = x2 – 5 –2x2 + 2x = 0 –2x(x – 1) = 0 x1 = 0 or x2 = 1, but x 1, so x = 0 is the only possible solution.
EXAMPLE
41
Solution
2
5x x Solve + 6 = 0. x+ 2 x+ 2
x + 2 0 ; x –2 Let t =
x , then the equation becomes x+ 2
t2 – 5t + 6 = 0 (t – 3)(t – 2) = 0 t1 = 2 or t2 = 3. Quadratic Equations
85
Now, solve for x: x =2 x+ 2
or
x =3 x+ 2
x = 2x+ 4
x = 3 x+ 6
x2 = 4
x1 = 3.
So the roots are –4 and –3.
Check Yourself 10 1. Solve
9 + 2 x = 4. 4x+ 3
2. Solve
x+ 1 x 3 12 = . x 2 x+ 2 x2 4
Answers 1 3 1. ; 4 2
2. no real solution
C. EQUATIONS INVOLVING RADICALS When the variable in an equation occurs in a square root, cube root, and so on, that is, when it occurs in a radical, the equation is called a radical equation. For example, the equations x+ 1 = 2 and
3
x – 1 – 5 = 0 are radical equations. Sometimes a suitable operation will
change a radical equation to an equation that is linear or quadratic. To solve a radical equation, follow the procedure. 1. Isolate the radicals Isolating a radical means putting the radical on one side of the equation and everything else on the other side, using inverse operations. If there are two radicals in the equation, isolate one of the radicals. 2. Get rid of the radical sign Raise both sides of the equation to a power equal to the index of the isolated radical. 3. If there is still a radical sign left, repeat steps 1 and 2. 4. Solve the remaining equation 86
Algebra 8
5. Check for extraneous solutions When you solve a radical equation, extra solutions may come up when you raise both sides to an even power. These extra solutions are called extraneous solutions. In radical equations you check for extraneous solutions by putting the values you found into the original equation. If the left side of the equation does not equal the right side then you have an extraneous solution.
Note 1. If a value is an extraneous solution, it is not a solution to the original problem. 2. It is very important to check your results in the original equation. In many equations, one of the results may not satisfy the original equation. However, sometimes it is possible that all results that you have found will be acceptable. EXAMPLE
42
Solution
Solve the radical equation
2 x+ 5 = 7.
Here the radicand is already alone; we do not need to isolate it. So take the square of both sides: 2x + 5 = 49 ; 2x = 44 ; x = 22. Now let us check to see if x = 22 is an extraneous solution: 2 x+ 5 = 7 2 22 + 5 = 7 49 = 7 7 = 7.
Since the last statement is true, x = 22 is not an extraneous solution. Therefore, there is one solution to this radical equation, x = 22. EXAMPLE
43
Solution
Solve
2 x 5 + x = 4.
First we isolate the radical: 2 x 5 = 4 x.
Now take the squares of both sides to eliminate the square root: 2x – 5 = 16 – 8x + x2. The new equation is x2 – 10x + 21 = 0 (x – 3) (x – 7) = 0 x = 3 or x = 7. Let us check the results in the original equation: Quadratic Equations
87
x=7;
2 .7 5 + 7 = 4
x=3;
2 3 5 +3 = 4
9 +7 = 4
1+ 3 = 4
3+7 = 4
4=4
10 = 4
This is true, so x = 3 is a solution.
This is false, so 7 is an extraneous solution. Hence, the only solution to the equation is x = 3. EXAMPLE
44
Solution
Solve
10 x+ 56 2 x +8 = 4 .
In this question there are two radical expressions. We can isolate only one expression, so it is better to isolate the more complex one. So we have 10 x+ 56 = 2 x+ 8 + 4.
Take the squares of both sides: 10 x+ 56 = (2 x+ 8) + 2 4 2 x+ 8 +16 .
This is a new equation involving radical expressions. Follow the same steps again to isolate the second radical. 8 x+ 32 = 8 2 x+ 8 x+ 4 = 2 x+ 8 x2 + 8 x+ 16 = 2 x+ 8 x2 + 6 x+ 8 = 0 (x+ 4) (x+ 2) = 0
x = –4 ,
x = –2
Now, check these results in the original equation. x = –4 ;
2 (–4) + 8 10 (–4) +56 = 4 0 16 = 4 4 = 4
This is true! x = –2 ;
2 (–2) + 8 10 (–2) + 56 = 4 4 36 = 4 2 6 = 4 4 = 4
This is true! Hence, both –4 and –2 are solutions to the equation. 88
Algebra 8
EXAMPLE
45
Solution
Solve the equation 5 + 3 x+ 3 = 3.
5 + 3 x+ 3 = 3 3
x+ 3 = 2 (by taking the cube of both sides)
x+ 3 = 8
x = –11. We do not need to check for extraneous solutions because this is an odd power. Therefore, –11 is the only solution to the equation.
EXAMPLE
46
Solve the equation
4x+ 1 + x+ 2 = 10 x+ 5 .
( 4 x+1 + x+ 2 )2 = ( 10 x+ 5) 2
Solution
4x+ 1+ 2 (4x+ 1)(x+ 2) + x+ 2 = 10 x+ 5 2 (4 x+1)( x+ 2) = 5 x+ 2 (2 (4 x+1)(x+ 2))2 (5 x+ 2)2 4(4 x+1)(x+ 2) 25 x2 + 20 x+ 4 16 x2 + 36 x+ 8 = 25 x2 + 20 x+ 4 9x2 16 x 4 = 0 (9 x+ 2)( x 2) = 0 x=
2 or x = 2 9
Now, check these results in the orijinal equation. x=
2 ; 9
x=2 ;
8 2 20 1 16 5 +1 + + 2 = + 5 = + = ; 9 9 9 9 9 3 8 +1 + 2 + 2 = 20 + 5 ;
5 =5
Both statements are true, so both x = Quadratic Equations
5 5 = 3 3
2 and x =2 are solutions. 9 89
Check Yourself 11 Solve the equations. 2x 4 = x 2
1.
5x+ 3 = 4
4.
3x2 2 x+15 + 3x2 2 x+ 8 = 7
2.
3.
x+ 2 4 x+ 8 = –3
Answers 1 13 1. 2. 2, 4 3. 7 4. , 1 3 5
D. EQUATIONS INVOLVING AN ABSOLUTE VALUE On the real number line, the absolute value of x is the distance from the origin to the point x. For example, there are two points whose distance from the origin is three units, –3 and 3. So the equation |x| = 3 has two solutions, 3 and –3. Let us first remember the mathematical definition of absolute value. absolute value of a function
Definition
For all real numbers x, f x , for f x 0 f ( x) = – f x , for f x < 0.
We can use this information to begin solving equations involving one or more absolute values. EXAMPLE
47
Solution
The absolute value of a number is never negative. |a| 0
EXAMPLE
48
Solution 1
Solve the equation |x – 2| = 5. Case 1 __________________
Case 2 _________________
x–20; x2 x–2=5 x=7
x–2<0; x<2 –(x – 2) = 5 x = –3
So the solutions are –3 and 7. Solve |2x – 3| = x + 1. Case 1 __________________ 2x 3 0 ; x
Case 2 ____________________
3 2
2x 3 < 0 ; x <
2 x 3 = x+ 1 x= 4 3 since 4 > , x = 4 is a solution. 2
So the solutions are 4 and 90
2 . 3
3 2
(2 x 3) = x+ 1 2 2 3 x= since < , 3 3 2 2 is a solution. x= 3
Algebra 8
Solution 2
Let us take the square of both sides: 4x2 – 12x + 9 = x2 + 2x + 1
|f(x)|2=f 2(x)= f(x). f(x)
3x2 – 14x + 8 = 0 (x – 4) (3x – 2) = 0. So the solutions are 4 and
EXAMPLE
49
Solution
2 . 3
Solve x2 – 2|x + 2| + 1 = 0. Case 1 _______________________
Case 2 ________________________
x + 2 0 ; x –2 x2 – 2(x + 2) + 1 = 0 x2 – 2x – 3 = 0 (x – 3)(x + 1) = 0 x = 3 or x = –1
x + 2 < 0 ; x < –2 x2 – 2(–(x + 2)) + 1 = 0 x2 + 2x + 5 = 0 = –16 Since < 0, there are no real roots.
Both 3 and –1 are greater than –2, so the solutions to the equation are –1 and 3.
EXAMPLE
50
Solution
Solve the equation |x2 – 5x + 3| = 3. Case 1 _______________________
Case 2 ________________________
x2 – 5x + 3 0 x2 – 5x + 3 = 3 x2 – 5x = 0 x(x – 5) = 0 x = 0 or x = 5 These values satisfy the inequality, so they are solutions.
x2 – 5x + 3 < 0 –(x2 – 5x + 3) = 3 x2 – 5x + 6 = 0 (x – 3)(x – 2) = 0 x = 3 or x = 2 These values also satisfy the inequality, so they are also solutions.
Hence, the solutions to the equation are 0, 2, 3, 5.
EXAMPLE
51
Solution
2 2 Solve the equation x 3x+ 1 = x 4x+ 4 .
Remember that x2 – 4x + 4 = (x – 2)2. Hence, the expression becomes x2 3x+ 1 = (x 2)2 = x 2
For any real number x, x2 = x .
Quadratic Equations
x2 3x+ 1 = x 2 . 91
Case 1 _______________________
Case 2 ________________________
x–20; x2
x–2<0; x<2
2
–(x2 – 3x + 1) = x – 2
x – 3x + 1 = x – 2 x2 – 4x + 3 = 0
x2 – 2x – 1 = 0 x1 = 1 – ñ2, x2 = 1 + ñ2
(x – 3)(x – 1) = 0
Since (1 + ñ2) is greater than 2,
x = 3 or x = 1
x = (1 – ñ2) is the only solution.
Since 1 2, x = 3 is the only solution. Hence, the solutions to the equation are (1 – ñ2) and 3.
EXAMPLE
52
Solution 1
Solve the equation |2x – 3| = |x – 7|. Let’s take the square of both sides. Then, 4x2 – 12x + 9 = x2 – 14x + 49 3x2 + 2x – 40 = 0 (3x – 10)(x + 4) = 0 x1 = –4 or x2 =
10 . 3
Therefore, both –4 and Solution 2
2x – 3 = x – 7
or
x = –4 If |f(x)| = |g(x)| then f(x) = g(x).
So both –4 and
10 are solutions to the equation. 3
2x – 3 = –(x – 7) 10 x= 3
10 are solutions to the equation. 3
Check Yourself 12 Solve the equations. 1. |2 – x| = 2x + 1
2. x|x + 1| – 2 = 2x
Answers 1 1. 2. –2, –1, 2 3 Algebraic symbols are used when you do not know what you are talking about. 92
Algebra 8
The following problem is posed and solved in the Chiu Chang Suan-Shu, a Chinese mathematical treatise which is over two thousand years old. A tree of height 20 m has a circumference of 3 m. An arrowroot vine winds seven times around the tree before it reaches the top. What is the length of the vine? Can you solve this puzzle?
EXERCISES
3 .4
A. Writing Equations in Quadratic Form 1. Solve the equations. a. x4 – 13x2 + 36 = 0 4
2
B. Equations Involving Products and Quotients
2
b. 3x – 8x – 3 = 0 c. (x2 – 9)2 – 4(x2 – 9) + 3 = 0 4
1 2 6 + 2 = 2 x 3x+ 3 x 3x+ 4 x 3x+ 5 e. (x + 1)(x + 2)(x + 3)(x + 4) = 120
d.
2
d. (x + 5) – 6(x + 5) – 7 = 0 2
3. Solve the equations. a. (16x3 – x)(x2 – 6x + 5) = 0
3 4x e. 2(3 4 x) + 25 = 0 5
c. x3 + 4x2 – 24 = 0
f.
d. x3 – 5x2 + 9x – 45 = 0
x+ 2 4 x+ 2 = 6
b. (x2 + 8x)(x2 + 8x – 6) = 280
3
g. x3 7 x 2 8 = 0
e.
2
1 7 1 h. x+ x+ = 2 x 2 x
i.
x2 + 2 8x + 2 =6 x x +2
2. Solve the equations. 1 a. 3 x2 + 2 x 2
1 7 x+ = 0 x 2
f. (x2 – 5x + 6)2 – (2x2 – 5x + 1)2 = 0 g. (6x2 – 5x – 4)2 + (10x2 – 29x + 21)2 = 0 h.
3x2 9x 12 2 =3 2 x 3x
i.
3x+ 4 6(x 2) =1 x2 3x+ 4
j.
4 x2 x 2 8 + + 4 + = 0 2 x 9 3 x 3
2
b. (x – 2x – 5) – 2(x – 2x – 3) – 4 = 0 c.
3 = 3 x x2 1+ x+ x2
Quadratic Equations
2x+ 3 3x 2 = 4x 1 3x+ 2
93
C. Equations Involving Radicals 4. Solve the equations. a.
D.Equations Involving an Absolute Value 6. Solve the equations.
x 1 1= 0
a. |x| = x + 2
b. ñx + x = 5 c.
3
d.
f.
3 x +1 = x 2
g.
3
j.
d. |x2 – 4x| = 5 e. |x2 – 2x + 3| = 6
x 1 = x 1 x 3+ x 8 =5
i.
c. x |x – 1| = 2
x 1 = 2
e.
h.
b. |2x – 5| = x – 2
f. x2 + |x – 1| + 1 = 0 g. x2 + |2x – 1| + 3 = 4x + 2
x2 x +6 2 = 0
x +16 x = 2
h.
x2 2 x +1 = 2 x 3
i.
x2 + x2 6 x +9 = 0
7. Solve the equations.
a. |x| + x2 = 0
x2 6 x x2 6x 3 = 5
b. ( x +1)( x – 1) =
4x – 1+ 2 x + 3 =1
1 2
c. (x + 1)2 – 2|x + 1| + 1 = 0
5. Solve the equations.
d. |x – x2 – 1| = |2x – 3 – x2| a.
x +1 9 x = 2 x 12
e. b.
2 x +5 + 5 x +6 = 12 x + 25
x2 4 x + 3 x2 + x 5
=1
f. |x + |3 – 2x|| = 3 – x c. d.
x x +1+ x +9 x + 4 = 0 2
2 x + 6 x +1 = x +1
Mixed Problems 8. Solve the equations.
e.
1 x2 3x = x +1
f.
x2 +1+ x2 8 =1
g.
3x2 +5 x +8 3 x2 +5 x +1 =1
h. 1+ x + 2 = 2 x + 2 +5 94
a.
x2 2x +1 =1 x
b. x2 + x2 6 x +9 = 2 x + 3 2 c. x + x +1 = 2 x 4x + 4
d. x2 x + 2 = x2 + 4x + 4 + x Algebra 8
Sometimes we need to solve two or more equations simultaneously. A set of equations like this is called a system of equations. There are no concrete rules that we can follow to solve systems of equations, but let us look at some general strategies.
EXAMPLE
53
Solve the system of equations. x+y=5 xy = 6
Solution
We can write y = 5 – x. Now substitute this value of y in the second equation. x(5 – x) = 6 x2 – 5x + 6 = 0 (x – 2)(x – 3) = 0 x = 2 or x = 3 If x = 2 then y = 3. If x = 3 then y = 2. Therefore, the solutions of the system are (2, 3) and (3, 2).
EXAMPLE
54
Solve the system. x2 + y2 = 65 x y = 28
Solution 1
Multiply the second equation by 2 and then add and subtract the resulting equations. x2 + y2 = 65 2x y = 56 + ________________________
x2 + y2 = 65 2x y = 56 –______________________
x2 + 2xy + y2 = 121
x2 – 2xy + y2 = 9
(x + y)2 = 121
(x – y)2 = 9
x + y = 11
x – y = 3
Therefore, we have four cases. Quadratic Equations
95
Case 1 __________
Case 2 __________
Case 3 __________
Case 4 __________
x + y = 11
x + y = 11
x + y = –11
x + y = –11
x–y=3 + ____________
x – y = –3 + _____________
x–y=3 + _____________
x – y = –3 + _____________
2x = 14
2x = 8
2x = –8
2x = –14
x=7
x=4
x = –4
x = –7
y=4
y=7
y = –7
y = –4
The solutions of the system are therefore (7, 4), (4, 7), (–7, –4), (–4, –7).
Solution 2
x2 + y2 = 65 x y = 28 y=
28 , x 0, substitute y in the first equation. x 2
28 x2 + = 65 x
x2 +
28 2 = 65 x2
x4 65 x2 + 784 = 0 (x2 49)( x2 16) = 0 x2 = 49 or x2 = 16 x = 7 or x = 4
By substituting these x values in y =
28 , we get the y values. x
The solutions of the system are (7, 4), (–7, –4), (4, 7), (–4, –7).
EXAMPLE
55
Solve the system. x2 = 13x + 4y y2 = 4x + 13y
96
Algebra 8
Solution
Let us subtract the two equations side by side. x2 – y2 = 9x – 9y (x – y)(x + y) = 9(x – y) x – y = 0 or x + y = 9 Case 1 ______________________________________________________ x–y=0; x=y Let us substitute x in the equation y2 = 4x + 13y. y2 = 4y + 13y y2 – 17y = 0 y1 = 0 or y2 = 17 x1 = 0 or x2 = 17 Therefore we have two solutions, (0, 0), (17, 17). Case 2 ______________________________________________________ x+y=9; x=9–y Let us substitute x in the equation y2 = 4x + 13y. y2 = 4(9 – y) + 13y y2 – 9y – 36 = 0 (y – 12)(y + 3) = 0 y3 = 12 or y4 = –3 x3 = –3 or x4 = 12 Therefore we have two more solutions, (–3, 12), (12, –3). Hence, the solutions of the system are (0, 0), (17, 17), (–3, 12), (12, –3).
EXAMPLE
56
Solution
Solve the system. x+ y x y 5 x y + x+ y = 2 2 2 x + y = 20
x – y 0 ; x y and x + y 0 ; x –y. Let us multiply both sides of the equation by 2(x – y)(x + y). 2(x + y)2 + 2(x – y)2 = 5(x – y)(x + y) 2x2 + 4xy + 2y2 + 2x2 – 4xy + 2y2 = 5x2 – 5y2 –x2 + 9y2 = 0.
Quadratic Equations
97
2 2 x + 9y = 0 Now, we have the system . x2 + y2 = 20
Let us add these two equations side by side. 10y2 = 20 y2 = 2 ; y = ñ2 x2 = 18 ; x = 3ñ2 Hence, the solutions of the system are (3ñ2, ñ2), (–3ñ2, ñ2), (3ñ2, –ñ2), (–3ñ2, –ñ2).
Check Yourself 13 Solve the systems of equations. x2 y = 2 1. x+ y = 4
x2 + 3xy 10y2 = 0 2. x2 + 2xy y2 = 28
Answers 1. (2, 2), (–3, 7)
2. (4, 2), (–4, –2), (5ñ2, –ñ2), (–5ñ2, ñ2)
One person’s constant is another person’s variable.
EXERCISES
3 .5
1. Solve each system of equations.
2. Solve each system of equations.
9x2 y2 = 44 b. 3x y = 11
x2 + y2 + 6x+ 2y = 0 a. x+ y+ 8 = 0
3x+ xy y2 = 0 b. 2 x2 3xy+ y2 = 0
x = 3 y c. y2 x = 39
1 1 1 + = d. x y 6 x+ y = 25
6 x + xy y = 0 c. 3x2 xy y2 = 0
1 1 x+ y + x y = 2 d. 4 3 x+ y + x y = 7
x2 3y2 = 13 e. xy = 4
x2 + y2 = 18 f. xy = 9
x+ y xy = 2 e. xy(x+ y ) = 48
x3 y3 = 8 f. x y = 2
x+ y+ xy = 5 g. xy+ x y = 13
x2 + 3x 4y = 20 h. 2 x 2 x+ y = 5
x+ y 2(x y) =1 x+ y g. x y 2 2 x 5xy+ 2y = 4
3x2 2 xy y2 = 7 h. x2 + xy+ 8y2 = 14
x y = 12 a. x y = 108
98
2
2
Algebra 8
CHAPTER REVIEW TEST
3A
1. Find the solutions of the quadratic equation 2x + 5x + 3 = 0. A)
1 and 3 2
D)
B) 3 and
1 and –3 2
1 2
C) 1 and
E) 1 and
3 2
C) 2
A) 8 and –6
B) 4 and –3
D) –3 and 8
C) –3 and 4
E) 8 and –3
6. What is the sum of the roots of the equation x4 – 13x2 + 36 = 0?
m + n = 5. Find m. B) –1
Find the values of h and k.
1 2
2. –1 is a root of x2 + mx + n = 0 and
A) 1
5. The sum and product of the roots of the equation 2x2 – hx + 2k = 0 are 4 and –3, respectively.
2
D) –2
E) 3
3. Which one of the following is false for the equation x2 – 7x + 1 = 0? A) The roots are not integer. B) The roots have the same sign.
A) 13
B) 18
C) 1
D) 0
E) –13
7. Which one of the following equations has roots –2 and 3? A) 2x2 – x – 6 = 0
B) x2 + x – 6 = 0
C) x2 – x – 6 = 0
D) x2 + x + 6 = 0
E) 2x2 + x – 6 = 0
C) The roots are rational. D) The roots are both positive. E) The sum of the roots is positive.
8. The sum of a number and its reciprocal is 4. Find the discriminant of x2 – 2ñ5x + 4 = 0. A) 2
B) 4
Chapter Review Test 3A
C) 36
D) 16
E) 24
Find the number. A) 1
B) –1
C)
4 3
D) –
1 2
25 . 12
E) –3 99
9. What are the roots of the equation
13. What is the solution of the equation x = ñx + 6?
1 3 + 2 = 0? x2 x
A) 1 and
1 2
D) –
A) –9 1 and 1 2
B)
C)
1 and – 1 2
B) 4
C) –4
D) 16
E) 9
1 4 and 2 3
E) –1 and 1
14. Which one of the following is not a solution of the system 5 x2 6 xy+ 5 y2 = 29 ? 7 x2 8 xy+ 7 y2 = 43
A) (2, 3)
B) (–2, –3)
D) (3, 2)
10. One of the roots of the equation ax2 + bx + c = 0
C) (–2, 3)
E) (–3, –2)
is twice the other root. What is the relation between the coefficients a, b, and c? A) 4b2 = 4c
B) 2b2 = 9ac
D) b2 = 8ac
C) 2b2 = 9a
E) 9b2 = 2ac
15. Two technicians can complete a mailing job in 12 hours while working together. Alone, the first technician can complete the mailing job seven hours faster than the other technician. How long will it take each technician to complete the mailing alone?
11. The diagonal of a rectangle is 5 cm, and the area
A) 14 and 21
is 12 cm2. Find the perimeter of the rectangle. A) 12
B) 24
C) 10
D) 14
B) 21 and 28
D) 11 and 18
C) 7 and 14
E) 15 and 22
E) 15
16. The equations 6x2 – 18x + 12 = 0 and mx2 – nx + 1 = 0 have the same solution. Find m and n. A)
12.
x – 1 x+ 1 + 1 = 0. Find the value of 4x.
A) 5 100
B) 4
C) 0
D) 16
E) –4
3 1 and 2 2
D)
1 3 and 2 2
B)
3 1 and 2 2
E)
C)
3 1 and 2 2
3 1 and 2 2 Algebra 8
CHAPTER REVIEW TEST
3B
1. x1 and x2 are roots of the equation
5. The longest side of a right triangle is 3 cm less
x2 + (2 – x1)x + 7x2 = 0. What is x1 + x2? A) –2
B) 5
C) 7
D) 10
E) –14
than twice the length of the medium side. The shortest side is 3 cm. Find the lengths of the other two sides. A) 8 and 15
B) 7 and 14
D) 4 and 5
C) 8 and 10
E) 6 and 8
2. Which one of the following is true for the equation |x|2 + |x| – 6 = 0?
6. A triangle has area of 2 cm2. The base of the
A) There is only one root.
triangle is 3 cm longer than its height. Find the length of the base and the height of the triangle.
B) The product of the roots is –3. C) The product of the roots is 4.
A) 5 and 2
D) The sum of the roots is 0. E) The sum of the roots is 1.
B) 4 and 1
D) 7 and 4
C) 6 and 3 E) 8 and 5
7. Which one of the following equations has roots 3. The sum of the squares of the roots of the equation x2 + 2hx = 3 is 10. Find the value of h. A)
1 2
B) –3
C)
3 2
D)
1 2
E) –1 or 1
1 – ñ3 and 1 + ñ3? A) x2 – 2x – 2 = 0
B) x2 + 2x – 2 = 0
C) x2 – x – ñ3 = 0
D) x2 – 2x + 2 = 0
E) x2 – 6 = 0
8. The sum and product of two numbers are 21 and 4. Find the discriminant of 2x – 2x + 5 = 0. 2
A) –36
B) 4
Chapter Review Test 3B
C) 25
D) –16
104, respectively. What is the bigger number? E) 9
A) 16
B) 22
C) 13
D) 14
E) 19 101
9. What are the roots of the equation
1
2
x x 3 +7 – 6 = 0? x+ 1 x+ 1
A)
3 and 2 4
D)
B)
A) 9
1 and 2 2
1 and –3 2
1
13. What is the solution to the equation x 2 + x 4 = 2?
C)
B) 1
C) –4
D) 4
E) 16
1 4 and 2 3
E) –1 and 2
14. Which one of the following is not a solution of the system x2 + y2 = 26 ? x2 y2 = 24
A) (1, –5)
B) (–5, –1)
D) (5, 1)
10. Which number is a solution to the equation
C) (–5, 1)
E) (5, –1)
x2 + |x – 1| – 3 = 0? A) 3
B)
2 2
C)
3 2
D) 0
E) –1
15. A pipe can fill a pool four hours faster than another pipe. The slower pipe filled the pool for seven hours, then the other pipe was opened. The pipes then filled the pool in two hours, working together. How long would each pipe take to fill the pool alone?
11. x – y = 12 and x y = 108 are given. Find x + y. A) ±20
B) ±24
C) ±28
D) ±30
E) ±32
A) 14 and 18 hours
B) 8 and 12 hours
C) 11 and 15 hours
D) 9 and 13 hours
E) 10 and 14 hours
16. The roots of the equation x3 – 3x + 7 = 0 are p, 12. x – 5 = x – 3 is given. Find the value of x. A) 5 102
B) –7
C) 1
D) 7
E) –4
q, and r. What is the product (p + 1)(q + 1)(r + 1)? A) –8
B) –9
C) 7
D) 8
E) 9 Algebra 8
A. CARTESIAN PRODUCT AND ANALYTIC PLANE 1. Ordered Pairs Everyone knows that in a football match different scores, like 3 – 1 and 1 – 3, have a different meaning. When the score is written as 3 – 1, it is clear that the home side is the winner. But when it is written as 1 – 3, we understand that the away side is the winner. A similar approach is also valid for ordered pairs (3, 1) and (1, 3), that is, they don't have the same meaning. Definition
ordered pair Given that a and b are any two elements, (a, b) is called an ordered pair where a is the first component and b is the second component. For example, below we have ordered pairs where components are numbers, names, colors, etc.: ordered pair
first component second component
(28, December)
28
December
(René, Descartes) (–9, 1) (black, white) (170 cm, 70 kg) (Descartes, René)
René –9 black 170 cm Descartes
Descartes 1 white 70 kg René
Note that the pairs (René, Descartes) and (Descartes, René) are not the same since their components are in different order.
Note The ordered pair (a, b) is not the same as (b, a) since they are written in different orders.
What is the difference between the two clocks above?
104
Algebra 8
EXAMPLE
1
Solution
Give three examples for the ordered pair (x, y) supporting the equation 2x – y = 4. Let x = 0, then y = –4. Let x = 5, then y = 6. Let x = –1.5, then y = –7. So three such ordered pairs may be (0, –4), (5, 6), (–1.5, –7). Here, note that we can find an infinite amount of such ordered pairs and although (0, –4) is an answer, (–4, 0) is not!
EQUALITY OF ORDERED PAIRS Two ordered pairs are equal if, and only if, corresponding components are equal to each other. That is, (a, b) = (c, d) if, and only if, a = c and b = d.
EXAMPLE
2
Solution
Given that (2x – 1, 7) = (3, 3y – 8), find x and y. (2x – 1, 7) = (3, 3y – 8) 2x – 1 = 3 and 7 = 3y – 8 x = 2 and y = 5
Note (x1, x2, x3) is named as an ordered triple. (x1, x2, x3, x4) is named as an ordered quadruple. (x1, x2, ..., xn) is named as an ordered n-tuple.
Check Yourself 1 1. How many different ordered pairs are given below? (5 hours, 20 minutes) (28, December, 2003) (20 minutes, 5 hours) (Italy, Rome) 2. Given that (6, 2x – y) = (x + y, –3), find x and y. 3. Complete the following ordered pairs (x, y) so they support y + 1 = 3x. (2, ?)
(?, 0)
(0, ?)
Answers 1. 3; (28, December, 2003) is an ordered triple! 2. 1; 5 3. (2, 5), (1/3, 0), (0, –1) Functions
105
2. Set Notation A set is a collection of objects. Each one of these objects that form a set is called an element of that set. For example, the months of a year form a set with 12 elements. All even numbers form another set with an infinite amount of elements. All dogs with seven legs is another set with no elements. Think about the students in your class. Is it a set? How many elements does it have?
set of strawberries!
Notation
1. We usually name sets with capital letters like A, B, C, etc. 2. If a1, a2, a3, ..., an are elements of a set A, we list all elements of this set as A = {a1, a2, a3, ..., an} and denote the number of elements by n(A). 3. The symbol means “is an element of ”. The symbol means “is not an element of ”. For example if A = {3, 5, 10}, then 3 A and 4 A. 4. The symbol means “intersection” and the symbol means “union”. For example if A = {1, 2, 3} and B = {2, 5}, then A B = {1, 2, 3, 5} since the union is the set of all elements that are either in A or B, and A B = {2} since the intersection is the set of all elements that are both in A and B. 5. To denote an empty set, that is a set with no elements, we use or { }. 6. The symbol means “is a subset of ”. For example if B = {–2, –1, 3, 7, 10} and C = {–1, 3, 10}, then C B since each element in set C is also element of set B.
EXAMPLE
3
Given that A = {1, 4, 5, 7} and B = {all odd numbers between 2 and 8}, a. find n(A) and n(B) . b. find A B and A B. c. is it correct that 9 A B? d. is it correct that A B = ?
Solution
We know that A = {1, 4, 5, 7}. If we list the elements of B, we get B = {3, 5, 7}. a. b. c. d.
106
n(A) = 4 and n(B) = 3. A B = {1, 3, 4, 5, 7} and A B = {5, 7}. Clearly 9 is not an element of A B. So 9 A B is correct. Since n(A B) = 2, A B = is incorrect. Algebra 8
Note
Certain letters are reserved for important number sets. These are , , , and : is the set of real numbers. is the set of rational numbers. is the set of integers. is the set of natural numbers. Note that all these sets contain an infinite amount of elements, so it is impossible to list them.
Notation
A = [a, b] denotes the set of all real numbers between a and b making it inclusive. A = (a, b) denotes the set of all real numbers between a and b making it exclusive. A = [a, b) denotes the set of all real numbers between a and b, where b is excluded. A = (a, b] denotes the set of all real numbers between a and b, where a is excluded. Instead of A = [a, b] we can also use the notation A = {all x such that a x b} or A = {x | a x b}. Similar notation can be used for the other sets described above. EXAMPLE
4
[a, b] a
b (a, b)
a
b [a, b)
a
b (a, b]
a
b
Given that A = [–2, 3) and B = (0, 4), find A B and A B.
Solution A
A -2
0
3
-2
0
B
B 0
4
0
AÈB -2
0
3
4
AÇB 4
0
3
Here we cannot list all of the elements of the given sets since they contain an infinite amount of elements. A B contains elements that are in A or in B so A B = [–2, 4). A B contains elements that are both in A and B so A B = (0, 3). Functions
107
Check Yourself 2 1. Given that A = {1, 2, 5, 9, 12} and B = {all x that are less than 14 and divisible by 3} find n(B), A B, A B. 2. Given that A = [–5, 3] and B = (2, 4], find A B and A B. Answers 1. 4, {1, 2, 3, 5, 6, 9, 12}, {9, 12} 2. [-5, 4], (2, 3]
3. Cartesian Product cartesian product
Definition
Let A and B be two non-empty sets. The set of all ordered pairs, whose first component is from A and whose second component is from B, is called the cartesian product of A and B and is denoted by A B. For example, if A = {1, 2, 3} and B = {x, y}, then A B = {(1, x), (1, y), (2, x), (2, y), (3, x), (3, y)} and A B is read “A cross B”,
B A = {(x, 1), (x, 2), (x, 3), (y, 1), (y, 2), (y, 3)}. Here, A B and B A are clearly different sets. The only common property between them is
not “A times B”.
that they have the same number of elements, which is equal to the product of number of elements of A and B. Note that A B = B A only when A and B are equal sets. NUMBER OF ELEMENTS OF A CARTESIAN PRODUCT Given two sets A and B, n(A B) = n(B A) = n(A) n(B). EXAMPLE
5
Solution
Given that M = {a, b, c} and N = {1, 2, 3, 4}, find the number of elements of M N and list them. n(M N) = n(M) n(N) = 3 4 = 12 M N = {(a, 1), (a, 2), (a, 3), (a, 4), (b, 1), (b, 2), (b, 3), (b, 4), (c, 1), (c, 2), (c, 3), (c, 4)}
The term Cartesian is used in the name of the French mathematician and philosopher René Descartes.
Listing the elements of a cartesian product, as shown above, is called the list method. Sometimes a set may have an infinite amount of elements which will result in an endless list. For example, think about making a list for A B where A = {yes, no} and B = {all even numbers}. Even though there are a finite number of elements, we may want to see the whole representation rather than the list. For these cases the coordinate method is the most efficient way to show the representation. To represent all the elements of a cartesian product by the coordinate method we choose the horizontal axis for the first component and the vertical axis for the second component.
108
Algebra 8
EXAMPLE
6
Given that C = {black, white} and D = {1, 2, 3, 4, 5, 6}, represent C D by: a. the list method.
Solution
b. the coordinate method.
a. By the list method, we have C D = {(black, 1), (black, 2), (black, 3), (black, 4), (black, 5), (black, 6), (white, 1), (white, 2), (white, 3), (white, 4), (white, 5), (white, 6)}. b. By the coordinate method, we have
black white
D 6 5 4 3 2 1 C
CD
Here note that the horizontal axis denotes elements of C and the vertical axis denotes elements of D. Since n(C D) = 12, we have 12 points plotted.
EXAMPLE
7
Solution
A chessboard is a cartesian product of {A, B, C, D, E, F, G, H} and {1, 2, 3, 4, 5, 6, 7, 8}.
Given that A = {1, 2, 3} and B = {all x such that 3 x 5}, represent A B and B A. Since A B and B A contain an infinite amount of elements, using the list method is not possible. But by coordinate method we can represent all of the elements as shown below: B 5
A
3
3 2 1 1 2 3
A
A´B
3
5
B
B´A
How many points are plotted on each plane? Functions
109
EXAMPLE
8
Solution
Given that A = (4, 7) and B = [1, 3] represent A B.
By the coordinate method we have B 3 1 4
7
A
A´B
Note that the vertical boundaries of the rectangle are not included in A B.
4. Analytic Plane René Descartes (1596-1650)
French mathematician and philosopher, René Descartes believed science and mathematics could explain and predict events in the physical world. Descartes developed the Cartesian coordinate system for graphing equations and geometric shapes. Modern maps use a grid system that can be traced back to Cartesian graphing techniques. It is said (although the story is probably a myth) that Descartes came up with the idea for his coordinate system while lying in bed and watching a fly crawl on the ceiling of his room.
To graph the cartesian products whose elements are ordered pairs of real numbers, we need a coordinate system. The rectangular or cartesian coordinate system consists of a horizontal number line, the x-aaxis or the abscissa, which we label as x, and a vertical number line, the y-aaxis or the ordinate, which we label as y. The plane on which such a coordinate system is constructed is called an analytic plane or xy-p plane. Axes divide the analytic plane into four parts which are called quadrants. The intersection point of axes is called the origin. ILLUSTRATION OF ANALYTIC PLANE y
2
4 quadrant 3 2 origin 1 nd
-4 -3 -2 -1 -1 -2 rd -3 quadrant -4
3
1quadrant st
(1,2) 1 2 3 4
x
4 quadrant th
Just as every real number corresponds to a point on the number line, every pair of real numbers corresponds to a point on the analytic plane. For example the pair (1, 2) corresponds to the point that lies one unit to the right of the origin and two units up. The first component, which lays on the x-axis, is 1. The second component, which lays on the y-axis, is 2. The point where the components meet is called the plot. Locating this point on an analytic plane is named plotting or graphing the plot. 110
Algebra 8
EXAMPLE
9
Solution
Plot the pairs (0, –2), (3, 1), (–4, 2), (1, 0) on an analytic plane. y
Note that the first component is on the x-axis and the second component lays on the y-axis.
(-4,2)
4 3 2 1
(3,1) (1,0) -4 -3 -2 -1 -1 1 2 3 4 -2 (0,-2) -3 -4
x
The coordinate system that maps use stems from the Cartesian coordinate system.
Check Yourself 3 1. Given that A = {1, 2} and B = {a, b, c, d}, represent A B and B A by the list method and the coordinate method. 2. Given that M = {all x such that 1 < x 5} and N = {all x such that 2 x 7}, express M N by the coordinate method. Is it possible to use the list method? 3. Plot the pairs (1, –2), (0, 1), (3, 3), (–4, –1) on an analytic plane. Answers B d c b a
1. {(1, a), (1, b), (1, c), (1, d), (2, a), (2, b), (2, c), (2, d)} {(a, 1), (a, 2), (b, 1), (b, 2), (c, 1), (c, 2), (d, 1), (d, 2)} 2.
1 2
3.
N
3
7 -4
1
Functions
5
a b c d
x
1 -1
; no
A
2 1
y
1
2 M
A
3
-2
111
B
B. RELATIONS 1. Representation of a Relation If A and B are two non-empty sets, then every non-empty subset of the cartesian product A B is a relation defined from A to B.
B Canada Russia Turkey China Ottowa Beijing Moscow Ankara
For example, if A = {Ottowa, Beijing, Moscow, Ankara} and B = {China, Turkey, Russia, Canada} we can describe a relation, say , from A to B in the following way: = {all pairs (x, y) such that x is the capital of y} and can be listed as follows: {(Ottowa, Canada), (Beijing, China), (Moscow, Russia), (Ankara, Turkey)}. Here is a subset of A B. Note that we can choose any subset of A B to find another relation. Since each subset is a collection of ordered pairs, a relation can simply be defined as the collection of ordered pairs.
Definition
A
Above all elements of A B is illustrated. Red ones are elements of relation .
relation A relation is a set of ordered pairs. The set of all the first components is called the domain and the set of all the second components is called the range of the relation. For example, {(0, 5), (1, –3), (–1, –3), (2, 0), (–3, 5)} is a relation since it is a set of ordered pairs. Its domain is {–3, –1, 0, 1, 2} and its range is {–3, 0, 5}. The relation above is represented by a list. This relation can also be represented by a table, by a map, or by a graph as seen below:
1st component 2nd component
1st component
5
·0
1
–3
·1
·5
–1
–3
· -1
· -3
·2
·0
0
–3
5
representation by a table 112
component 5
0
2
2nd
2nd component
1
-1 -3
2
· -3
representation by a map
1st component
-3
representation by a graph Algebra 8
EXAMPLE
10
month
season
January
Winter
September
Fall
April
Spring
March
Spring
June
Summer
October
Fall
February
Winter
May
Spring
Given the following relations represented by the table above, answer the following: a. How is the first component related with the second component? b. Represent it by a list. c. Represent it by a map. d. Represent it by a graph. e. Find the domain. f. Find the range. Solution
Before we find the solution, to get rid of confusion, let us assume that we are on the northern hemisphere since seasons are different on the southern hemisphere of the earth. a. The first component is the month, and the second component is the season which contains the given month. b. {(January, Winter), (September, Fall), (April, Spring), (March, Spring), (June, Summer), (October, Fall), (February, Winter), (May, Spring)}
· Fall · Spring · Summer
season Summer Spring Fall Winter May
· Winter
d.
October February
· January · September · April · March · June · October · February · May
season
April March June
month
January September
c.
month
e. The domain is {January, September, April, March, June, October, February, May}. f. The range is {Winter, Fall, Spring, Summer}. Functions
113
EXAMPLE
11
Given the domain {–2, –1, 0, 1, 2} and the relation containing ordered pairs of the form (x, y) such that x < y 2 and y , a. represent the relation by a list. b. represent the relation by a graph. c. find the range.
Solution
a. Choose x = –2 from the domain. Related y values should support –2 < y 2 and y . So for x = –2, we have y = –1, 0, 1, 2. That gives us the elements (–2, –1), (–2, 0), (–2, 1) and (–2, 2). We proceed in the same way for x = –1, 0, 1, 2 and get the final list as {(–2, –1), (–2, 0), (–2, 1), (–2, 2), (–1, 0), (–1, 1), (–1, 2), (0, 1), (0, 2), (1, 2)}. b.
y 2 1 -2 -1 -1
1
x
c. The range is {–1, 0, 1, 2}. How can we represent this relation by a map? EXAMPLE
12
Given the domain [–1, 2] and the relation containing ordered pairs of the form (x, y) such that y = x + 1, a. if possible, represent the relation by a list. b. represent the relation by a graph. c. find the range.
Solution
To draw a line, 1. select two numbers for x and find the y value for each. 2. plot those two points of the form (x, y). 3. connect them with a straight line and extend the line on each side.
a. Since the domain contains an infinite amount of elements it is impossible to list all of the elements of that relation. b. We just sketch the line y = x + 1, but note that we take the part of the line whose x values are between –1 and 2 (remember the domain!).
y
3
-1
2
x
c. As it is seen on the graph y can take any value between 0 and 3, inclusive, therefore the range is [0, 3].
Note If the domain of a relation contains an infinite amount of elements we cannot represent it by a list, table or map. 114
Algebra 8
EXAMPLE
13
The relation between blood types is represented by the map on the right.
0
a. Explain the relation verbally. b. Represent the relation by a list.
A
B
c. Find the domain. d. Find the range.
Solution
AB
a. All blood types can give blood to the same type. Additionaly type 0 can give blood to any type and type AB can get blood from any type. b. {(0, 0), (0, A), (0, B), (0, AB), (A, A), (A, AB), (B, B), (B, AB), (AB, AB)} c. The domain is {0, A, B, AB}. d. The range is also {0, A, B, AB}.
Check Yourself 4 1. Represent the relation {(Monday, sunny), (Tuesday, rainy), (Wednesday, sunny), (Thursday, cloudy), (Friday, cloudy)} by mapping and graphing. 2. Represent the relation having ordered pairs of the form (x, y) such that y2 = x in the domain {0, 1, 4} by listing and graphing. 3. Given the domain [–3, 4] and the relation containing ordered pairs of the form (x, y) such that y = –2x + 3 represent the relation by graphing and find its range. Answers · sunny
cloudy
· rainy
rainy sunny
· cloudy
2. {(0, 0), (1, –1), (1, 1), (4, –2), (4, 2)} 2 1 -1 -2
Wednesday Thursday Friday
· Monday · Tuesday · Wednesday · Thursday · Friday
Monday Tuesday
1.
3.
y 9
y x 1 2 3 4
4 x
-3 -5
Functions
; [–5, 9] 115
2. Inverse of a Relation As we learned previously, a relation is a set of ordered pairs. The inverse of a set of ordered pairs is obtained by interchanging the places of the first and the second components. Since the first and the second components are interchanged when we are finding the inverse of a relation, the domain and the range are also interchanged.
EXAMPLE
14
Given the relation {(–2, 3), (0, 1), (–4, 0), (2, 3)} containing ordered pairs of the form (x, y), a. list the inverse relation. b. find the domain and range of the inverse relation. c. graph both of these relations.
Solution
a. We simply change places of components to find {(3, –2), (1, 0), (0, –4), (3, 2)} b. The domain of an inverse relation is {0, 1, 3} and its range is {–4, –2, 0, 2} c. Let us plot the set of points {(–2, 3), (0, 1), (–4, 0), (2, 3)} and {(3, –2), (1, 0), (0, –4), (3, 2)} on the same analytic plane using different colors: y y=x
4 3 2 1 -4 -3 -2 -1
-1 1 2 3 4 -2 -3 -4
x
Note that each point and its inverse is symmetric with respect to the line y = x.
GRAPH OF INVERSE OF A RELATION The graph of inverse of a relation is symmetric to the original graph with respect to the line y = x. 116
Algebra 8
EXAMPLE
15
Plot the graph of the inverse of the relation whose graph is given below. y
x
Solution
Since the graph of inverse of a relation is symmetry of the original graph with respect to the line y = x, we first draw the axis of symmetry, that is y = x, and then the reflection of the relation. So we get the following graph: y
y=x
x
Note that the domain and the range of the relation and its inverse contain an infinite amount of elements.
EXAMPLE
16
Solution
Given the relation containing ordered pairs of the form (x, y) such that y = x + 2, find the rule for the inverse relation. If the relation was given by a list it would be enough to interchange x and y. Note that here it is impossible to list the elements of the relation since it contains an infinite amount of elements. However we can apply the same technique directly on the given rule for relation. The rule for relation is y = x + 2. If we interchange x and y we have x = y + 2. Leaving y on one side alone we get the rule for the inverse relation as y = x – 2.
Note If the rule for a relation is known, then the rule for its inverse is obtained by interchanging x and y. Functions
117
Check Yourself 5 1. Given the relation {(1, 2), (–4, 7), (1, 0)}, find the inverse relation, its domain and the range. 2. Plot the graph of the inverse of the relation whose graph is given below. y
x
3. Given the relation containing ordered pairs of the form (x, y) such that y = 3x – 1, find the rule for the inverse relation. Answers 1. {(2, 1), (7, –4), (0, 1)}; {0, 2, 7}; {–4, 1}
2.
y
3. y =
x +1 3
x
118
Algebra 8
EXERCISES
4 .1
A. Cartesian Product and Analytic Plane 1. Find the unknowns in the following ordered pairs.
5. For the given sets A and B list A A, B B, A B, B A. a. A = {all x + such that x2 < 15}, B = {all x such that x2 – 1 = 0}
a. (3x + 5, 4) = (2, y) b. (x – 2y, 3x + y) = (1, 3)
b. A = {months of winter in northern hemisphere}, B = {2, 5, 6}
2. Let A = {2, 3, 4}, B = {3, 7, 8, 9}, C = {1, 3, 5, 9, 10}. a. Find A B.
b. Find n(A C).
c. Find B C.
d. Find A B C.
e. Find A B C.
f. Find n((A B) C).
6. For the given sets A and B, plot A B on an analytic plane. a. A = {1, 2, 3}, B = {4, 6} b. A = {all x such that 1 < x < 4} and B = {2, 3, 4} c. A = {all x such that –2 < x 3} and B = {all x such that 2 x 4}
3. For the given sets A and B find A B and A B. a. A = {all x such that –3 < x 1}, B = {all x such that –1 < x < 2} b. A = {all x such that x > 2}, B = {all x such that –3 x < 5}
B. Relations 7. List the following relations which have ordered pairs in the form (x, y). a. y = x2, the domain is {0, 1, 2, 3}.
4. Using the following find A and B.
b. 2x – 3y = 1, the domain is –3 < x < 4 such that x .
a. A B = {(1, a), (1, b), (1, c), (2, a), (2, b), (2, c)} b.
B
8. Represent the following relations having ordered
4
pairs of the form (x, y) by mapping.
2
a. {(1, b), (2, d), (3, c), (5, e)} 1 2
Functions
A
b. x – y = 1, the domain is {1, 2, 3, 4, 5}. 119
9. Represent the following relations having ordered pairs of the form (x, y) by a graph. a. {(1, 3), (2, 1), (1, 2), (2, 3)}
Mixed Problems 12. Plot all pairs (x, y) that satisfy the equation y = x – 1 on an analytic plane.
b. y = 2x + 1, the domain is [0, 3]. c. y = x, the domain is . d. y x – 1, the domain is .
13. In this figure a relation is represented by mapping. List the elements of the inverse relation.
10. Find the inverse relation, domain and range of the inverse relation for the following relations having ordered pairs in the form (x, y).
a d
b c
a. {(winter, December), (fall, September), (summer, June), (spring, March)} b. y = 2x – 3, the domain is [–20, 34] c. y = x2 + 2, the domain is [–3, 3]
14. How many elements does A B have if A denotes days in a week and B denotes months in a year?
11. Graph the inverse relation for the following: a.
y
15. Prove that n(A B) = n(A) + n(B) – n(A B).
(0,2)
x
(-3,0)
b.
y
16. Given that A = {1, 2, 3},
(2,2) (0,0)
(4,0) x
B = {a, b}, and C = {blue, red}, list A B C. How can you show A B C by the coordinate method?
(2,-2)
c.
y (-1,4)
(2,4)
17. Given that A = {–1, 0, 3}, B = [–5, 5], and
(0,0)
120
(3,0) x
C = [1, 4), plot the difference of A B from A C. Algebra 8
Many board games use principles of the cartesian product and the 10 9 cartesian coordinate system. The Game of Battleship is a navy game 8 where two players try to sink each other's hidden ships. On a piece of 76 paper two players draw 10 × 10 tables, the first of which represent 54 the player's ocean and the second which represents their opponent's 3 2 ocean. Horizontal grids are named by letters and vertical grids are 1 A B C DE F GH I J named by numbers. Each player should place five ships in his own C3 = HIT hidden ocean as follows: “Carrier”(5 squares), “Battleship”(4 squares), G7 = MISS “Cruiser”(3 squares), “Submarine”(2 squares), and “Destroyer”(1 square). Ships may be placed in any horizontal or vertical position - but not diagonally. Taking turns, players call out their “shots” by telling an ordered pair denoting a location in their opponent’s ocean and attempting to “hit” the opponent’s ships in order to sink them. When a shot is called, the opponent immediately tells whether if it is a “hit” or “miss”. If the shot is a “hit”, the opponent tells the player what kind of ship was hit. Players note shots of their opponent in their own view and their shots in the enemy’s view to decide a further strategy of play. Strategy and some luck must be combined to be the first to locate and sink all 5 of your opponent’s ships in order to win the game.
The salvo game version is recommended for more experienced players. It differs mainly in how many shots are taken in a turn by each player. Each player at the start takes a salvo of 5 shots in his turn. Whenever a player has had one of his ships sunk, he loses one shot for his next salvo. As his ships are removed from the game, the shots for each salvo are reduced. The advanced salvo game offers a challenge for the expert player. It is played as salvo, except after a salvo of shots is called, the opponent simply announces how many hits were made - but not where or on what ships. Try to remember the board games you know. Do they have any relation with the cartesian Functions 121 coordinate system? What can be the components of an ordered pair?
A. DEFINING A FUNCTION 1. Definition There are many variables around us. Some of them are closely related to each other and some of them are not. For instance, the number of students in a classroom and the quantity of oxygen in the same classroom are closely related. But the number of students in a classroom and the temperature outside are not related at all. Functions are used to show this relation between variables. Using that relation we can estimate the results for possible cases. Below, the numbers on the right are related to the numbers on the left: 11 24 39 4 16 You can easily guess the rule that relates the number on the left to the number on the right. It is: “square the number”. So the relation converts a number x to another number x2. We can symbolize this as: x x2 So, If x = 10, “the square of the number” is 100
(1st sentence)
If x = –4, “the square of the number” is 16
(2nd sentence)
If x = 0.5, “the square of the number” is 0.25
(3rd sentence)
Obviously, a way of expressing the result of the rule as shown above is not very practical or mathematical. Writing it in the form 10 100 is also not good since it is not clear which rule we are using. That is, the meaning of “” can be confusing. (it may be “add 90 to the number” or “multiply the number by 10” as well). In order to talk about this rule, in our case “square the number”, we should name it as f. When we apply this rule to x, we get x2. So f is the rule that converts x into x2. Symbolically, f(x) = x2. Let us rewrite the above sentences once more:
122
f(10) = 100
(1st sentence)
f(–4) = 16
(2nd sentence)
f(0.5) = 0.25
(3rd sentence) Algebra 8
We can write each of the numbers above on the “left” and “right” as ordered pairs like (10, 100), (–4, 16), (0.5, 0.25). Note that for each number on the “left”, which is the first component, there corresponds just one number on the “right”, which is the second component. We know that the mathematical relation is a set of ordered pairs. Whenever the first component of an ordered pair is associated with exactly one second component we name that relation a function.
function
Definition
A function f is a rule that assigns to each element x in set A exactly one element y or f(x) in set B. Set A is called the domain and set B is called the range of the function f. We name x as the independent variable or the argument, and y as the dependent variable since the value of y depends on x. A function can also be
Euler's abstract thought of as a set of definition of function ordered pairs whose in 1755 in his Institutiones Calculi first components are Differentialis: If some quantities so depend on other quantities that if the latter are changed the former undergo change, then the former quantities are called functions of the latter. This denomination is of broadest nature and compromises every method by means of which one quantity could be determined by others. If, therefore, x denotes a variable quantity, then all quantities which depend on x in any way or are determined by it are called functions of it.
EXAMPLE
17
B = {(1, 2), (2, 6), (3, 9)}
FUNCTION
Different x-values all different. The set of all the first components NOT A FUNCTION of the ordered pairs is the domain of the Same x-values function. The set of all the second components of the ordered pairs is the range of the function. Consider the set, A = {(cat, dog), (chicken, duck), (cat, mouse)}. The set A would not be a function because the first component, cat, is paired with 2 different second components. Consider the set B = {(1, 2), (2, 6), (3, 9)}. This is a function since each first component has only one second component paired with it. The domain of B is the set {1, 2, 3} and the range of B is the set {2, 6, 9}. C = {(1, 3), (1, 4), (4, 6)} would not be a function since 1 has two components paired with it, 3 and 4.
C = {(1, 3), (1, 4), (4, 6)}
State whether the following relations define a function or not. a. {(0, 2), (0, 3), (1, 6), (2, 4), (3, 5)} b. {(–3, 1), (–1, –1), (0, 1), (1, 3), (2, –2)} c. A relation having ordered pairs of the form (radius, area of circle) d. A relation having ordered pairs of the form (name, surname) e. {(1, z), (2, d), (4, f)} with the domain {1, 2, 3, 4}
Functions
123
Solution
a. In order to have a function for each value in the domain we should have exactly one element assigned in the range. Since 0 is assigned both to 2 and 3, this is not a function.
·0 ·1 ·2 ·3
·2 ·3 ·4 ·5 ·6
b. The domain is {–3, –1, 0, 1, 2}. Since there is only one value · -2 · -3 assigned for each value in the domain, this is a · -1 function. Note that in the image on the right –3 and 0 are · -1 ·0 both assigned to 1, but this does not prevent it from being a ·1 ·1 function. The elements from the domain may be assigned to ·3 ·2 the same value from the range. The important point is that each element of the domain must not have more than one different value from the range. c. For each given radius there is exactly one possible circle area. This relation is a function. d. Two people with the same name can have different surnames. This relation is not a function. e. As in the previous examples if nothing else is stated as the domain, the set of all the first components is the domain. For this example it would be obvious to think that the domain is {1, 2, 4}. This relation is a function if we didn't see the phrase “with domain {1, 2, 3, 4}”. But the definition of a function states that every element in a domain must be assigned by exactly one element. We can see that the element “3” from the domain remains unassigned. Therefore this relation is not a function.
Note Any function is a relation but any relation is not always a function.
To each person there corresponds just one birth date. Have you ever heard of a person without a birth date? Or a person with two birth dates?
124
To each satellite there corresponds just one planet. Have you ever heard of a satellite without a planet? Or a satellite with two planets?
Algebra 8
Notation
If the letter f represents a function, then the notation f(x) means “apply the rule f to the number x” and we read f(x) as “f of x”. EXAMPLE
18
Given the function with the rule “multiply a number by 3 and add 5”, formulize it and find its value when the number is a. –3
Solution
b. –1
c. 0
d. 2
Clearly, we have a long rule for a mathematician. If we denote the number by x and the rule by f, then we have f(x) = 3x + 5 and we need to find f(–3), f(–1), f(0), f(2). a. Substituting –3 in the place of x, f(–3) = 3 (–3) + 5 = –4. b. Substituting –1 in the place of x, f(–1) = 3 (–1) + 5 = 2. c. Substituting 0 in the place of x, f(0) = 3 0 + 5 = 5.
f(x) does not mean “f times x”.
d. Substituting 2 in the place of x, f(2) = 3 2 + 5 = 11. EXAMPLE
19
Given the function f(x) = |2x – 3|, a. find f(1).
Solution
b. find x, if f(x) = 0.
a. f(1) = |2 1 – 3| = |–1| = 1 b. We are looking for x value(s) for which f(x) = 0. That means |2x – 3| = 0.
“| |” stands for absolute value. Absolute value of a number is itself if it is nonnegative, its negative if it is negative. e.g. |25| = 25, |–0.7| = 0.7
Solving this equation we get that x = 1.5.
A common way to consider functions is through the idea of a function machine. Imagine this machine as having an input, where items are entered, and an output, where results are obtained. If a number is dropped into the input, a function machine will output a single value.
Functions
125
EXAMPLE
20
Solution
Given the function f(x) = x2 – 3x, find f(–2) + f(4). To find f(–2) + f(4) we use the formula f(x) = x2 – 3x twice by substituting –2 and 4 in the place of x. f(–2) = (–2)2 – 3(–2) = 10
In general, f(a) + f(b) f(a + b).
f(4) = 42 – 3 4 = 4 So, f(–2) + f(4) = 10 + 4 = 14.
Notation
In mathematics, there are different notations used for functions. The expressions below are the most frequent notations and have the same meaning: f(x) = x2 y = x2 Since this is just a matter of notation we may also use letters like g or h to denote a function and t, s, u, v to denote its argument. For the previous example if we had chosen g as the function name and u as the variable, then the formula would have been g(u) = u2 – 3u which would definitely give us the same result. EXAMPLE
21
Solution
Given the function g( x) =
2 x +1 , find g(2), g(2a), g(a + 3), g(–x), g(anything). x2 1
Substituting 2 in the place of x, g( 2) =
2 2 +1 5 = . 22 1 3
Substituting 2a in the place of x, g( 2a) =
2 2a +1 4a +1 = . ( 2a)2 1 4a2 1
Substituting a + 3 in the place of x, g( a + 3) Substituting –x in the place of x, g( x) =
2 ( a + 3) + 1 2 a +7 = 2 . ( a + 3)2 1 a + 6a + 8
2 ( x)+1 2 x+1 = 2 . ( x)2 1 x 1
Substituting anything in the place of x, g( anything ) =
2 ( anythi ng ) +1 . (anything )2 1
Notation
f : A B means function has domain A and range B. This notation is rarely used. When it is not used, it means that the domain is all possible real numbers for which the function is defined and the range is all possible corresponding values. EXAMPLE
126
22
Given the function g : such that g(u) = u + 5, find g(0), g(–2), g(–0.5). Algebra 8
Solution
Substituting 0 in the place of u, g(0) = 0 + 5 = 5. Substituting –2 in the place of u, g(–2) = –2 + 5 = 3. We cannot substitute –0.5 in the place of u since the function has integers as its domain (note that –0.5 ), so g(–0.5) is undefined. What would you say about g(–0.5) if the domain of function was ? Expressing a function with the help of one formula may not always be possible. Sometimes we need two, three or even more formulae to describe a function. We face these kind of situations frequently and name such functions as piecewise defined functions.
Note A piecewise function is defined so that for any value in the domain there corresponds exactly one formula. EXAMPLE
23
Solution
Given that f(2x + 1) = x + 7, find f(3). We cannot substitute 3 in the place of x. If we do so, we get f(2 3 + 1) = 3 + 7 which gives f(7) = 10. But we need f(3) not f(7)! To find it we will look for the value of x which will make the inside of the expression f(2x + 1) equal to 3. Solving 2x + 1 = 3, we find x = 1. Substituting 1 in the place of x, we get f(2 1 + 1) = 1 + 7 which gives f(3) = 8.
EXAMPLE
24
1 Given that f ( ) = x 1, find x
a. f(2). Solution
b. f(x).
a. We must use the same method of the previous example. To find f(2) we must find the 1 . In other words we x 1 1 must solve = 2, which means x = . Substituting this value in the given formula we 2 x have
value of x which will give 2 when put in the place of the expression
1 1 1 f ( 1 ) 1 or f (2) . 2 2 2 1 1 1 1 a, so x . Rewriting the formula f ( ) x 1 in terms of a we have f ( a) 1. x a a x Since this is just a matter of notation instead of letter “a” we can choose any other letter 1 we like. Since we must find f(x), let us use “x” in place of “a”. So f ( x) 1. x Note that since we now have the formula for f(x) we are able to find any value of f 1 without repeating the procedure in part a. Try to find f(2) directly using f ( x) 1. x What will you get?
b. Let
Functions
127
Check Yourself 6 1. Is the relation having ordered pairs in the form of (month, its season) a function? What about the relation with ordered pairs in the form of (season, a month of it)? 2. Given the function f(x) = x2 – 5x, find f(0), f(3), f(–a), f(x + 1), f(a) + f(b). x2 x if x > 0 , find f(–2), f(0), f(5). 3. Given the function f ( x) = if x 0 4 4. Given that f(2x – 3) = x2 – x, find f(1).
Answers 1. yes, no 2. 0, –6, a2 + 5a, x2 – 3x – 4, a2 – 5a + b2 – 5b 3. 4, 4, 20 4. 2
2. Applied Problems Functions are widely used for modelling real life data. In modelling functions, we must be able to translate the verbal description into mathematical language. We do this by assigning symbols to represent the independent and dependent variables and then finding the function that relates these variables. Once we model a real life situation, we are able to estimate the result for any specific value.
25
EXAMPLE
The area of a rectangle is 40 m2. a. Express its perimeter P as a function of its width. b. Using the required function find the perimeter of the rectangle if its width is 5 meters.
Solution
b a Given a rectangle with sidelength a and b Perimeter = 2(a + b) Area = a b
EXAMPLE
128
26
40 m. We x know that the perimeter of a rectangle is twice the sum of its length and width. So the 40 perimeter can be expressed as P( x) = 2( x + ). x
a. Let x m be the width of the rectangle. Since its area is 40 m2 , its length is
b. Simply we need to find P(5)= 2(5+ meters.
40 ) = 26. The perimeter of the rectangle is 26 5
On a certain route an airline carries 8000 passengers per month, each paying $50. The airline wants to increase the fare. However, the market research estimates that for each $1 increase in fare, it will lose 100 passengers. Express the airline's monthly income as a function of increase in fare. Algebra 8
Solution
Let x be the increase in fare. Then the new fare = 50 + x, the number of passengers = 8000 – 100x (each $1 increase is a decrease of 100 in passengers). So if they increase the fare by x, the new monthly income will be (new fare) × (number of passengers) = (50 + x)(8000 – 100x) Let f be the income function. So f(x)= (50 + x)(8000 – 100x) or f(x) = –100x2 + 3000x + 400000.
EXAMPLE
27
A telephone company does not charge any money for calls that lasts at most 15 seconds, $0.2 per minute for calls that last at most 4 minutes and $0.15 multiplied by unit time after 4th minute. Express the call cost as a function in terms of minutes.
Solution
Let x be the call length in minutes and f(x) be the cost function. Clearly x > 0. When x 0.25 (15 seconds = 0.25 minutes), f(x) = 0. When 0.25 < x 4, f(x) = 0.2x. When x > 4, besides 0.2 × 4 which is the cost of first 4 minutes, the caller will pay $0.15 multipled by extra time after the 4th minute. So, f(x) = 0.8 + (x – 4)0.15. Since we have three different cases, the cost function will be expressed piecewisely: 0 f ( x) = 0.2 x 0.15 x +0.2
if 0 < x 0.25 if 0.25 < x 4 if x > 4
Check Yourself 7 1. The area of a circle is found by multiplying the square of its radius by . Write the formula for the area of the circle function S(r) where r is the radius. 2. A gardener has 140 m of fencing for her rectangular vegetable garden. Express the area as a function of one side length. Answers 1. r2 Functions
2. –x2 + 70x 129
3. Graph of a Function a. Plotting Ordered Pairs Visualization makes great use to get familiar with a specific function. The best way to visualize a function is to graph it. The graph of a function f is the collection of ordered pairs (x, f(x)) or (x, y) such that x is in the domain of the function.
ILLUSTRATION OF THE GRAPH OF A FUNCTION y y = f(x) f(x)
(x, f(x))
x
x
Keep in mind that x is the directed distance from the y-axis, and f(x) is the directed distance from the x-axis.
EXAMPLE
28
Given the graph of a function on the right, a. find
y
f ( 2)+ 2 f (3) . f (0)
b. find x, if f(x) = 0.
1 -2
2
3 x
-1 y = f(x)
Solution
a. In this example we have no formula to use but only the graph. To find the given value we should first find each of f(–2), f(3) and f(0). f(–2) means the y-coordinate of the point whose x-coordinate is –2. Clearly the answer is 0. With the same procedure we find that f(3) = –1 and f(0) = 1. Therefore,
f ( 2) + 2 f (3) 0 + 2 ( –1) = –2. f (0) 1
b. To find the values of x for which f(x) = 0, we should find all the points whose y-coordinate is 0 and then take x-coordinates of those points as an answer. Clearly, there are two such points (–2,0) and (2,0) so the required x values are –2 and 2. 130
Algebra 8
EXAMPLE
29
Solution
Plot the graph of f(x) = x – 2 with the domain {–2, –1, 0, 1, 2, 3}. In this example the domain contains a finite number of elements. Substituting –2, –1, 0, 1, 2 in the place of x, we get f(–2) = –4, f(–1) = –3, f(0) = –2, f(1) = –1, f(2) = 0, f(3) = 1 which gives us the ordered pairs (–2, –4), (–1, –3), (0, –2), (1, –1), (2, 0), (3, 1). So we get the following graph: y 2 1 -4 -3 -2 -1 -1 -2 -3 -4
EXAMPLE
30
Solution
If domain is not given with the formula, then it is the set of all possible real numbers for which the function is defined.
Graph of f(x) = mx + n is a line.
Functions
y = f(x)
1 2 3 4
x
Plot the graph of f(x) = x – 2. Note that the difference of this function from the one in the previous example is that its domain is any real number for which the function is defined. So we must have an infinite amount of points plotted on our graph. Let us evaluate f(x) for some values of x (see the table on the right). This will give us the following line as graph: y y = f(x)
2 1 -4 -3 -2 -1 -1 -2 -3 -4
1
2 3 4
x
f(x)
–2
–4
–1
–3
0
–2
1
–1
2
0
3
1
x
131
b. Intercepts of a Function An x-intercept is the first component of the ordered pair (a, 0) where a is any real number.
If the graph of a function f crosses the x-axis, then the function has an x-intercept. To find the x-intercepts, we need to find all the x-values that support the equation f(x) = 0. Since an equation may have more than one solution, the graph of a function can cross the x-axis more than once.
Note The x-intercept(s) of a function are also called the zeros or the roots of the function. A function can have more than one x-intercept.
x-intercepts are on the x-axis.
EXAMPLE
31
Find the zeros of the following functions: a. f(x) = 2x + 10
Solution
b. g(x) = x2 – 9
a. Zeros of the function support the equation f(x) = 0 which means 2x + 10 = 0. So at x = –5 function has a zero. b. Solving g(x) = 0 which means x2 – 9 = 0, we get x = 3 and x = –3.
A y-intercept is the second component of the ordered pair (0, b) where b is any real number.
Similarly, if the graph of a function f crosses the y-axis, then the graph of the function has a y-intercept. To find the y-intercepts, we need to let x = 0, and then solve for y. Since we are talking about a function, f(0) can have at most one value. That means that the graph of a function can cross the y-axis at most once. Note that we used the phrase “at most” since it is possible to have no y-intercept if 0 is not in the domain of function.
132
Algebra 8
Note A function can have at most one y-intercept.
y-intercepts are on the y-axis.
EXAMPLE
32
Solution
Find the y-intercept of the function f ( x) =
x+ 3 . x2 +1
To find the y-intercept we substitute x = 0. That gives us f (0) =
0+ 3 = 3. 02 +1
c. Vertical Line Test for Functions By definition, a function needs to have at most one y-value assigned to each x-value. That is, a vertical line can cross the graph of the function at most once for any x-value.
VERTICAL LINE TEST FOR FUNCTIONS A set of points in the coordinate plane is the graph of a function if, and only if, no vertical line crosses the graph at more than one point. EXAMPLE
33
Which of the following are graphs of functions? Why or why not? a. b. c. y y y
x
Functions
x
d.
y
x
x
133
Solution
a. As demonstrated below, no vertical line intersects the graph at more than one point. So the graph belongs to a function. y
x
b. Since we can find a vertical line which intersects the graph at more than one point, this graph does not belong to a function. y
x
c. Since we can find a vertical line which intersects the graph at more than one point, this graph does not belong to a function. y
x
d. Since no vertical line intersects the graph at more than one point, this graph belongs to a function. y
x
134
Algebra 8
EXAMPLE
34
Solution
Plot the graph of a function whose x-intercepts are –3, 2, 4 and whose y-intercept is 3 if the graph passes through the point (1, 1.5). y
Our graph should cross the x-axis at –3, 2, 4 and the y-axis at 3. Moreover, no vertical line should intersect the graph at more than one point and the point (1,1.5) should be on the graph. On the right we have a possible graph. Note that we can plot an infinite amount of different graphs using these data.
y=f(x)
3 1.5 -3
4
1 2
x
Check Yourself 8 1. Plot the graph of the function f(x) = 2x + 4. 2. Find the x-intercept(s) of the function f(x) = 3x – 1. 3. Find the y-intercept of the function g(x) = x2 – 2x + 7. 4. State whether the following graphs belong to functions or not. Explain your answer. y
y
x
y
x
x
5. Plot the graph of any function with x-intercepts –2, 1 and y-intercept 4 so that the graph passes through the point (3, –1). Answers y
1.
2.
1 3
y
3. 7 4. yes, no, yes 5.
4
4
-2
1
-2 x
3 x
-1
Functions
135
4. Domain of a Function If a function f does not model data or come with conditions, its domain is the largest set of real numbers x for which f(x) is a real number. For example, an area of a circle with radius x can be modelled such that f(x) = x2. Since the radius must be positive, the domain is the set of all positive numbers. The domain of a function may be stated explicitly. For example, if we write f(x) = x + 3
x [2, 5),
then the domain is the set of all real numbers x such that 2 x < 5. You can think of candies in the jar as x values and the domain.
Or if we write f ( x) =
x2 4 x + 3 2 x +1
f : .
then the domain is the set of all integers. If the function is given by an algebraic expression and the domain is not stated explicitly, then the domain is the set of all real numbers for which the expression is defined as a real number. For example, the function f(x) = óx – 2 is defined when x – 2 0. So its domain is the set of all real numbers more than or equal to 2, that is x [2, ). Sometimes to find the domain of a function, we find all values that make the function undefined and throw them out of the set of real numbers. 1 is not defined at x = 0, so its domain is any real number x x except 0, that is, x (–, 0) (0, ) or x R \ {0}.
For example, the function f ( x) =
Notation
To express the domain of a function f(x), we write “x …” or “D( f ) = …”. RULES FOR FINDING THE DOMAIN OF A FUNCTION 1. The polynomial functions have any real number as their domain. 2. The rational functions have any real number except the ones that make the denominator zero as their domain. 3. Even degreed root functions (square root, fourth degree root, etc.) have any real number that makes the expression under the root non-negative as their domain. 4. Odd degreed root functions (cubic root, fifth degree root, etc.) have any real number as their domain. 5. If a function contains the combination of the above functions, then the domain is found by taking the intersection of all conditions. 136
Algebra 8
EXAMPLE
35
Find the domain of the following functions: a. f(x) = x2 + 2x – 8
Solution
b. f ( x) =
1 x+ 2
c. f ( x) = 2 x +1
d. f ( x) = 3 x 7
a. The polynomial functions are defined for any real number x. So D( f ) = . b. We have a rational function denominator of which should not be equal to zero which means x + 2 0, that is x –2. So D( f ) = (–, –2) (–2, ). c. The square root is defined when the expression under it is nonnegative. 1 1 That means 2x + 1 0, therefore, x . So, D( f ) =[ , ). 2 2
d. The cubic root is defined for any real number x. So D( f ) = .
EXAMPLE
36
Find the domain of the following functions: 1 a. f ( x) = 2 x 2x c.
Solution
f ( x) = 4 (12 3 x)( x2 +7) +
1 x + 2x 2
b.
f ( x) =
d. f ( x) =
1 x 5 + 3 x2 2 x 6x 1 – 4 2 x +7 + x3 – 5 x 3– x
a. Since we have a fraction, the denominator should be non-zero. That means x2 – 2x 0 or x(x – 2) 0. That gives x 0 or x 2. So D( f ) = (–, 0) (0, 2) (2, ). b.
1 means 6 – x 0. 6x matter what x is.
x 5 means x – 5 0.
3
x2 2 x always gives real number no
6 x 0 . To find the domain we should solve the system x 5 0 So D( f ) = [5, 6) (6, ).
c.
4
(12 3 x)( x2 +7) means (12 – 3x)(x2 + 7) 0.
To find the domain we should solve the system
1 means x2 + 2x 0. x2 + 2 x
(12 3 x)( x2 +7) 0 12 3 x 0 or since x2 +7 is always positive. 2 x( x + 2) 0 x + 2 x 0
So, D( f ) = (–, –2) (–2, 0) (0, 4]. d.
1 1 – 4 2 x +7 + x3 – 5 x gives a real number when 0 and 2x + 7 0. 3– x 3– x
3 – x > 0 To find the domain it is enough to solve the system . 2 x +7 0 So, D( f ) = [–3.5, 3). Functions
137
EXAMPLE
37
Write any function by using the following domains: a. D( f ) = (–, 4]
Solution
b. D( f ) = [1, 2)
a. Any real number x that is less than or equal to 4 will work for our function, that is x 4 or 4 – x 0. If we think 4 – x as inside of a square root function, this will give us the domain D( f ) = (–, 4]. So f ( x) = 4 x is such a function. b. If we think the problem backwards x [1, 2) should be a solution that we get from a x 1 x – 1 0 system of inequalities which can be a condition for the function or x < 2 x – 2 < 0
f(x) = 2004 x – 1+
1 x–2
.
Note that these are not the only possible answers.
Check Yourself 9 1. Find the domain of the function f ( x) = 3 x2 + 2 x +8. 2. Find the domain of the function g( x) = x 1+
x+ 4 . x2 4
3. Write any function by using the domain D( f ) = (–2, 4]. Answers 1.
138
2. [1, 2) (2, )
3. f ( x) = 4 – x x+ 2
Algebra 8
B. PROPERTIES OF FUNCTIONS 1. Equal Functions EXAMPLE
38
Solution
Are the functions f ( x) =
x2 1 and g(x) = x + 1 the same? Why? x 1
x2 1 = x +1. But this doesn’t mean that f(x) = g(x). To understand this x 1 fact it is enough to see that D( f ) = (–, 1) (1, ) and D(g) = . Since D( f ) D(g), the
We know that
functions f ( x) =
x2 1 and g(x) = x + 1 are not the same. x 1
EQUALITY OF FUNCTIONS Two functions f(x) and g(x) are equal if, and only if, 1. f(x) = g(x), 2. D( f ) = D(g). EXAMPLE
39
Solution
Are the functions f(x) = (x0.5)4 – (x0.25)4 and g(x) = x2 – x equal? Why? Clearly, f(x) = (x0.5)4 – (x0.25)4 = x2 – x = g(x) However, note that in the original 4 f(x) = (ñx )4 – (ñx )4, which means D( f ) = [0, ).
Since D(g) = D( f ), functions f and g are not equal.
2. Even and Odd Functions Let us consider the function f(x) = x2. Clearly, f(–2) = f(2) = 4 f(–5) = f(5) = 25. It is not hard to realize that the function f gives the same value for any number and its negative in its domain. So we can say that f(–x) = f(x) for any x D( f ). These kind of functions are called even functions. Let us consider the function f(x) = x3. Clearly, f(–2) = –f(2) = –8 f(–5) = –f(5) = –125. For this function we can say that f(–x) = –f(x) for any x D( f ). These kind of functions are called odd functions. Functions
139
Definition
even, odd functions A function f is even if, for each x in its domain, f(–x) = f(x). A function f is odd if, for each x in its domain, f(–x) = – f(x).
EXAMPLE
40
Classify whether the following functions are even or odd. a. f(x) = 3x4 – 4
b. f(x) = 5x3 – 2x d. f ( x) =
Solution
x2 +1 3x
c. f ( x) =
2 2x 7
e. f ( x) = x + 2
We should evaluate f(–x) and decide if it is the same of the given function f(x) or its negative. a. f(–x) = 3(–x)4 – 4 = 3x4 – 4 = f(x). This function is even. b. f(–x) = 5(–x)3 – 2(–x) = –5x3 + 2x = –(5x3 – 2x) = – f(x). This function is odd. 2 2 = 2( x) 7 2 x 7
c. f (– x) =
This function is neither even nor odd since f(–x) f(x), nor is f(–x) – f(x). x2 +1 ( x)2 +1 x2 +1 = = = f ( x). This function is odd. 3( x) 3x 3x
d. f (– x) =
e. f (– x) = (– x)+ 2 = x + 2. This function is neither even nor odd.
Note A function is even, odd or neither. Generally functions that we face are neither even nor odd. EXAMPLE
41
Classify whether the following functions are even or odd. a.
Solution
f ( x) =
x3 2 x xx
a. f ( x) =
b.
f ( x) =
2x 1 2 x +1
c.
f ( x)
x7 x 1
x7 x 1
( x)3 2( x) x3 + 2 x ( x3 2 x) x3 2 x = = = = f ( x). ( x) x x x x x xx
This function is even. Here note that |–x| = |x| since the absolute value of a number is equal to the absolute value of its negative. 1 1 2x 1 x x 2 1 2 1 2x 2 x 1 b. f ( x) = x = = 2 x= =– x = f ( x). This function is odd. x 1 1+ 2 2 +1 1+ 2 2 +1 +1 2x 2x x
c. f ( x) =
x 7 x +1
+
x +7 x 1
=
x +7 ( x 1)
+
x 7 ( x +1)
x +7 x 7 = + = f ( x). x 1 x +1
This function is odd. 140
Algebra 8
If f is even, then the points (x0, y0) and (–x0, y0) are on the graph of the function f, which means the graph of an even function is symmetric with respect to the y-axis. GRAPH OF AN EVEN FUNCTION y
y=f(x)
y0
-x0
x0
x
If f is odd, then the points (x0, y0) and (–x0, –y0) are on the graph of the function f which means the graph of an odd function is symmetric with respect to the origin. GRAPH OF AN ODD FUNCTION y y=f(x)
y0 -x0
x0
x
-y0
EXAMPLE
42
Classify whether the following functions are even or odd: a.
b.
y
x
Solution
c.
y
x
d.
y
y
x
x
a. The graph is symmetric with respect to the y-axis making this function even. b. The graph is symmetric with respect to the origin making this function odd. c. Using the vertical line test we can easily state that the graph does not belong to a function. So we cannot talk about the function being even or odd. d. The graph has no symmetry with respect to the y-axis or the origin. The function is neither even nor odd.
Functions
141
EXAMPLE
43
Complete the following graphs if they belong to an even or an odd function. a.
b.
y
y
x
Solution
x
In order to complete the graph for an even function just imagine the y-axis as a mirror and draw the reflection of each point on the other side. In order to complete the graph for an odd function connect each point with the origin and extend that line to the other side of the origin until you have the same distance. a.
y
b.
y
x
y
y
x
x
x
Check Yourself 10 1. Are the functions f ( x) = ( x 3)( x + 4) and g( x) = x 3 x+ 4 equal? Explain, why. 2. Classify whether the following functions are even or odd: f(x) = x7 + 4x
f ( x) =
x2 + 2x4 x4
f(x) = 3x + 3–x – 2|x|
3. Complete the graph on the right if it belongs to
y
a. an even function. b. an odd function.
x
Answers 1. no 2. odd, neither, even
3. a.
y
b. x
142
y
x
Algebra 8
3. Increasing, Decreasing and Constant Functions It is very useful to know where the graph of a function rises and where it falls. The graph shown below rises, falls, then rises again as we move from left to right, that is as the argument gets larger. y
D
B
y=f(x) C
A a
b
c
d
x
It rises from A to B, falls from B to C, and rises again from C to D. The function f is said to be increasing when its graph rises and decreasing when its graph falls. If there is no rise or fall in the graph, then we say the function is constant. Definition
increasing, decreasing, constant function A function f is increasing on an interval I if f(x1) < f(x2) for any x1 < x2 in I. A function f is decreasing on an interval I if f(x1) > f(x2) for any x1 < x2 in I. A function f is constant on an interval I if f(x1) = f(x2) for any x1 < x2 in I. GRAPH OF INCREASING, DECREASING, CONSTANT FUNCTION y
y
y=f(x)
y=f(x)
y y=f(x)
f(x1) x1
x2
f(x1)
f(x2) x
An increasing function
Functions
f(x2) x1
f(x1) x2 x
A decreasing function
x1
f(x2) x2
x
A constant function
143
EXAMPLE
44
Investigate the following functions for increase and decrease: a. f(x) = 4
Solution
b. f(x) = 2x + 5
c. f(x) = –x + 4
Let us plot their graphs to see the rise and fall. y
y
y 5
y=4
4
4
4
-2.5 x
x
x y = x + 4
y = 2x + 5
a. f(x) = 4 is constant on D( f ). b. f(x) = 2x + 5 is increasing on D( f ). c. f(x) = –x + 4 is decreasing on D( f ).
Note A line in the form y = mx + n has m as its slope.
A linear function is always 1. increasing if its slope is positive, 2. decreasing if its slope is negative, 3. constant if its slope is zero.
EXAMPLE
45
State the intervals on which the function, whose graph is given on the right, is increasing, decreasing or constant.
y
-5
Solution
-2
1
x
Note that when the question is about the intervals of increase and decrease we are interested in the argument, that is, the x-axis. Looking at the graph we see that there is a rise until x = –5 (note that since the beginning of the graph is not fixed by a point, this rise begins at minus infinity), there is a constant behaviour until x = –2, there is a fall until x = 1, and finally there is a rise “until” plus infinity. Mathematically our answer is as follows: The function is increasing on (–, –5] and [1, ), decreasing on [–2, 1], constant on [–5, –2]. Note that we find the largest possible interval of increase and decrease. So the intervals have closed brackets whenever possible.
144
Algebra 8
EXAMPLE
46
Solution
Prove that f ( x) =
5 is decreasing on (–, –0.5). 2 x +1
In this problem we will use the definition. Let x1 < x2 < –0.5, then we must show that f(x1) > f(x2) on (–, –0.5). f ( x1 ) f ( x2 ) =
10( x2 x1 ) 5 5 = 2 x1 +1 2 x2 +1 (2 x1 +1)(2 x2 +1)
x2 x1 > 0 10( x2 x1 ) Since x1 < x2 < 0.5, 2 x1 +1< 0 . So, > 0. x (2 1 +1)(2 x2 +1) 2 x2 +1< 0
It means that f(x1) – f(x2) > 0 or f(x1) > f(x2). Thus, f ( x) =
EXAMPLE
47
Solution
5 is decreasing on (–, –0.5). 2 x +1
Investigate f(x) = x2 for increase and decrease. Note that f(x) is an even function. So let’s investigate it in [0, ). If 0 x1 < x2, then f(x2) – f(x1) = x22 – x12 = (x2 – x1)(x2 + x1). x2 x1 > 0 . So (x2 – x1)(x2 + x1) > 0. Since 0 x1 < x2, then x2 + x1 > 0 It means that f(x2) – f(x1) > 0 or f(x1) < f(x2).
Thus, f(x) = x2 is increasing on [0, ). Since f(x) is even, it is decreasing on the other side of the y-axis (Think about the symmetry with respect to the y-axis). So the function is increasing on [0, ), decreasing on (–, 0].
Check Yourself 11 1. Find a if f(x) = (a – 3)x + 4 is constant on D( f ). 2. State the intervals on which the function whose graph is given on the right is increasing, decreasing or constant. 3. Prove that f ( x) =
y 2 3
5
x
4 is increasing on (2, ). 2x
4. Investigate f(x) = x3 for increase and decrease. Answers 1. 3
2. increasing on [2, 3], decreasing on (–, 2], [5, ), constant on [3, 5]
3. consider f(x1) – f(x2) when 2 < x1 < x2 Functions
4. increasing on 145
4. Reading Graph of a Function Graphs give us the opportunity to see the general details of a function at one glance. In order to get the necessary information it is important to read a graph properly. We read a graph the same way that we would read a book, from left to right. The behaviour of the function is determined by the y-values, but the intervals are reported in terms of the x-values.
Notation
1. If a point is plotted by a full circle, it means that the point is included in the graph. 2. If a point is plotted by an empty circle, it means that the point is not included in the graph. 3. If the endpoint of a graph is a full or empty circle, it means the part of the graph on that side stops at that point; otherwise it means that the graph continues up to infinity.
EXAMPLE
48
Answer the following using the graph below: y
y=f(x)
2 -6 -7
-4
5 -3
-1
6
x
-3
a. Find f(–6), f(0), f(ñ3), f(–5), f(–4). b. Find the domain and the range of f. c. Find the x and the y-intercepts of f. d. Solve f(x) = –3. e. Solve f(x) 0. f. Find the intervals on which f is increasing, decreasing and constant. g. Is f even or odd? h. Find the minimum and the maximum value of f . 146
Algebra 8
Solution
a. We just find the corresponding y coordinate on the graph for each given x value, that is for –6, 0, ñ3, –5, –4. Clearly, f(–6) = –1, f(0) = –3, f(ñ3) = –3. f(–5) and f(–4) are undefined since no part of the graph has x coordinate being equal to –5 or –4. b. Graphically finding the domain means writing all of the values of x for which a vertical line intersects the graph of f . So, D( f ) = (–, –6] (–4, ). Similarly finding the range means writing all of the values of y for which a horizontal line intersects the graph of f . So, E( f ) = [–3, ). c. Finding the x-intercepts means finding the values of x for which the graph crosses the x-axis. So, the x-intercepts are –7, –3 and 6. Finding the y-intercept means finding the value of y for which the graph crosses the y-axis. So, the y-intercept is –3. y y=f(x)
d. To solve f(x) = –3 means finding values of x for which y is equal to –3. We see that there are an infinite amount of solutions: x [0, 5].
5 x
-3
e. To solve f(x) 0 graphically, you must ask for which values of x is the graph of f above or at the same level with the line y = 0. So, the answer is x (–, –7] [–3, 0) [6, ).
f. Recall that intervals are reported in terms of the x-values and for increase, decrease or being constant we write the largest possible interval.
y
-7
y=f(x)
6
-3
x
y y=f(x) -6
-4
5 x
f is increasing on (–4, 0) and [5, ) f is decreasing on (–, –6] f is constant on [0, 5]
g. f is neither even nor odd since its graph has no symmetry with respect to the y-axis nor the origin. h. The maximum value of f is the highest possible y value that the graph can reach. In our graph since it moves as high as possible we say that the maximum value doesn’t exist or the maximum value is plus infinity. The minimum value of f is the lowest possible y value that the graph can reach. In our graph the minimum value is –3. Functions
147
EXAMPLE
49
Answer the following using the graph below: y 3 2 y=f(x)
1 2 -6
-4
-2 -1
-1
3
5
7
x
-3
a. Find the domain and the range of f. b. Find the x and the y-intercepts of f. c. Solve f(x) = –1. d. Solve –3 < f(x) –1. e. Solve f(x) > 2. f. Solve f(x) 2. g. Find the intervals on which f is increasing, decreasing and constant. h. Is f even or odd? i. Find the minimum and the maximum value of f . j. When does f have its minimum and maximum value? Solution
a. D( f ) = [–6, 7) and E( f ) = [–3, 3]. b. The x-intercepts are –1, 3 and the y-intercept is –1. c. x {0, 2}. d. x [0, 1) (1, 2]. e. x [–6, –2). f. x [–6, –2] {5}. g. f is increasing on [1, 5], decreasing on [–4, 1] and [5, 7), constant on [–6, –4]. h. f is neither even nor odd. i. The maximum value of f is 3 and the minimum value is –3. j. f has its maximum value when x [–6, –4] and has its minimum value when x = 1.
148
Algebra 8
Graphs are widely used in many areas like statistics, economics, engineering, medicine, meteorology, etc. Interpreting information from them in real situations is the same as reading a graph. In fact those graphs belong to the functions that are the models of daily situations.
EXAMPLE
50
Describe a situation that can be modelled by the graph shown. Heart Rate
Time
Solution
EXAMPLE
51
The graph shows the heart beat rate of a person in a time interval. It increases for a period of time and then decreases back to its initial condition. The person may have dealt with a sport activity during the interval of increase and after finishing that activity, the heart beat rate decreased to its normal condition.
Answer the following questions using the graph below: a. In what period did the car sales occur? b. How many cars were sold in April? c. In which month were 15 cars sold? d. In which month were the sales the best? e. In which month were the sales the worst? f. In which two months were the sales same? g. In which months did the sales increase?
Car Sales 35 30 25 Number 20 of cars 15 10 5 0
Jan Feb Mar Apr May Jun Month
h. In which months did the sales decrease? Solution
a. The first six months of the year, that is from January to June, inclusive. b. 25
c. March
d. May
e. June
f. January and February
g. April and May
h. March and June Functions
149
Check Yourself 12 Answer the following using the graph below: y
y=f(x)
(5,1)
(-4,1) (2,0)
x
(0,0)
(1,-3)
1. Find the domain and the range of f . 2. Find the x the and the y intercepts. 3. Find the intervals on which f is increasing and decreasing. 4. Solve f(x) 1. Answers
150
1. D( f ) = (–, 2) (2, 5), E( f ) = [–3, )
2. 0; 0
3. increasing on (1, 2) and (2, 5), decreasing on (–, 1)
4. x [–4, 2) (2, 5)
Algebra 8
EXERCISES
4 .2
A. Defining a Function 1. State whether the following relations are functions or not. a. {(1, x), (2, y), (3, x)} with the domain {1, 2, 3} b. {(a, b), (a, c), (b, a), (c, a)} with the domain {a, b, c} c. {(1, 1), (2, 2), (3, 1), (4, 1)} with the domain {1, 2, 3, 4, 5}
2. Express the following rules in a function notation.
2 x +1 5. Given that f ( x) = 3 3x
find
if x >1 if x =1, if x <1
f (2)+ f (1) . f ( 1)
6. Plot the graphs of the following functions within the given domain. a. f(x) = –x2, D( f ) = {–2, –1, 0, 1, 2} b. g(x) = 2 – x, D(g) = {–1, 0, 1, 2, 3}
a. Multiply by 5, then add 2.
c. h(x) = ñx, D(h) = {0, 1, 4, 9}
b. Square the number then subtract twice the number.
d. f(x) = 2x + 1, D( f ) =
c. Divide by two, then add three times its cube.
7. Find the x and the y intercepts of the following functions.
3. Express the following rules in words. a. f(x) = 2x – 4 c. h( x) =
b. g( x) = x +1
x 3x +1
4. Find the required values for the following functions. 2
a. If f(x) = 2x – 5x, find f(1), f(–3), f(–x), – f(x), f(ñx), f(x + h). b. If g( x) =
a. f(x) = 3x – 7 b. f(x) = (x – 6)(x + 1) c. f ( x) =
2x 4 x +5
d. f ( x) = x + 4 5
8. State whether the following graphs belong to a function or not. a.
b.
y
3x 1 , find x2 + 2
c.
y
x
y
x
g(0), g(4), g(3x), 2g(–x), g(x2), g(2a + b). u 2 +1 c. If h( u ) = 3 , find u + 2u
x
d.
e.
y
f.
y
y
h(–1), h(0), h(2u), h(x + 1), h(u2), h2(u). (notation h2(u) is used in the place of (h(u))2 which means
x
x x
the square of the function.)
Functions
151
9. Find the domain of the following functions. h. f ( x) =
a. f(x) = 2x + 1
5 x x 1
2
b. f(x) = x – 2x + 5 c. f ( x) =
x +1 x4
d. f ( x) =
x 2 x 9
i.
f ( x) =
j.
f ( x) =
1 3 2 5x x2 7 x +10 x2 +6 x 9
e. f ( x) = 3x 6 f.
f ( x) =
g. f ( x) =
1 2 x x+ 3 1 x 5 6x
h. f ( x) = x 5 + 30 3 x i.
f ( x) =
j.
f ( a) =
1 5x 1
+ 1 x
1 + 1+ a 3 2 a a 1 3
3u 4 1 k. f ( u ) = + + u3 u 2 3u 10
11. Write a formula for the function by using the
following domains. a. [3, 5) b. [–4, –2) (–2, 0) c. [–2, 1] {3}
12. For which values of x are the following functions equal? a. f(x) = 2x – 3 and g(x) = 5x + 1 b. f(x) = x + 22 and g(x) = x2 + x + 6 c. f ( x) =
10. Find the domain of the following functions.
a. f ( x) = 15 2 x 8 x2 1 x+ 4 b. f ( x) = 2 + x x2 5 3x
c. f ( x) = d. f ( x) =
x+ 3
B. Properties of Functions 13. Classify whether the following functions are even, odd or neither. a. f(x) = x3 + x
b. f(x) = x5 + x3 – 1
c. f(x) = 5 – x2 + x4
d. f ( x) =
x3 x2 x +1 4x + 3 x 2x 8
e. f ( x) =
2
g. f ( x) =
e. f ( x) = x2 +6 x +5 + x2 + x + 2 f.
f ( x) = x2 (1 x)( x 2)
g. f ( x) =
152
1 ( x 1)2 (3x x2 )
3x 1 and g(x) = 3 x2
1 x 3 4
3
x 2x x2 +1
f.
f ( x) =
h. f ( x) =
x 1 x2 +1 x 1+ x2 10 x 10 x 10 x +10 x
if x 0
i.
x + 2 f ( x) = x 2
j.
3x2 +1 if x > 0 f ( x) = 2 3x 1 if x < 0
if x < 0
Algebra 8
14. Given their graphs classify whether the following
16. Plot the graph of a function by the following intervals of increase and decrease.
functions are even, odd or neither. a.
b.
y
c.
y
e.
y
f.
y
x
x
b. Increasing on [–5, 2] and [6, 9], decreasing on [2, 6] and [9, 11]
x
x
x
d.
a. Increasing on (–, 2] and [5, ), constant on [2, 5]
y
c. Increasing on (–, 3] and [7, ), decreasing on [4, 5], constant on [3, 4] and [5, 7]
y
x
17. Prove that:
15. Complete the following graphs to get an even function and to get an odd function. a.
y
(-4,6)
a. f(x) = 3 – 2x is decreasing on . 4 b. f ( x) = is increasing on (2; ). 2x c. f ( x) =
21x 9 1 is increasing on (–; ). 3x 1 3
d. f ( x) =
4x + 31 is decreasing on (–7; ). x +7
e. f(x) = x4 – 8x is increasing on [2; ). (-7,4) (0,0)
18. Given that a f(x) = x f(x) + bx + 2 and
x
(-2,-2)
g(x) = x + 5, find g(ab) if f is a constant function.
19. f(x) = ax3 + bx + 1, a 0, b 0 and f(10) = 4,
b.
y
find f(–10).
(4,4)
(0,4)
20. Answer the following using the graph below: y
x
3
(7,-4)
y = f(x) 5
-6
(5,-6)
-2
3
6
x
a. Find the domain and the range.
c.
y
b. Find the intervals on which the function is increasing and decreasing.
(2,4) (7,2)
c. Find the x and the y intercepts.
(-2,0) (0,-2)
(5,0)
x
d. Solve f(x – 2) = 0. e. Solve f(x) < 0.
Functions
153
21. Answer the following using the graph below: y
23. Two ships leave the port at the same time, one sailing east at a rate of 9 km/h and the other sailing south at 12 km/h. If t is the time (in hours) after their departure, express the distance d between the ships as a function of t.
y = f(x)
2 -10
-6
-5
-3
3
5
6
10 x
a
-7
a. Find the domain and range of f(x). b. Solve f(x) = 0.
b
d
c. Find f(0). d. Find f(–10).
N
e. Solve f(x) 0. f. Solve f(x) > 2.
24. The freezing point of water is 0° C, or 32° F, and
g. Solve f(x) = 2.
the boiling point is 100° C, or 212° F.
h. Is f(x) even or odd?
a. Express the Fahrenheit temperature F as a linear function of the Celsius temperature C.
i. Write down the intervals where f(x) is increasing, decreasing or constant. j. Find the minimum and the maximum value of f(x). k. Find all the x values for which f(x) = –7. l. Find all the x values for which f(x) (–7; 2).
b. What temperature increase in F corresponds to an increase in temperature of 1° C?
25. A cross section of a rectangular pool with the
dimensions 60 m by 20 m is shown in the figure. The pool is being filled with water at a rate of 10 m3 / min. 60 1
Mixed Problems x 1 22. Given that f ( x) = x2 2 x 3 x + 4 find f ( f ... f (0)...). 2004 times
154
h
3
15
if x > 3 if 1< x 3, if x 1
a. Express the volume V of the water in the pool as a function of time t. b. Express V as a piecewise function of the depth h at the deep end for 0 h 2 and then for 2
26. The graph below gives the weight of a certain person as a function of age. Describe in words how this person's weight has varied over time. What do you think happened when this person was 40 years old?
31. If f(x) = 2x9 – 3x7 + ax – 3 and f(11) = 2, find
100 Weight (kg)
x +1 = x, find f(–1), f(3), f(0). x 1
30. If f
f(–11).
75 50 25 10 20 30 40 50 60 70
Age (years)
32. Find the intervals on which the function
f(x) = x2+|5x+1|+5 is increasing and decreasing.
27. The graph gives a salesman’s distance from his home as a function of time on a certain day. Describe in words what the graph indicates about his travels on this day.
1 1 33. If f x = x3 3 , find f(–x).
x
x
Distance from home (km)
34. If f(x) = ax2 + bx + c and 8 A.M.
10
NOON
2
4
6 P.M.
Time (hours)
f(x – 1) + f(x) + f(x + 1) = x2 + 1, find f(2).
28. Find the range of the following functions. a. f(x) = 4 – x2
b. f ( x) = x +1+ 3
1 c. f ( x) = 2 x +6
x3 d. f ( x) = x+ 4
29. If f(x) + f(x + 1) = 2x + 3 and f(2) = 3, find
f(99).
Functions
35. If f : + , f(25) = 1, f ( x) =
2 f ( x +1)+1 , 2
find f(3).
1 36. If g( x)+ 2 g = x, find g(x) in terms of x.
x
155
A. BASIC OPERATIONS Two functions f and g can be combined to form new functions f + g, f – g, f g, and f / g in a similar way that we add, subtract, multiply, and divide the real numbers. For example, we define the function f + g as (f + g)(x) = f(x) + g(x). This new function is called as a sum of functions f and g. Its value at a given x value is found by adding the value of f and g at that x value. The operation is similar for the case of the difference, the product and the quotient. Clearly, the new function is defined only when f and g is defined, that the domain of the new function is also the intersection of the domains of f and g. In case of the quotient the values that make denominator equal to zero must be excluded from the domain. SUM, DIFFERENCE, PRODUCT AND QUOTIENT OF FUNCTIONS Let f and g be two functions with domains A and B respectively. Then
EXAMPLE
52
Solution
Operation
Definition
Domain
Addition
( f + g)(x) = f(x) + g(x)
AB
Subtraction
(f – g)(x) = f(x) – g(x)
AB
Multiplication
(fg)(x) = f(x)g(x)
AB
Division
(f/g)(x) = f(x)/g(x)
A B, g(x) 0
Given that f(x) = x2 – 4 and g(x) = x + 3, find a. f + g, f – g, fg, f / g and their domains.
b. (f + g)(1), (f – g)(1), (fg)(1), (f / g)(1).
a. Note that D( f ) = and D(g) = . So we have ( f + g)( x) = f ( x)+ g( x) = x2 – 4 + x + 3 = x2 + x – 1 with the domain , ( f – g )( x) = f ( x) – g( x) = x2 – 4 – ( x + 3) = x2 – x – 7 with th e domain , ( fg)( x) = f ( x)g( x) = ( x2 – 4)( x + 3) = x3 + 3 x2 – 4 x – 12 with the domain , ( f / g )( x) =
f ( x ) x2 – 4 = with domain ( , 3) (–3, ) since – 3 makes t he g( x) x+ 3
denominator of the new function equal to zero. 2 b. ( f + g)(1) = f (1)+ g(1) = 1 – 4 + 1+ 3 =1
( f – g )(1) = f (1) – g(1) = 12 – 4 – (1+ 3) = –7 ( fg )(1) = f (1)g(1) = (12 – 4)(1+ 3) = –12 ( f / g )(1) = f (1) / g(1) = (12 – 4 ) /(1+ 3) = –3 / 4 156
Algebra 8
EXAMPLE
53
Solution
Given that f(x) = 2, g(x) = 3x2 – 11, h(x) = 2 x 6 and m( x) = find m(5) and the domain of m.
m( x) = m(5) =
f ( x)+ g( x) + f ( x)g( x), h( x)
f ( x) + g( x) 2 + 3x2 – 11 3 x2 9 + f ( x)g( x) = + 2(3x2 – 11) = +6 x2 22. h( x) 2x – 6 2x 6 3 52 11 2 56
+6 5 2 22 =160.
The domain of function m can directly be found by considering the final formula for m(x). The only condition that we should deal with is 2x – 6 > 0. So, D(m) = (3, ).
B. COMPOSITION OF FUNCTIONS Aside from the four basic operations there is a very important way of combining functions to get a new function. We call this a composition. This operation is illustrated as follows: Here we see two different functions f and g. Their common point is that the range of one is the domain of the other. Because of that fact the domain of the function g is linked with the range of the function f. Consider x = –2: g(–2) = 3 f(g(–2)) = f(3) = 0
f(g(x)) means g is applied first, f is applied second.
Read f(g(x)) as “f of g of x”.
In other words, –2 finally maps to 0. Note that to find the final mapping we first use g and then f. If we name that final mapping as h, then h(–2) = 0 where h(x) = f(g(x)). That final mapping h is another function and is the composition of f and g as shown below on the right:
f
g Range of g 2 0 1 3 Domain of g
Range of f
2
1
3
0
5
1
Domain of f f(g(x)) h 2 0 1 3
Domain of h
1 0 1 Range of h
As an algebraic example if f(x) = ñx and g(x) = 3x + 1, we know that f(x) is the rule “take the square root of the number x” and g(x) is the rule “multiply the number x by 3 and add 1”. In that case f(g(x)) means you must “first multiply the number x by 3 and add 1, then take the square root of the result” so we can define h( x) = f ( g( x)) = g( x) = 3x +1. Here we say that h(x) is a composition of f(x) and g(x), that is h(x) is a composite function. The domain of f(g(x)) is the set of all x in the domain of g such that g(x) is in the domain of f.
Functions
157
Composite functions are composed of “bridges” that link the domain of “first” with the range of “last”.
composition
Definition
f(g(x)) is defined as composition of f and g. It is also denoted by (f g)(x) or simply f g. fog
g x
f g(x)
f(g(x))
Arrow diagram for (f o g)(x) EXAMPLE
54
Solution 1
Given that f(x) = x2 + 3x, g(x) = x – 4, find f(g(x)) and g(f(x)). f(g(x)) = g(x)2 + 3g(x) = (x – 4)2 + 3(x – 4) = x2 – 8x + 16 + 3x – 12 = x2 – 5x + 4 g(f(x)) = f(x) – 4 = (x2 + 3x) – 4 = x2 + 3x – 4 So f(g(x)) = x2 – 5x + 4 and g(f(x)) = x2 + 3x – 4. What can you say about f(g(x)) and g(f(x))? Is it true that they are the same?
Solution 2
f(g(x)) = f(x – 4) = (x – 4)2 + 3(x – 4) = x2 – 5x + 4 g(f(x)) = g(x2 + 3x) = (x2 + 3x) – 4 = x2 + 3x – 4
Note In general, f(g(x)) g(f(x)). 158
Algebra 8
EXAMPLE
55
Let f(x) = x + 1, g(x) = x3. Find: a. (f g)(2) b. (g f)(1) c. (f f)(5) d. (g g)(–1)
Solution 1
a. (f g)(x) = f(g(x)) = g(x) + 1 = x3 + 1 (f g)(2) = 23 + 1 = 9 b. (g f)(x) = g(f(x)) = (f(x))3 = (x + 1)3 3
(g f)(1) = (1 + 1) = 8 f g means g is applied first, f is applied second and in general f g g f.
Composite functions are formed of functions that are inside another function.
c. (f f)(x) = f(f(x)) = f(x) + 1 = x + 2 (f f)(5) = 5 + 2 = 7 d. (g g)(x) = g(g(x)) = (g(x))3 = x9 (g g)(–1) = (–1)9 = –1
Solution 2
Note that these values can be calculated without finding the formula. a. In case of ( f g)(2) we may first calculate g(2) = 8 and then f(8) = 9 to find the answer since ( f g) (2) = f(g(2)) = f(8) = 9. b. (g f )(1) = g( f(1)) = g(2) = 8 c. ( f f )(5) = f( f(5)) = f(6) = 7 d. (g g)(–1) = g(g(–1)) = g(–1) = –1
EXAMPLE
56
Solution 1
Let f ( x) =
x , g(x) = x5 and h(x) = x2 + x. Find f g h. x 1
Here we have a composition of three functions. ( f g h)( x) = f ( g( h( x))) =
Solution 2 Functions
g( h( x)) ( h( x))5 ( x2 + x)5 = = 2 . 5 g( h( x)) – 1 ( h( x)) – 1 ( x + x)5 1
( f g h)(x) = f(g(h(x))) = f(g(x2 + x)) = f((x2 + x)5) =
( x2 + x)5 . ( x2 + x)5 – 1 159
EXAMPLE
57
Solution
Write the function h( x) = 5x + 3 as a composition of two functions. Let us read what the formula tells us to do: “first add 3 to five times the number x and then take the square root of all”. Let f(x) = 5x + 3 and g(x) = ñx. Note that f is applied first, g is applied second. So h(x) = (g f)(x). The composite function form is never unique. For example, consider the previous example: If n is any nonzero integer, we could choose f(x) = (5x + 3)1/n and g(x) = xn/2. Thus, there are an infinite amount of composite function forms. Generally, our aim is to choose a formula such that the expression for the function is simple, as in the above example.
EXAMPLE
58
Solution
Given f(x) = 3x + 1 and g(x) = 2x – m such that (f g)(x) = (g f )(x), find g(
1 ). 10
( f g)(x) = f(g(x)) = 3 g(x) + 1 = 3(2x – m) + 1 = 6x – 3m + 1 (g f )(x) = g( f(x)) = 2 f(x) – m = 2(3x + 1) – m = 6x + 2 – m
1 Since ( f g)(x) = (g f )(x), 6x – 3m + 1 = 6x + 2 – m which gives m = – . 2 1 1 7 So we have g(x) = 2 x + . Therefore, g( ) = . 2 10 10 EXAMPLE
59
Solution
If the rule for function f g h is “first take the square root of a number plus twenty-five, second divide the new number plus 2 by itself, third take the cube of the newest number” and p(x) = 3x2 + 6x. Find (f g h)(x) and (h p f)(–2). First of all let us find formulae for f, g and h. Since f g h means h is applied first, g second and f third we have h( x) = x + 25, g( x) =
x+ 2 , f ( x) = x 3. x 3
x + 25 + 2 According to the rule, ( f g h)(x) = . x + 25 Now let us find (h p f)(–2):
(h p f )(–2) = h(p( f(–2))) = h(p(–8)) = h(144) = 13. 160
Algebra 8
EXAMPLE
60
Solution
Given that f(g(x)) = 4x – 1 and g(x) = x + 2, find f(x). Since f(g(x)) = 4x – 1 and g(x) = x + 2 we have f(x + 2) = 4x – 1. Let x + 2 = a. Then x = a – 2. Substituting x = a – 2 in f(x + 2) = 4x – 1 we get f(a – 2 + 2) = 4(a – 2) – 1 or f(a) = 4a – 9. Since this is just a matter of notation in the final formula instead of a we choose x to get f(x) = 4x – 9.
Check Yourself 13 x +1 and g(x) = x, find f + g, f – g, fg, f / g. 2x 4 2. Given that f(x) = x2 and g(x) = 3x + 4, find (f g)(1) and (g f)(1).
1. Given that f ( x) =
3. Write the function h(x) = (2x – 7)5 as a composition of two functions such that h(x) = (f g)(x). Answers 2 2 2 1. 2 x 3x +1 , 2 x +5 x +1 , x + x , x2+1 2x 4 2x 4 2x 4 2x 4x
Functions
2. 49, 7 3. f(x) = x5, g(x) = 2x – 7
161
C. INVERSE OF A FUNCTION 1. One-to-one Functions Consider the two functions represented by a map as shown below: f
g
·1
·4
·1
·2
·5
·2
·3
·6
·3
·4 ·5
Each number in the domain of f takes a different value from the range. But it is not the same for the function g since 1 and 2 take the same value in the range. We name a function one-to-one if each element in the domain corresponds to exactly one element from the one and g is not one-to-one. range. So here f is one-tto-o To understand whether a given formula belongs to a one-to-one function or not we need an algebraic tool. The following definition for one-to-one function meets our needs: one-tto-o one function
Definition
one if, for each x1 x2 in its domain, f(x1) f(x2). A function f is one-tto-o EXAMPLE
61
Which one of the following is a one-to-one function? a. f(x) = 2x – 1
Solution
b. f(x) = x2
Let us use the definition for one-to-one function to find out whether they support the definition or not: a. Let x1 x2. Then f(x1) = 2x1 – 1 and f(x2) = 2x2 – 1. Let’s assume that f(x1) = f(x2), so 2x1 – 1 = 2x2 – 1 which means x1 = x2. But this is wrong since we let x1 x2! So our assumption that f(x1) = f(x2) is also wrong. That means when x1 x2, we have f(x1) f(x2). Therefore, the function is one-to-one. b. Clearly, if we choose –2 2 we get f(–2) = f(2) = 4. Although we can find an infinite amount of such examples, just one of them is enough to decide that function is not one-to-one.
162
Algebra 8
We know that the graph of a relation is a function if any vertical line one if crosses the curve at most once. Similarly, a function is one-tto-o any horizontal line crosses the curve at most once. For example: y
y
x
x
Graph of a function which is not one-to-one
Graph of a one-to-one function
HORIZONTAL LINE TEST FOR THE ONE-TO-ONE FUNCTIONS A function is one-to-one if, and only if, no horizontal line crosses its graph at more than one point. Note that in the previous example we could have found the answers without using the definition but just by plotting their graphs and applying the horizontal line test. EXAMPLE
62
Which one of the following graphs belong to one-to-one functions? a.
b.
y
x
Solution
c.
y
x
x
y
y
y
x
y
x
x
a. This graph does not belong to a one-to-one function. b. This graph does not belong to a function, so we can not talk about one-to-one function. c. This graph belongs to a one-to-one function.
Note A function that is only increasing or decreasing in its domain is one-to-one. Functions
163
Check Yourself 14 3x 1 is a one-to-one function. 5 2. Draw a graph which belongs to a one-to-one function.
1. Prove that f ( x) =
3. Draw a graph which does not belong to a one-to-one function. Answers 1. Compare f(x1) and f(x2) when x1 x2. 2. Consider a function which is either increasing or decreasing. 3. Consider a function which is sometimes increasing and sometimes decreasing.
2. Definition for The Inverse of a Function As mentioned earlier in this book, many mathematical relations can be modelled as functions. For example,
C = f(r) = 2r
V = g(a) = a3
The circumference of a circle is a function of the radius r.
The volume of a cube is a function of the edge a.
In many cases we are interested in reversing this correspondence determined by a function:
r m C
C 2
The radius of a circle is a function of the circumference C.
a n V 3 V
The edge of a cube is a function of the volume V.
As illustrated above, reversing the relation between two quantities produces a new function. 164
Algebra 8
Recall that a function is a relation that assigns to each element in the domain exactly one element from the range.
f x0
y0
Domain
Range
In the figure above we see an arrow diagram for function f. So “Using the rule f we say that x0 becomes y0”. We symbolize this as “f(x0) = y0”.
–1
Read f (x) as “f inverse of x”.
Now consider this question: Using the rule f what gives us y0? Of course the answer is x0. So “Using the rule f to get y0 we need x0”. We need a new notation so that “x0” will be our answer. And we symbolize this fact as “f –1(y0) = x0”.
1 . f ( x) 1 The reciprocal is f ( x)
f –1(x) doesn’t mean
f 1
written as (f(x))–1.
x0
y0
Range
Domain
Note The domain of a function is the same as the range of its inverse. The range of a function is the same as the domain of its inverse.
The domain and the range change their places in an inverse of a function just as the server and the opponent change their places in a table tennis game.
Functions
165
EXAMPLE
63
Given the function f = {(0, 2), (–1, 4), (4, 6), (5, 5)}, a. find f –1.
Solution
b. find f –1(4).
a. Interchange the x and y-coordinates of each ordered pair of f to find f –1: f
–1
= {(2, 0), (4, –1), (6, 4), (5, 5)}
b. To find the value of f –1(4), notice that f –1(4) is the second coordinate when the first coordinate is 4 in the function f –1. So f –1(4) = –1. EXAMPLE
64
Solution
If f(x) = 2x – 3, find f –1(5). If we substitute 5 in place of x we find f(5) = 2 5 – 3 = 7 but not f –1(5)! Finding f –1(5) means finding the x value for which f(x) = 2x – 3 gives 5 as a result. 2x – 3 = 5, so x = 4 or f –1(5) = 4. If f –1 also supports the definition for a function, we will call f –1 as the inverse of f. But this is not always possible. For example, consider the relation below: · -3
f
· -1
· -2 · -1
·0
·1
·1 ·2
·3
Domain
Range
Clearly, this is a function since each element is assigned to exactly one element. Now let’s consider its inverse: · -3
f 1
· -1 ·0 ·1
· -2 · -1 ·1
·2
·3
Range
Domain
As we see f –1(1) = –3 or f –1(1) = 0 is not possible for a function (just one element must be assigned for each element in the domain). So although f is a function, f –1 is not a function. That means the inverse of f doesn’t exist.
Note It is not true that every function has an inverse. For the inverse of a function to be defined, it is necessary that different elements in the domain always give different values. As we know such functions are called one-to-one functions. 166
Algebra 8
As another example let us consider the following function graph of which is given below: y y = f(x)
x
To find the inverse, we use the same procedures that we used for relations. Drawing the reflection with respect to y = x we get the following picture (below left): y
y
y = f 1(x)
y = f 1(x)
y = f (x) x
x
y=x
We know that a set of points in the coordinate plane is the graph of a function if, and only if, no vertical line crosses the graph at more than one point. But that rule doesn’t hold for f –1 (look at the picture above right). That means the inverse of f doesn’t exist. Note that the inverse we draw will be a function if f has no horizontal line crossing the graph at more than one point, that is if f is one-to-one. We know that the range of a function becomes the domain of its inverse. We also know that no element from the domain of a function must be left unassigned. Sometimes the range of a function is explicitly so that it is larger than the function’s its real range, that is some elements of the range are not used. In that case when we talk about the inverse some elements in the domain of inverse will be unassigned. This will result in an absence of the inverse function. For example, if f(x) = x + 2 such that f: , although the range seems to be it is in fact (Note that when we put any integer in x + 2, we always get other integers). To guarantee that we will not face such functions we must be sure that any element in the range is assigned by an element from the domain. Such functions are called onto functions. CRITERIA FOR EXISTENCE OF INVERSE OF A FUNCTION A function has an inverse if, and only if, it is one-to-one and onto. If a function is given by only formula or graph (where no explicit range is given), it is naturally an onto function. So there is no need to think about this detail to define its inverse. Functions
167
Now we can give the formal definition for the inverse of a function: inverse function
Definition
Let f be a one-to-one and onto function with the domain A and the range B. Then its inverse function f –1 has the domain B and the range A such that f(x) = y f –1(y) = x. By definition the inverse function f –1 undoes what f does. That is, if we take x, apply f, and then apply f –1, we arrive back at x where we started. Similarly, f undoes what f –1 does. That is why f and f –1 are the inverses of each other.
Note f(f –1(x)) = x and f –1(f(x)) = x. EXAMPLE
65
Solution
Prove that f(x) =x3 and g(x) = 3 x are inverses of each other. When we find f(g(x)) = ( 3 x )3 = x, we can see that f(x) and g(x) are the inverses of each other. Note also that g(f(x)) = x.
EXAMPLE
66
Solution
If f(x) = x + 2 and f –1(x) = x – a, find a. We know that f(f –1(x)) = x. So f(x – a) = x or (x – a) + 2 = x. That gives a = 2.
3. Finding The Inverse of a Function Given the graph of a function, to find the inverse we use the following procedure: 1. Verify that the function is one-to-one by applying the horizontal line test. 2. Take symmetry of the graph of the function with respect to the line y = x. EXAMPLE
67
Answer the following using the graph on the right:
y
(7,11) y = f(x)
a. Does f have an inverse? Why?
(6,6)
b. Draw the graph of the inverse function f –1. c. Find f –1(-6), f –1(-3), f –1(0), f –1(6), f –1(11).
(0,-3)
(4,0)
x
–1
d. Find the domain and the range of f . Solution
(-10,-6)
a. Applying the horizontal line test we can see that f is one-to-one. So it has an inverse y
x
168
Algebra 8
b. We draw the line y = x and reflect f with respect to it to draw the graph of f –1: y (11,7) (0,4)
(6,6)
1
y = f (x)
(-3,0) x
(-6,-10)
c. f –1(6) = –10, f –1(–3) = 0, f –1(0) = 4, f –1(6) = 6, f –1(11) = 7. d. f –1 can take any x between –6 and 11, inclusive, so the domain is [–6, 11]. The value y of f –1 can be any number x between –10 and 7, inclusive, so the range is [–10, 7].
Given the formula for a function, to find the inverse we use the following procedure: 1. Verify that the function is one-to-one. 2. Solve the equation y = f(x) for x and interchange x and y in the end.
EXAMPLE
68
Given that f(x) = 2x + 5, answer the following: a. Does f have an inverse? Why? b. Find the formula of the inverse function f –1. c. Find f –1(19), f –1(1). d. Find the domain and the range of f –1.
Solution
a. f is a linear function which is always increasing. That means it is one-to-one and so it has an inverse. y5 . b. Let y = 2x + 5. Then x = 2 Interchanging x and y we have the inverse function as y =
x5 x5 or f –1(x) = . 2 2
1 5 x5 19 5 1 = 2. we get f 1(19) = = 7 and f (1) = 2 2 2 Note that we could also solve equations 2x + 5 = 19 and 2x + 5 = 1 to find f –1(19) and
c. Using the formula f 1( x) =
f –1(1), respectively. d. Clearly, D( f –1) = and E( f –1) = . Functions
169
FINDING THE INVERSE OF A FUNCTION To plot the graph of the inverse of a function, reflect the graph of the function with respect to the line y = x. To find the formula for the inverse of a function, solve the equation y = f(x) for x and interchange x and y. EXAMPLE
69
If possible draw the graphs of inverses for the following functions: a.
b.
y
y y = g(x)
(0,4)
y = f(x)
(-6,0)
Solution
(-3,3)
(2,3)
(0,0)
x
x
a. Clearly (horizontal line test), f is one-to-one. So the inverse is the symmetry of f with respect to the line y = x: y y = f 1(x)
y = f (x) (0,4) (-6,0)
(4,0)
x
(0,-6)
b. g is not one-to-one (horizontal line test). We cannot draw the graph of the inverse function since it doesn’t exist. EXAMPLE
70
Given that the following one-to-one functions, find their inverses. a. f(x) = 4 – 3x
Solution
b.
f ( x) =
2x – 6 3
c. g(x) = x3 – 5
Since we know that the functions are one-to-one, we can find their inverses directly a. Let y = 4 – 3x. Then x = b. Let y =
4– y 4– x 4– x . Interchanging x and y: y = or f –1( x) = . 3 3 3
2x – 6 3y +6 3x +6 3 x +6 . Then x = . Interchanging x and y: y = or f –1( x) = . 3 2 2 2
c. Let y = x3 – 5. Then x = 3 y +5. Interchanging x and y: y = 3 x +5 or g –1( x) = 3 x +5. 170
Algebra 8
EXAMPLE
71
If possible find the inverses of the following functions: a.
Solution
f ( x) =
2x + 4 3x 5
a. We can verify that f ( x) =
b.
f ( x) = 3 2 x 1+7
c. f(x) = x2 – 2x
2x + 4 is one-to-one as follows: 3x 5
Let x1 x2. Then assume that f(x1) = f(x2). So, 2 x1 + 4 2 x2 + 4 = 3x1 5 3x2 5 6 x1x2 10 x1 +12 x2 20 = 6 x1x2 +12 x1 10 x2 20 22 x2 = 22 x1 x2 = x1.
But this is not correct since we let x1 x2. This means that if x1 x2, then f(x1) f(x2). So f is one-to-one. Now let’s find its inverse: f ( x) = y =
2x + 4 3x 5
3xy 5y = 2x + 4 3xy 2 x = 4+5 y x(3y 2) = 4+5 y x=
4+5 y 3y 2
Interchanging x and y we have y =
4+5 x 4+5 x or f 1( x) = . 3x 2 3x 2
b. We leave the verification of the fact that f is one-to-one to student and proceed to finding its inverse. y = 3 2 x 1+7 y 7 = 3 2x 1 ( y 7)3 = 2 x 1 ( y 7)3 +1 =x 2 ( x 7)3 +1 ( x 7) 3 +1 or f 1( x) = . 2 2 c. Choosing x = 0 and x = 2 we can realize that f will give the same value. So f is not one-to-one. Proof of this using definition is left to the student in exercises. As a result f does not have an inverse.
Interchanging x and y we have y =
Functions
171
Check Yourself 15 1. If f(2) = 3, f(5) = 6, f(6) = –1, find f –1(–1), f –1(3), f –1(6). 2. If possible draw the inverse of f graph of which is given on the right. 3. Find the inverse of f ( x) =
y
(-3,2)
2x + 3 . 7
(0,0)
Answers 1. 6, 2, 5
y = f(x)
x (1,-3)
y
2. (3, 1)
3. f 1( x) =
(0, 0)
x
7x 3 2
(2, 3)
The functions which are not one-to-one can have their domains restricted so that they become one-to-one. As a result their inverses will then be functions.
EXAMPLE
72
Given that f(x) = x2 + 1, a. find the largest possible domain so that the inverse exists. b. find f –1(x) using the new domain.
Solution
a. The inverse exists if we have a one-to-one function. Note that for each value of x and its negative we have the same value of y. So, if we ignore the negative x values, we will have a one-to-one function. That means to have an inverse, domain must be restricted to [0, ). b. Let y = x2 + 1 and we will try to solve the equation for x: x2 = y – 1, so x = y – 1. If we interchange x and y we have y = x – 1. Note that the range of the inverse function is the domain of the original function, that is [0, ). So y = x – 1. Therefore, 1 the inverse function is f ( x) = x – 1.
172
Algebra 8
EXAMPLE
73
Solution
Prove that f(x) = x2 – 2x – 2 has an inverse if D( f ) = [1, ) and then find its inverse. f(x) = x2 – 2x – 2 = x2 – 2x + 1 – 3 = (x – 1)2 – 3. Here (x – 1)2 is equal to the same number for x – 1 = x0 and x – 1 = –x0 for any x0 > 0. This fact prevents f(x) = (x – 1)2 – 3 from being a one-to-one function. But if we guarantee that x – 1 is never negative, then the function will be one-to-one. And that is possible when x – 1 0, x [1, ). To find the inverse, we solve y = (x – 1)2 – 3 for x: y = ( x 1)2 3 y + 3 = ( x 1)2 y + 3 =| x 1| Since x 1 0 we have
y+ 3 = x 1 or
y+ 3 +1= x.
Interchanging x and y we have f 1( x) = x + 3 +1.
EXAMPLE
74
Solution
x +1 –1 3 If f 1 = x , find f (2) + f(8). x
Let us find f –1(2). We don’t have the formula for f –1(x) but for f 1(
x +1 ). The expression x
inside the brackets must be equal to 2 since we are looking for f –1(2). When we solve
x +1 = 2, we find x = 1. x
x +1 ) = f 1(2) =13 1. x 1 x +1 ) = x3 Finding f(8) is simple. Note that f is inverse of f –1. So if we need f(8), the result of f ( x x +1 x +1 must be equal to 8, that is f 1( will be f(8) for the corresponding x value. ) = 8, where x x
That means when x = 1, f 1(
So, x3 = 8, x = 2 and
x +1 2+1 3 = = . x 2 2
3 5 Finally, f 1(2)+ f (8) =1+ = . 2 2 Functions
173
EXAMPLE
75
Using the graph below find (f g–1 f –1)(5) + (f g f)(–3). y y = f(x)
1 1
x y = g(x)
Solution
Note that we don’t have the graph of any inverse function and we don’t need them at all. To find the value of a function just find the y-value for the given x-value on the graph. To find the value of an inverse function just find the x-value for the given y-value on the graph. (f g–1 f –1)(5) = f(g–1(f –1(5))) = f(g–1(4)) = f(0) = 2 (f g f)(–3) = f(g(f(–3))) = f(g(0)) = f(4) = 5 So (f g–1 f –1)(5) + (f g f)(–3) = 2 + 5 = 7.
EXAMPLE
76
Solution
If (f g)(x) = 5x – 2 and f(x) = 4x – 3, find g–1(x). (f g)(x) = f(g(x)) = 4 g(x) – 3 = 5x – 2 Solving the last equation for g(x) we have g( x) = To find the inverse, we solve y =
5x +1 4y 1 for x: x = 4 5
1 Interchanging x and y we have g ( x) =
EXAMPLE
77
Solution
4x 1 . 5
Given f(x) = 2x – 7, find f –1(4x + 1). Let us find f –1(x): f ( x) = 2 x – 7 = y y +7 x= 2 x +7 f –1( x) = 2 Now let us find f –1(4x + 1): f –1( 4x +1) =
174
5x +1 4
(4 x +1)+7 = 2x+ 4 2 Algebra 8
EXAMPLE
78
Given that f(x) = 3x + 1 and g(x) = 2x + 2, find a. (f g)–1
Solution
b. g–1 f –1
a. Let us find f g. y = (f g)(x) = f(g(x)) = 3(2x + 2) + 1 = 6x + 7 To find the inverse, x = (f g)–1(x) =
y–7 6
x–7 . 6
b. Let us find f –1 and g–1. y = f(x) = 3x + 1 To find the inverse, x = f –1(x) =
y –1 3
x –1 . 3
y = g(x) = 2x + 2. To find the inverse, x =
y–2 2
x–2 . 2 x–1 –2 x–7 (g–1 f –1)(x) = g–1( f –1(x)) = 3 = 2 6
g–1(x) =
Note that we have (f g)–1(x) = (g–1 f –1)(x). Is it always correct? Why?
Check Yourself 16 1. Given that f(x) = x2 – 4, restrict the domain so that the inverse exists and find its formula. 2. Given that f(2x + 1) = x – 5, find f –1(0). 3. Given that f(g(x)) = 2x + 1 and g(x) = x + 4, find f –1(x). 4. Given that f(x) =
x –1 and g(x) = 3x, find (f g)–1 and g–1 f –1. 4
Answers 1. [0, ), Functions
x+ 4
2. 11 3.
x +7 2
4.
4x +1 4 x +1 , 3 3 175
EXERCISES
4 .3 5. Find the required values using the given data:
A. Basic Operations 1. Find f + g, f – g, fg, f / g for the following functions.
a. f ( x) =
2
a. f(x) = x + 1, g(x) = x – 1
3x +5 x+ 2 and g( x) = , x +1 x 1
( f g f )(0) = ?
b. f(x) = x3 + 3x2, g(x) = x2 + 5x
x +1 2 b. f(2x + 1) = 3x – 1 and g = x +1, x 1 ( f g)(–1) = ?
c. f ( x) = x + 3, g( x) = x + 2 d. f ( x) = x 1, g( x) = x +1
c. f (3x 1) =
2. If
1 f ( x) = 3x + 4, g( x) = x 2 + x, h( x) = , x
( f g)(1) = ?
find
2 x +1 if x > 2 d. f ( x) = and if x 2 x 2
the following functions. b. f h ( 1) g
a. (f + g)(3)
if x > 0 x g( x) = if x 0 x ( f g f )(–2) = ?
c. (hg – f)(4)
B. Composition of Functions
6. If f(x) = x3 – 3x2 + 3x – 1 and g( x) =
3. Find f f, f g, g f, g g for the following
solve ( f g)(x) = 0.
functions. a. f ( x) =
x +1 , g( x) = x2 + x x 1
(g f )(x) = 3f(x) – 1,
b. f(x) = x – 2x + 1, g(x) = x + 1
find a such that ( f + g)(a) = 19.
8. If f ( x) =
4. Express h(x) in terms of f(x) and g(x) such that h(x) = (f g)(x).
C. Inverse of a Function functions or not. a. {(x, 1), (y, 1), (z, 2), (k, 3)}
7
x4 b. h( x) = 5
b. {(1, 1), (2, 2), (3, 3), (4, 1)} 3
2
c. h(x) = (2x – x) – (2x – x) + 4 176
1 x , find f f ... f ( x). 1+ x 2004 times
9. State whether the following relations are one-to-one
a. h( x) = 2 x 4
2
2 x +1 , x+ 3
7. If ( f g)(x) = 5g(x) + 4 and
2
2x + 3 x+ 3 c. f ( x) = , g( x) = x 3x 2
x+ 3 x2 and g( x+1) = , 2x 1 x 1
c. {(a, b), (b, a), (c, d), (d, c)} Algebra 8
10. State whether the following functions are one-to-one
13. Plot the graphs of inverse functions for the
or not.
following functions, if possible:
a. f(x) = x2 – 1, D( f ) = {0, 1, 2, 3}
a.
b.
y
y
(4,5)
b. f(x) = x2, D( f ) = +
(0,4)
c. f(x) = x2 d. f(x) = 3x + 5 e. f ( x) =
x
x 1 x2
x
(2,-2)
(-8,-3)
c.
y
(-5,-5)
(5,7)
d.
11. State whether the following graphs belong to b.
y
(0,3)
x
(0,0)
(2,-3)
(-6,-3)
one-to-one functions or not. a.
y
(-6,7) (-5,4)
(4,1) (-2,-1)
(5,-5)
e.
y
(6,-5)
f.
y
y (0,5)
(-5,3)
x
x
x
x
(4,-2)
x
(6,-5)
c.
y
14. Find the required values using the given data for the functions with the inverse.
x
2 x +5 1 , f (3) = ? 5 b. f –1(x) = x2 – 3, D(f –1) = [0, ), f(–3)= ?
a. f ( x) =
12. Find the inverses of the following one-to-one functions: 3x + 4 x 1
a. f(x) = 6 – 4x
b.
f ( x) =
c. f(x) = 4x3 – 5
d.
f ( x) = 3 5 x 1+9
Functions
c. 2 f ( x) 1=
f ( x)+1 1 , f (2) = ? x2
d. f (3x +1) =
4 x 1 1 , f (3) = ? x2
3 e. f(x2 + x + 1) = x – 2, D( f ) = , , 4 f –1(0) = ? 177
15. Find the following using the given data for
18. Find the following using the graph of functions
functions with inverses:
y = f 1(x) y
2 x +1 1 a. f = x +1, f ( x) = ? x 1
b. f ( x 1) =
below:
3x 2 1 , f ( x) = ? x + 21
1
3x 5 1 c. f = x, f (2 x) = ? x +1
1
Mixed Problems
a. (g f –1)(–2) + ( f –1 g f )(2) b. ( f –1 g)(2) + ( f g)(3)
16. Find the following using the given data for
functions with inverses: a. f(x) = 3x + 1, g(x) = 2x + 3, (f g–1)(x) = ?
19. Prove that
1 x+ 2 b. f ( x) = , g( x) = , ( g 1 f )( x) = ? x 2 x +1
c. f 1( x) =
2 x +1 1 3x +1 , g ( x) = , ( f 1 g1 )( x) = ? x3 x 2
d. f 1( x) =
x 2 1 2 , g ( x) = , ( f g)1( x) = ? x x2
f. g( x) =
3x +1 , ( f g)( x) = 2 x+ 3, f 1( x) = ? x 1
g. g( x) =
2 x +1 , ( g 1 f )( x) = 2 x 1, f 1( x) = ? x
17. Find the required values using the given data for functions with inverses:
20. If f and g are even functions such that f(0) = g(4) = –2, g(0) = f(4) = 4, f(–2) = g(–2) = 0, find
a. f(x) = x + 1, g(x) = 3x – 4, ( f
b. f(x) = x – 2, g(x) = 2x + 1, ( f g–1)(3) = ? x+ 3 , g( x) = 6x +1, ( f g)1(2) = ? c. f ( x) = x2
2x + 3 1 2 x +5 d. f = x, g = 3x 2, x +1 x+ 3 ( f g )(3) = ? 178
21. Prove that if f and g are even, then f + g, f – g, fg,
f / g are also even.
22. Prove that if f and g are odd, then f + g, f – g are odd and fg is even.
g)(2) = ?
2
1
( g f )(4)+( f g)(–4)+( g f )(–4) . ( f g)(2)+( g f )(2)+( f g)(4)
–1
a. f(x) = x2 – 2x is not one-to-one. b. f(x) = 2x3 + 4 is one-to-one.
e. f(x) = 2x + 3, (g f )(x) = 4x – 5, g(x) = ?
x y = g(x)
23. Find the inverses of the following functions:
a. f(x) = x2 + 6x, D( f ) = (–, –3] b. f(x) = x2 – 8x + 5, D( f ) = [4, )
24. Prove that (g f )–1 = f –1 g–1.
Algebra 8
Cryptography is the art of secret writing. Its aim is to protect information by transforming it into a coded language which unwanted eyes are unable to use. This transformation, however, must be done in a reversible way so that individuals intended to view the information may do so. Some of the classical methods of cryptography are as follows (However, none of them are used today, because they are considered either insecure or impractical).
Simple Substitution: When we encode the message MATH IS COOL , it becomes NBUIAJTADPPM. Here, the rule is that every letter is substituted by the letter that follows it. That is, A by B, B by C, “space” by A, etc.
Mixing: Take a key sequence consisting of the first few natural numbers in mixed order, for example 3, 5, 1, 2, 4. The rule is that the first letter will move to the 3rd place, the second to 5th place, the third to 1st place, the fourth to 2nd place, the fifth to 4th place, and so on for the next five letters. So MATH IS COOL becomes THM A CIOS OL. Note that the “space” character is also treated as a letter.
Method of Vigenere: Pair each letter with a number with its order in the alphabet as follows: A B C D E F G H I J K L M N O P Q R S T U V W X Y Z 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 Functions
179
Our message MATH IS COOL becomes 13 1 20 8 27 9 19 27 3 15 15 12 Choose a key word, for example, FUNCTION which is 6 21 14 3 20 9 15 14. Now, add 6 to the first letter of our encoded message, 21 to the second, 14 to the third, and so on. The new message will be read as 19 22 34 11 47 18 34 41 9 36 29 15. If the number is more than 27, divide it by 27 and take the remainding number. Finally, the message will become SVGKTRGNIIBO. Find the pair for these numbers from the table above. The whole procedure is summarized in the table below. 13 1 20 8 27 9 19 27 3 15 15 12 6 21 14 3 20 9 15 14 6 21 14 3
convert the message using table. convert the keyword and write it as the same length as the original message.
19 22 34 11 47 18 34 41 9 36 29 15
add the top two rows.
19 22 7 11 20 18 7 14 9
divide the previous row by 27 and write the remainding number.
9
2 15
Note that all of these methods are functions that are defined from a limited domain to a range. We can invent even more complicated methods by applying a few of them together for the same message (Think about the composition of functions!). Here it is very important to know the rules. To decode a crypted message we just apply the rule in reverse order (Think about the inverse of a function!). Below is an encoded message. We first applied the Method of Vigenere with the key word CAT (each letter is paired by its order in the alphabet as “space” being the 27th letter), and then we applied the simple substitution.
GQHKTVXWFEVCSPM Try to decode it! What will happen if the rule is a function that has no inverse? What happens if we assign the same number to two different letters? If a few of the rules are applied in order, by which rule should we start the decoding procedure? Try to develop your own method of encryption. Algebra 8 180
Definition
quadratic function A function f(x) is a quadratic function if y = f(x) = ax2 + bx + c,
a0
where a, b, and c are real numbers. For example, f(x) = 2x2 + 3x + 1, y = x2 – 1, y = –ñ2x2 and f(x) = 2x – x2 are all quadratic functions. y = 3x – 4 is not a quadratic function, because a = 0. The domain of a quadratic function is the set of all real numbers. The graph of a quadratic function f(x) = ax2 + bx + c is called a parabola. a>0 y
a<0 y
axis of symmetry
V
c The sign of a has no effect on the size or shape of the parabola, the sign of a determines whether the parabola opens upward or downward.
x-intercept x1
x2
x1 x
x2 x
x-intercept c y-intercept
a>0
a<0
V
vertex
When a > 0, the parabola opens upward.
When a < 0, the parabola opens downward.
The point V in the parabolas above is called the vertex of the parabola. The vertex is the lowest or the highest point of the parabola. The vertical line drawn through the vertex is called the axis of symmetry. It divides the curve into two symmetrical halves. The points x1 and x2 are called the x-iintercepts of the parabola. They are the zeros of the function. The point c is called the y-iintercept of the parabola. 182
Algebra 8
A. GRAPHING y = ax2 How can we find the graph of the quadratic function y = ax2? If we have the function, we can plot the graph by making a table of values. To find the values, we substitute different values of x into the equation to obtain the corresponding y values. These x and y values provide the coordinates for points which we can plot to form the shape of the graph. Let us graph the function y = ax2. If a > 0, we get the table of ordered pairs opposite. Then we plot the points
x
–
–2
–1
0
1
2
+
y
+
4a
a
0
a
4a
+
on a graph and draw a parabola through them. y 4a 3a 2a a -2
-1
1
x
2
y = ax2, a > 0
We can see that the vertex of the parabola is at the origin (0, 0), and the axis of symmetry lies along the y-axis (the line x = 0). If a < 0, we get a different set of ordered pairs.
x
–
–2
–1
0
1
2
+
y
–
4a
a
0
a
4a
–
y -2
-1
a
1
2
x
2a 3a 4a
Quadratic Functions
183
1
EXAMPLE
Sketch the graphs of the functions. a. y = x2 , y = 2x2 and y = 1 x2 2 a.
Solution
As |a| increases, the parabola becomes narrower. As |a| decreases, the parabola becomes wider.
x
–
–2
–1
0
1
2
y = x2
4
1
0
1
4
y = 2x2
8
2
0
2
8
1 2 x 2
2
1 2
0
1 2
2
y= y
ax2 x2
b. y = x2 , y = 2x2 and y = 1 x2 2
1 x2 2 x2 2x2
y 4
a>1
3
x y
2
x2 ax2
1 -4
-3
-2
x
-1
1
–
–2
–1
0
1
2
y = –x2
–
–4
–1
0
–1
–4
–
y = –2x2
–
–8
–2
0
–2
–8
–
–
–2
–
1 2
0
–
1 2
–2
–
2
3
4
x
0
b.
y=
ax2 +
x
bx
y
y
b a
x a>0
y= –
b a
y
x
1 2 x 2
y
y
-4
-3
-2
-1
1
2
3
4 x
b x b a a a<0
x
2x2
184
x2 1 x2 2
Algebra 8
Check Yourself 1 Graph the functions. 1. y = 3x2
2. y = 1 x2 3
4. y = – 1 x2 3
3. y = –3x2
B. GRAPHING y = ax2 + bx + c 1. Vertex Point a>0
a<0
Then ax2 + bx+ c = a(x x1 )(x x2 )
b 2a
b 2a
b 2a
b 2a
Let x1 and x2 be zeros of the polynomial ax2 + bx + c.
b 2a
b+ b = a x x 2a 2a b b+ b 2 = a x2 x x 2a + 4a2 2a xb+ x + xb x b2 = a x2 + + 2 2 2a 4a 4a
bx b 2 = a x2 + + 2 a 4a 4a
b 2a
2
b = a x+ 2 4 a a = a(x h)2 + k where h =
If a quadratic function has two known x-intercepts, we can find the x-coordinate of the vertex point by averaging them. h=
x1 + x2 b =– 2 2a
b and k = . 2a 4a
This is called the completed square form of the quadratic function. COMPLETED SQUARE FORM The completed square form of a quadratic function is f(x) = a(x – h)2 + k. The vertex of a parabola of a function in complete square form is the point (h, k). b 4ac b 2 and k = f(h) or k = . 2a 4a Since (h, k) is the vertex of a parabola, then the x-coordinate of the vertex point is the equation of the axis of symmetry, x = h.
At the vertex point V(h, k), h =
Quadratic Functions
185
If a > 0, then the y-coordinate of the vertex is the minimum value of the function. If a < 0, then the y-coordinate of the vertex is the maximum value of the function. a>0
a<0 y
y
D 4a
maximum value of the function V(h,k)
b 2a x
The trajectory of a shell fired from a cannon is a parabola. To reach the maximum value range with a cannon, it is shown in calculus that the barrel of the cannon must be set at 45°.
EXAMPLE
2
D 4a
x
V(h,k) minumum value of the function
Find the coordinates of the vertex point of each function. Write the equation of the axis of symmetry, and determine the maximum or minimum value of the function. a. y = x2 – 2x – 3
Solution
b 2a
b. y = –x2 + 4x + 5
c. y = x2 + 3x + 1
a. V(h, k) is the vertex point. b 2 1 h 2a 2 k = f(h) = 12 – 2 1 – 3 = –4 So the vertex is V(1, –4). The equation of the axis of symmetry is x = 1. Since a = 1 > 0, the vertex is the minimum value and k = –4 is the minimum value of the function.
b. h =
b 4 = =2 2a 2 (–1)
k = f(h) = –22 + 4 2 + 5 = 9 So the vertex is V(2, 9). The equation of the axis of symmetry is x = 2. Since a = –1 < 0, the vertex is the maximum value and k = 9 is the maximum value of the function. 186
Algebra 8
c. h =
b 3 3 = = 2a 2 1 2 2
5 3 3 k = f (h) = + 3 + 1 = 4 2 2 3 5 So the vertex point is V , . 2 4
3 The equation of the axis of symmetry is x = . 2
Since a = 1 > 0, the vertex is the minumum value and y = the function. EXAMPLE
3
Solution
5 is the minimum value of 4
The line 3x + 1 = 0 is the axis of symmetry of the parabola y = –2x2 + mx – 1. Find the maximum value of the function. 3x + 1 = 0 ; x =
1 is the equation of axis of symmetry. 3
1 b m 4 h= = = ; m= 3 2a 2 (–2) 3
Since a = –2 < 0, the vertex is the maximum value. 2
4 1 7 1 k = f (h) = 2 1 = 3 3 9 3
So k =
EXAMPLE
4
Solution
7 is the maximum value. 9
The minimum value of the function y = x2 – 8x + n is –4. What is the value of n? Given k = –4 and h = k = f (h) ;
b 8 = = 4 ; h = 4. 2a 2 1
4 = 42 8 4 + n ;
n = 12
Check Yourself 2 Find the vertex of each parabola. 1. y = x2 – 10x + 20
2. y = 1 x2 + x – 8 2
3. y = 4 – x2
4. y = 4x + x2
Answers 1. (5, –5) 2. (–1, –8.5) Quadratic Functions
3. (0, 4)
4. (–2, –4) 187
2. Intercepts Consider the function y = f(x) = ax2 + bx + c,
a 0.
If x = 0, then y = c. The point (0, c) is called the y-intercept of the graph. We mean that the parabola intersects the y-axis at this point. If y = 0, then ax2 + bx + c = 0. Let x1 and x2 be roots of this equation. The points (x1, 0) and (x2, 0) are called the x-intercepts of the graph. We mean that the parabola intersects the x-axis at these points. There are three possibilities for the x-intercepts. 1. If > 0, the parabola intersects the x-axis at two distinct points. 2. If = 0, the parabola is tangent to the x-axis. 3. If < 0, the parabola does not intersect the x-axis. D<0 D=0
a<0 D>0
x D>0
a>0
D=0 D<0
EXAMPLE
5
Solution
Find the x- and y-intercepts of each function. a. y = x2 – 9
b. y = –4x2 + 5x
d. y = x2 + 3x + 7
e. y = 4x2 + 20x + 25
c. y = 3x2 + 4x – 2
a. x = 0 ; y = –9 So (0, –9) is the y-intercept. y = 0 ; x2 – 9 = 0 x1 = –3, x2 = 3 So (–3, 0) and (3, 0) are the x-intercepts. b. x = 0 ; y = 0 So (0, 0) is the y-intercept. y = 0 ; –4x2 + 5x = 0 ;
x1 = 0 , x2 =
5 4
5 So (0, 0) and , 0 are the x-intercepts. 4 188
Algebra 8
c. x = 0 ; y = –2 So (0, –2) is the y-intercept. y = 0 ; 3x2 + 4x – 2 = 0 = 42 – 4 3 (–2) = 40 x1 =
4 2 10 2 10 4 + 2 10 2 + 10 , x2 = = = 6 3 6 3
So (
2 10 2 + 10 , 0) and ( , 0) are the x-intercepts. 3 3
d. x = 0 ; y = 7 So (0, 7) is the y-intercept. y = 0 ; x2 + 3x + 7 = 0 = 32 – 4 1 7 = –19 < 0 So the parabola does not cut the x-axis, i.e. there are no x-intercepts. e. x = 0 ; y = 25 So (0, 25) is the y-intercept. y = 0 ; 4x2 + 20x + 25 = 0 = 202 – 4 4 25 = 0 x1 = x2 =
20 5 = 8 2
= 202 – 4 4 25 = 0 5 So the parabola is tangent to the x-axis at the point (– , 0). 2 EXAMPLE
6
The function y = x2 – (m + 2)x + 5 + m is given. Find the value(s) of m for which the parabola a. does not cut the x-axis.
b. is tangent to the x-axis.
c. intersects the x-axis at two distinct points. Solution
= (m + 2)2 – 4(5 + m) = m2 – 16 = m2 – 16 = 0 ;
m = 4
Let us consider the sign of . m ¥ m2 16
Quadratic Functions
4
4
+
¥
+ 189
a. If m (–4, 4), the parabola does not cut the x-axis. b. If m = 4, the parabola is tangent to the x-axis. c. If m (–, –4) (4, ), the parabola cuts the x-axis at two distinct points.
Check Yourself 3 Find the x- and y-intercepts of each parabola. 1. y = –x2 + 3x – 4
2. y = 3x2 – 6x + 3
3. y = 3x – x2
Answers 1. no x-intercepts, (0, –4)
2. (1, 0), (0, 3)
3. (0, 0), (3, 0)
3. Sketching a Graph To sketch the parabola y = ax2 + bx + c, follow the steps. 1. Check a. When a > 0, the graph has a minimum point and the parabola opens upward. When a < 0, the graph has a maximum point and the parabola opens downward. 2. Find the coordinates of the vertex point, V(h, k). 3. Find the x- and y-intercepts. 4. Sketch the graph of the parabola.
EXAMPLE
7
Solution
Graph the functions. a. y = x2 – 3x – 10
b. y = –2x2 – 8x
d. y = –3x2 + 12
e. y = x2 – 2ñ2x + 2
c. y = x2 + 4x + 5
a. a = 1 > 0, so the parabola opens upward. h=
3 b = , 2a 2
k = f (h) = (
y
3 2
-2
3 2 3 49 ) 3 10 = ; 2 2 4
3 49 ) is the vertex point. V( , – 2 4
x = 0 ; y = –10, (0, –10) is the y-intercept.
x
5
f(x)=x2 3x 10 -10 49 4
V
y = 0 ; x2 – 3x – 10 = 0 ; x = –2, x = 5 ; (–2, 0) and (5, 0) are the x-intercepts. 190
Algebra 8
b. a = –2 < 0, so the parabola opens downward. h=
V
8 = 2, 2 (–2)
k = f (h) = 2(–2) 8 (–2) = 8 ; 2
V(–2, 8)
y 8
f(x)=2x2 8x
x = 0 ; y = 0, (0, 0) is the y-intercept. y = 0 ; –2x2 – 8x = 0 ; x = 0, x = –4 ; (–4, 0) and (0, 0) are the x-intercepts.
-4
c. a = 1 > 0, so the parabola opens upward. h=
y
4 = 2, 2 1
5
k = f (h) = (–2)2 + 4 (–2) 2 + 5 = 1 ;
x
-2
f(x)=x2+4x+5
V(–2, 1)
x = 0 ; y = 5, (0, 5) is the y-intercept. 2
y = 0 ; x + 4x + 5 = 0 ;
1
V
x
-2
= 42 – 4 1 5 = –4 < 0 ; the parabola does not cut the x-axis. d. a = –3 < 0, so the parabola opens downward. h=
y
0 = 0, 2( 3)
k = f (h) = 3(0)2 +12 = 12 ;
V 12
V(0, 12) f(x)=3x2+12
x = 0 ; y = 12, (0, 12) is the y-intercept. y = 0 ; –3x2 + 12 = 0 ; x = 2 ; (–2, 0) and (2, 0) are the x-intercepts.
2
-2
e. a = 1 > 0, so the parabola opens upward. h=
y f(x)=x2 2ñ2x+2
2 2 = 2 2 1
k = f (h) = ( 2 )2 2 2 2 + 2 = 0 ;
x
V( 2, 0)
x = 0 ; y = 2, (0, 2) is the y-intercept.
2 V ñ2
x
2
y = 0 ; x – 2ñ2 + 2 = 0 ; x1 = x2 = ñ2 ; the parabola is tangent to the x-axis at the point (ñ2, 0). Quadratic Functions
191
EXAMPLE
8
Solution
Sketch the graph of the function y = –x2 – 2x + 8 for |x| 3. Find the range of the function. a = –1 < 0, so the parabola opens downward. h=
y
2 = 1, 2 (–1)
9 8
k = f (h) = (–1)2 2 (–1) + 8 = 9 ;
V(–1, 9)
5
x = 0 ; y = 8, (0, 8) is the y-intercept. y = 0 ; –x2 – 2x + 8 = 0 ; x1 = –4, x2 = 2 ; (–4, 0) and (2, 0) are the x-intercepts. |x| 3 ; –3 x 3 x = –3 ; y = –9 + 6 + 8 = 5, so (–3, 5) is on the parabola. x = 3 ; y = –9 – 6 + 8 = –7, so (3, –7) is on the parabola. So the range of the function is [–7, 9].
EXAMPLE
9
Solution
3
-4 2
-3 -1
x
-7
Find the values of p for which px2 + 4x + p is greater than zero for all real values of x.
px2 + 4x + p > 0 for all real numbers x.
y = px2+ 4x+p
This is possible only if p > 0 and < 0. (1)
p>0 2
2
4 – 4p < 0 (2)
x
(1) p > 0 (2) 42 – 4p2 < 0 ; 42 – 4p2 = 0 ; p = 2 p
¥
2
0
2
¥
(1)
+
+
(2)
+
+
system
So p (2, ).
Check Yourself 4 Sketch the graph of each function. 1. y = x2 – 5x + 4 192
2. y = –x2 + 4x – 3
3. y = 4x2 – 20x + 25
4. y = 3x2 – 2x Algebra 8
Answers 1. y
2.
3.
y
4
1
3
4.
y
y
25
x
2/3
-3 4 x
1
5/2
x
x
4. Shifting Graphs Let y = f(x) be a function. Vertical Shifting y = f (x ) + k
k > 0 shift the graph of y = f(x) k units upward. (see Figure 1) k < 0 shift the graph of y = f(x) |k| units downward. (see Figure 1)
Horizontal Shifting y = f (x+ h)
h > 0 shift the graph of y = f(x) h units to the left. (see Figure 2) h < 0 shift the graph of y = f(x) |h| units to the right. (see Figure 2)
Reflection y = –f(x)
Reflect the graph of y = f(x) in the x-axis. (see Figure 3)
Horizontal and Vertical Shifting y = f(x + h) + k Shift the graph of y = f(x) |k| units upward or downward and |h| units to the left or to the right. k>0
y
y
x2+k
h>0
x2 (x h)2
(x+h)2
x2 k
x2 k
0
x
y
-h
0
h
x
x2
x
0
-k x2
Figure 1 EXAMPLE
10
Quadratic Functions
Figure 2
Figure 3
Sketch the graph of the each function using the shifting method. a. y = x2 + 3
b. y = x2 – 3
c. y = (x – 3)2
e. y = –(x – 4)2
f. y = (x + 2)2 – 3
g. y = –(x – 2)2 + 1
d. y = (x + 3)2
193
Solution
a.
y
b.
x2+3
c.
y
x2
d.
y
(x+3)2
x2 (x 3)2
x2
y x2
x2 3
3 x
0
0
x
0
x
3
0
-3
x
-3
e.
f.
y
g.
y
y
(x+2)2 x2
x2
x2
(x+2)2 3
4
1
x
0
-2
x
0
x
2
-3
(x 4)2
x2
0
(x 2)2+1
y
Check Yourself 5 Sketch the graph of the each function by shifting. 1. y = x2 + 4 Answers 1. y
2. y = –(x + 1)2 2.
x2+4
3. y = (x – 2)2 – 2 3.
y -1
x2
x
(x+1)2
y
x2
4.
(x2)2
y 1
x
-1
4
x2
4. y = –(x + 2)2 + 1
(x2)22
4 2 -2
-2 (x+2)2+1
2
-3
x
-4
x (x+2)2
x2
5. Parabolas with Absolute Value (Optional) EXAMPLE
194
11
Sketch the graph of each function. a. y = |x2 – 3x + 2|
b. y = –x2 + 2|x| + 3
c. y = x|x + 2|
d. y = |2x2 – 12x + 16| + 2 Algebra 8
Solution
a. First graph the function y = x2 – 3x + 2.
y
Then, take above the part of the graph which is below
2
the x axis, symmetric with respect to x-axis. a = 1 > 0, so the parabola opens upward. h=
3 3 = , 2 1 2
f(x)=|x2 3x+2| 1 1/4 3/2
1
-1/4
2
x
2
3 1 3 k = f (h) = 3 + 2 = ; 2 4 2 1 3 V , is the vertex point. 4 2
x = 0 ; y = 2, (0, 2) is the y-intercept y = 0 ; x2 – 3x + 2 = 0 ; x = 1, x = 2 ; (1, 0) and (2, 0) are the x-intercepts. b. y = –x2 + 2|x| + 3 Case 1 ____________________________________________________________________ If x 0, |x| = x ; y = –x2 + 2x + 3 a = –1 < 0, so the parabola opens downward. h=
2 = 1, 2 (–1)
k = f (1) = 12 + 2 1+ 3 = 4 ;
V(1, 4) is the vertex point.
x = 0 ; y = 3, (0, 3) is the y-intercept. y = 0 ; –x2 + 2x + 3 = 0 ; x = –1, x = 3 ; (–1, 0) and (3, 0) are the x-intercepts. Case 2 ____________________________________________________________________ If x < 0, |x| = –x ; y = –x2 – 2x + 3
y 4
a = –1 < 0, so the parabola opens downward.
3
2 h= = 1, 2 (–1) k = f (–1) = (–1) 2 2 (–1) + 3 = 4 ;
f(x)=x2+2|x|+3
-3
-1
1
x
3
V(–1, 4) is the vertex point.
x = 0 ; y = 3, (0, 3) is the y-intercept. y = 0 ; –x2 – 2x + 3 = 0 ; x = –3, x = 1 ; (–3, 0) and (1, 0) are the x-intercepts. Quadratic Functions
195
c. y = x|x + 2| Case 1 ________________________________________________________________________ If x + 2 0, x –2, |x + 2| = x + 2 ; y = x2 + 2x a = 1 > 0, so the parabola opens upward. h=
2 = 1, k = f (–1) = (–1) 2 + 2 (–1) = 1 ; 2 1
V(–1, –1)is the vertex point.
x = 0 ; y = 0, (0, 0) is the y-intercept. y = 0 ; x2 + 2x = 0 ; x = –2, x = 0 ; (–2, 0) and (0, 0) are the x-intercepts. Case 2 ________________________________________________________________________ If x + 2 < 0 ; x < –2, |x + 2| = –(x + 2) ;
y
2
y = –x – 2x 2
a = 1 > 0, so the parabola opens upward.
f(x)=x|x+2|
1
2 h= = 1, 2 (–1)
-3
-2
-1
x
-1
k = f (–1) (–1) 2 2 (–1) 1 ;
V(–1, 1) is the vertex point. x = 0 ; y = 0, (0, 0) is the y-intercept. y = 0 ; –x2 – 2x = 0 ; x = –2, x = 0 ; (–2, 0) and (0, 0) are the x-intercepts. d. y = |2x2 – 12x + 16| + 2
y 2
First graph the function y = |2x – 12x + 16|. Then shift this graph 2 units up.
18
f(x)=|2x2 12x+16|+2
16
a = 2 > 0, so the parabola opens upward. h=
y=|2x2 12x+16|
12 = 3, 22
k = f (h) = 2(3)2 12 3 +16 = 2 ; V(3, –2)is the vertex point.
x = 0 ; y = 16, (0, 16) is the y-intercept.
4
y = 0 ; 2x2 – 12x + 16 = 0 ;
2
x = 2, x = 4 ; (2, 0) and (4, 0) are the x-intercepts.
196
2 3 4 -2
x
Algebra 8
Check Yourself 6 Sketch the graph of each function. 1. y = |x2 – 6x + 5|
2.
y = –x|x| + 4
Answers 1. y
2.
y
5 4
4 2 1
3
5
x
x
C. EQUATION OF A PARABOLA We have learned how to construct a parabola if we are given its equation. But if we are given the graph of a parabola, how can we write its equation? There are three different approaches to finding the equation of a parabola. The approach we use depends on the information we know. 1. If we know the vertex point of the parabola, V(h, k) then we use the formula y = a(x – h)2 + k to write the equation of the parabola. We need to know another point on the parabola to write the equation. 2. If we know the x-intercepts of the parabola then we use the formula y = a(x – x1)(x – x2) to write the equation of the parabola. We need to know another point on the parabola to write the equation. 3. If we know any three points on the parabola, then we use the formula y = ax2 + bx + c to write the equation of the parabola. EXAMPLE
12
Write the equation of each parabola. a.
b.
y 2
3
x -3
Quadratic Functions
y
y=g(x)
4
y=f(x) 2
c.
y
y=h(x)
2 1
x
3
4
x
197
Solution
a. The vertex point of the parabola is V(2, 2), so y = a(x – 2)2 + 2. Also, the parabola passes through the origin (0, 0), so this point satisfies the equation of the parabola. 0 = a(0 2)2 + 2 ; 0 = 4 a+ 2 ;
a=
1 2
1 y = (x 2)2 + 2 2 1 y = x2 + 2 x 2
This is the equation of the parabola. b. Let y = ax2 + bx + c. Since the points (0, 3), (–3, 3) and (1, 4) are on the graph of the parabola, these points satisfy the equation of the parabola. (0, 3), 3 = c
b b2 4ac 2a
(–3, 3), 3 = 9 a 3 b+ 3 ; (1, 4), 4 = a+ b+ 3 ;
9a 3b = 0 a+ b = 1
9a 3b = 0 (1) (2) a+ b = 1 From (1) and (2),
1 3 a= , b= . 4 4
So the equation is y =
1 2 3 x + x+3 . 4 4
c. Let y = a(x – 3)(x – 4) (0, 2) is on the parabola. 2 = a(0 3)(0 4) ;
a=
1 6
1 y = (x 3)( x 4) 6 So the equation is y = 198
1 2 7 x x+2 . 6 6 Algebra 8
EXAMPLE
13
x1 and x2 are the roots of the parabola below. Find the value of x1 + x2. y y=g(x)
3
-6 x1
x2 2
x
-2
Solution
g(2) = g(–6) = 3
y B
A
3
C
y=g(x)
AB = BC, so the x-coordinate of the vertex point is h=
x2 2
-6 x1
x
-2
6 + 2 = 2. 2
h =
x + x2 b ; 1 = 2 ; 2a 2
x1 + x2 = 4.
T
EXAMPLE
14
Solution
A parabolic concrete bridge support needs to pass through the points (–50, 0), (0, 30), and (50, 0). Write the equation of the parabola formed by the bridge.
Let y = ax2 + bx + c. Since the points (–50, 0), (0, 30), and (50, 0) are on the graph of the parabola, these points satisfy the equation of the parabola. (0, 30), c = 30 (–50, 0), 0 = 2500a – 50b + 30 ;
250a – 5b = –3
(1)
(50, 0), 0 = 2500a + 50b + 30 ;
250a + 5b = –3
(2)
From (1) and (2), a=
–3 , b = 0. 250
So the equation is y = –
Quadratic Functions
3 2 x +30 . 250 199
EXAMPLE
15
The shape of the famous “Gateway to the West” arch, which is the spirit of the City of St. Lois can be modeled by a parabola. The equation for the parabola is y = –
1 2 21 x + x. 150 5
a. Sketch the graph of the arch's equation on a coordinate axis. b. What is the distance from one side of the arch to the other? Solution
a. a = –
1 < 0, so the parabola opens downward. 150 21 5
h= –
2 ( –
1 ) 150
y 661.5
V
= 315,
1 21 315 2 + 315 150 5 = 661 .5 V(315, 661.5)
k = f (315) = –
630 315
x
x = 0 ; y = 0 (0, 0) is the y-intercept. 1 2 21 x + x=0 ; 150 5 x = 0, x = 630 ; (0, 0) and (630, 0) are the x-intercepts. y=0 ; –
b. From the x-intercepts, we can see that the distance along the bridge is 630 m. EXAMPLE
16
Solution
A farmer has 120 m of fencing. He wants to put a fence around three sides of a rectangular plot of land, with the side of a barn forming the fourth side. Find the maximum area the farmer can enclose. What dimensions give this area? Let x represent the width of the plot. Then, since there are 120 m of fencing, x + x + length = 120 ; length = 120 – 2x. The area of the plot is A = (120 – 2x) x = 120x – 2x2. We can see that a < 0, so the graph of the function of the area has its maximum value at the vertex point. To make the area as large as possible, let us therefore find the vertex of the graph of the function A = –2x2 + 120x. –b 120 =– = 30. This is the width of the plot at its maximum The x-coordinate of the vertex is 2a –4 area. Now we have the solution: A = –2 302 + 120 30 = 1800 m2 and the dimensions of the plot are 30 m by (120 – 2 30)m, i.e. 30 m by 60 m.
200
Algebra 8
EXAMPLE
17
Solution
Find the maximum vertical distance d between the parabola and the line in the figure. The distance between the vertex point of the parabola and the midpoint of the AB line segment in the figure will be the maximum distance.
y
b –4 = =1 So h = – 2 a –4
3
k = f (1) = 5 ,
5 4
f(x)=x 2 d
2
V (1, 5).
B
1
Let us find the coordinates of point A and point B.
-2
-1
2
–2 x + 4x+ 3 = x – 2
-1
1
3
2
4
5
x
-2
2 x2 – 3x – 5 = 0 A
5 (2 x – 5)(x+1) = 0 ; x = , x = –1 2
-3 -4
5 1 A(–1, –3) and B( , ) 2 2
A
5 1 –1+ –3 + 2, 2 ) = ( 3 , –5 ) M( 2 2 4 4 So d = (
f(x)=2x2+4x+3
M
B ( 5 , 1) 2 2
(1, 3)
3 –5 1 625 626 626 – 1)2 +( – 5) 2 = + = = . 4 4 16 16 16 4
Check Yourself 7 Write the equation of each parabola. 1.
2.
y
y
3. 3
y
4
3
x
-1
3
2
-3
x
x
-2
-4
-1
Answers 1. y =
1 2 2 x – x–1 3 3
2. y = –
1 2 3 x + x–3 4 4
3. y = –
1 2 x – x+ 2 4
The things of this world cannot be made known without a knowledge of mathematics.
Quadratic Functions
201
P ARABOLIC R
EFLECTORS
Parabolic curves are used in the design of lighting systems, telescopes, and radar antennas, mainly because of the reflective property you can see in the figures below. axis
axis
F
Figure a . A reflecting telescope: light rays parallel to the axis are concentrated at the focus.
Figure b . A parabolic flashlight: a light source at the focus sends out beams of light parallel to the axis.
Figure a shows the application of the reflective property of a parabola to create a reflecting telescope. The eyepiece of the telescope is placed at the focus F of a parabolic mirror. Light enters the telescope in rays that are parallel to the axis of the parabola. We know from physics that when light is reflected, the angle of incidence equals the angle of reflection. Hence, the parallel rays of light strike the parabolic mirror so that they all reflect through the focus, which means that all the parallel rays are concentrated at the eyepiece. This maximizes the light-gathering ability of the mirror. Flashlights and automobile headlights (see Figure b) simply reverse this process. A light source is placed at the focus of a parabolic mirror. The light rays strike the mirror with an angle of incidence equal to the angle of reflection, and each ray is reflected along a path parallel to the axis. As a result, the light emits a light beam of parallel rays.
Radar utilizes both of these properties. First, a pulse is transmitted from the focus to a parabolic surface. As with a reflecting telescope, parallel pulses are transmitted in this way. The reflected pulses then strike the parabolic surface and are sent back to be received at the focus.
202
Algebra 8
5 .1
EXERCISES
6. The equation of the axis of symmetry of the
A. Graphing y = ax2 1. Graph each set of functions in the same plane. a. y = 2x2, y – 4x2 = 0, 2x2 = 3y
parabola y = 2x2 + (m + 1)x – 4 is 4x – 3 = 0. Find the minimum value of the function.
b. y = –3x2, y + 4x2 = 0, –3x2 = 2y c. y = 3x2, 3y = x2, 2y + x2 = 0, y + 2x2 = 0
7. – 2. Determine whether each point lies on the graph
4 is the minimum value of the function 3
y = 3x2 + 2x – n. Find the value of n.
of the function 4y + 5x2 = 0. 45 a. 3, – 4
4 b. , 1 5
c. (–2, –5)
d. (4, 10)
8. The vertex point of the parabola y = –3x2 – 2(2k + 5)x + 4 is on the y-axis.
3. If the parabola y = px2 passes through the following
Find k.
points, find p. a. (–3, 3)
4 b. 2, 5
c. (3, –7)
4 d. , –2 5
2
4. Graph the function y = x if x 4. 2
9. Find the x- and and y-intercepts of each function. a. y = –x2 + 4x
b. y = (x – 3)2
c. y = 8(x + 7)2 + 4
d. y = x2 – 8
e. y = 2x2 + 5x + 3
f. y = –x2 – 4x – 5
g. y = 16x2 + 24x + 9 h. y = –
B. Graphing y = ax2 + bx + c
2 2 x –4 3
5. Find the coordinates of the vertex point of each function. Write the equation of the axis of symmetry and determine the maximum or minimum value of the function. a. y =
3 2 x 4
c. y = 3x2 + 2
d. y = –x2 – 1
e. y = –2x2 + 5x
f. y = 0, 7x2 + 0,8x
g. y = 3x2 – 4x + 3
h. y = –x2 + x + 1
i. y = 3(x + 1)(x – 4) j. y = –2(x + 4)2 k. y = –(x + 3) – 5 Quadratic Functions
Find the values of m for which the parabola a. does not cut the x-axis.
b. y = –3x2
2
10. The function y = –x2 + (m – 2)x + m – 3 is given.
2
l. y = (x – 1) + 3
b. is tangent to the x-axis. c. cuts the x axis at two distinct points.
11. The parabola y = mx2 – (m + 1)x + 2m – 1 passes through the point A(2, 9). Find the value of m. 203
12. Sketch the graph of each function.
22. Sketch the graph of each function by shifting.
a. y = 3x2 – 9
b. y = 2x2 + 5
a. y = x2 + 4
b. y = x2 – 2
c. y = –3(x + 4)2
d. y = 2(x + 7)2
c. y = –x2 – 2
d. y = –(x + 5)2
e. y = x2 – 2x – 3
f. y = –x2 – 4x
e. y = (x – 4)2
f. y = –(x – 3)2
g. y = x2 – 4x + 1
h. y = x2 – 5
g. y = –(x + 1)2 – 2
h. y = (x – 2)2 + 3
i. y = (x – 3)2 + 4
j. y = (x + 3)(4 – x)
13. Sketch the graph of the function y = x2 – 2x – 8 if –1 < x < 5.
23. Sketch the graph of each function. a. y = |x2 – 4x + 3|
b. y = –2x2 + 4|x| + 1
c. y = x|x – 2|
d. y = |x2 – 6x + 8| – 1
14. Sketch the graph of the function y = –2x2 + 4x + 1 if |x – 1| < 2.
C. Equation of a Parabola 15. Find the minimum and the maximum values of the function y = x + 4x + 7 if x [–4, 1]. 2
24. Write the equation of each parabola. a.
b.
y
y
3 2
16. Find the minimum or the maximum value of the
1
function if m + n = 16.
5
x
1
6 x
17. Find the minimum or the maximum value of the function y = m2 + n if m – 2 = n.
c.
d.
y
y 4 3
3
18. The parabola y = 2ax + bx – 3b passes through 2
the points (1, –3) and (2, –5). Find the values of a and b.
19. Write the equation of the parabola which is symmetric to y = 3(x + 1)2 with respect to the line x = 2.
20. Given f(x) = –3(x – 1)2 + 2, show that f(x) f(1).
-3
e.
x
f.
y 3
x
-1
y
5
3
x -3
-3
3
x
21. f(x) = x2 + 3mx + 4m and f(x) 0 are given. Find m. 204
-8
Algebra 8
25. Write the equation of the parabola through each
30. The figure shows the graph
set of points.
y
of the function
a. A(–1, 0),
B(0, 4),
C(1, 2)
y = mx2 – 7mx + 5.
b. A(0, –4),
B(2, 0),
C(4, –4)
If |AB| = 3, find m.
A
B
x
26. Write the equation of the parabola through each vertex point V and point A. a. V(0, –4), A(3, 5)
b. V(2, 0), A(4, 4)
31. Find the area of the
1 15 c. V(–1, –4), A(2, 5) d. V , , A(0, –4) 4 2
y
trapezoid OABC in the figure.
5 2
y=2x2+6x C
B A
O
27. In the figure,
y
3 |OB| = |AO|. Find c.
y=x2 4x c A
O
B x
32. Find the area of the
y
|OB| = 5 |AO| and V is the vertex point of the parabola. Find k.
k
C
V
O
A
A
x
2
B O
x
2
29. The figure shows the
y
graph of the function y = ax2 + bx + c. Find the value of
B
2
33. Find the area
y
square OABC in the figure. 24
28. In the figure,
-5
1
x
y
of the rectangle ABCD in the figure.
D
C y=x2 6x+5
O
A
B
x
a + b + c. Quadratic Functions
x
205
39. The figure shows the
Mixed Problems
34. x1 and x2 are the x-intercepts of the function
y = mx2 – 4x – 2m – 2. If x1 x1 < 0, find the possible values of m.
y
graph of the function y = –x2 + bx + 1. Find the value of a + b + c + d.
y=x2+bx+1
5 c
a
2
x
d
35. The perimeter of a rectangle is 32 cm. Find its
maximum possible area.
40. In the figure, 36. The function y = 5x – 3x + k is tangent to the 2
line y = 2. Find k.
y
the area of the triangle ABC is 6 cm2. Find k.
y=k(x2+5x+4) C A
B O
x
37. 3 is the minimum value of the function
y = mx2 – 2mx + 2m + 1. Find m.
38. The figure shows the
y
graph of the function
6
41. In the figure, the
y = –x2 + bx + c. of the point A.
2 A
206
B
maximum value of the 49 function is . 4 Find the area of the
Find the coordinates x
y C A -2
B O
5
x
triangle ABC. Algebra 8
An equation is a statement that says two expressions are equal. For example, a = b is an equation. An inequality is a statement that says two expressions may or may not be equal. For example, a > b, a b, and a < b are all inequalities. An equation usually has a finite number of solutions, but an inequality may have an infinite number of solutions. We can show the solutions of an inequality as an interval and as a graph. Look at the following examples of intervals and their graphs. Interval
Inequality
Graph
the interval (–, a)
x
a
the interval (–, a]
xa
a
the interval (a, )
x>a
a
the interval [a, )
xa
a
For any two real numbers a and b, exactly one of the following is true:
the open interval (a, b)
a
a
b
a < b, a = b or a > b.
the closed interval [a, b]
axb
a
b
the half-open interval [a, b)
ax
a
b
the half-open interval (a, b]
a
a
b
the interval (–, )
Trichotomy Property:
Notice in the table that an open circle on a graph shows an open interval, i.e. a line segment with an open endpoint (a point which is not included in the segment). A filled circle shows a closed endpoint (a point which is included in the segment).
EXAMPLE
1
Write each inequality using interval notation and show it as a graph. a. 1 x < 5
Solution
7
c. (–3, 4)
d. x 9
c. –3 < x < 4 1
a. The interval notation is [1, 5) and the graph is b. (7, )
208
b. x > 7
3
4
5
d. (–, 9]
. 9
Algebra 8
linear inequality
Definition
A linear inequality is an inequality that can be written in one of the forms ax + b > 0, ax + b < 0 ax + b 0, ax + b 0 for the real numbers a and b, a 0. If both sides of an inequality are multiplied or divided by a negative number, the direction of the resulting inequality must be reversed.
Property
For example, if we multiply both sides of the inequality a < b by –2, we obtain –2a > –2b. The order of the inequality is reversed. EXAMPLE
2
Solution
Solve the inequality 2x + 3 5x and graph this solution. 2x + 3 5x –3x –3 1x
1
Therefore, x [1, ), or
EXAMPLE
3
Solution
.
Solve the inequality 2 < 3x – 1 8 and graph this solution. 2 < 3x – 1 8 3 < 3x 9 1
1
3
.
For k > 0, the following statements are true.
Property
1. |x| k means –k x k. 2. |x| > k means x < –k or x > k. EXAMPLE
4
Solution
Solve the inequality |2x – 1| < 5 and graph this solution. |2x – 1| < 5 –5 < 2x – 1 < 5 –4 < 2x < 6 –2 < x < 3 Therefore, x (–2, 3), or
Quadratic Inequalities
2
3
. 209
Check Yourself 1 Solve the inequalities. 1. –x + 7 > 2x + 1 2. 11 x+ 3 17 4 3. |2x – 1| + 2 7 Answers 1. x < 2 2. –47 x 65 3. –2 x 3 FIND THE MISTAKE! a>4 4a > 16 4a – a2 > 16 – a2 a(4 – a) > (4 – a)(4 + a) a>4+a 0>4 Can you find the mistake in this working?
Sign Chart In general, to solve a linear inequality such as ax + b > 0 or ax + b 0 we need to know the sign of the polynomial ax + b, a 0. Look at the steps. First we find the zero of the polynomial: b ax+ b = 0 ; x = . a
Then we construct a sign chart. x ax+b
x0=
¥ ax+b has sign opposite to a
b a
¥ ax+b has the same sign as a
This sign chart shows us: b if x , , the sign of the polynomial is opposite to the sign of a, a b if x , , the sign of the polynomial is the same as the sign of a. a 210
Algebra 8
EXAMPLE
5
Solution
Solve the inequality 3x – 2 0. 2 Find the zero: 3 x 2 = 0 ; x = . 3 Draw the sign chart: x
x0=
¥
2 3
3x 2
¥
+
2 If x , , then 3 x2 is negative. 3 2 If x , , then 3 x2 is positive. 3 2 2 Since 3 x – 2 0, x is in the interval , , i. e. x , . 3 3 EXAMPLE
6
Solution
Solve the inequality 6 – 2x 0 by using a sign chart. 6 – 2x = 0 ; x = 3 x 6 2x
¥
3
+
¥
Since 6 – 2x 0, the sign is positive. So x (–, 3].
Check Yourself 2 Solve the inequalities by using a sign chart. x+ 14 x 12 1. 5 – 2x < 0 2. 3 6 8 Answers 5 1. x > 2. x –20 3. x –10 2
3. 6x + (x – 2)(x + 2) (x + 4)2
Mathematics, of course, is not the only cornerstone of opportunity in today’s world. Reading is even more fundamental as a basis for learning and for life. What is different today is the great increase in the importance of mathematics to so many areas of education, citizenship, and careers.
Quadratic Inequalities
211
A rectangular room has dimensions 30 m 12 m 12 m. A spider is in the horizontal center of one end wall, one unit away from the ceiling. A fly is in the horizontal center of the opposite wall, one unit away from the floor. What is the shortest distance the spider needs to travel (without leaving a surface) to get to the fly (which remains stationary)?
EXERCISES
6 .1
1. Write each inequality using interval notation. a. x > –2
b. x 6
c. x < –7
d. x 4
e. 2 x < 4
f. –1 < x 0
g. –5 < x <
1 2
h. 0 ,5 x
a. –2x + 3 < 4
3 2
2. Determine the sign of each polynomial. a. 4x + 1
b. –3x – 5
x 7 c. + 2 3
d. ñ3x – 6
212
3. Solve and graph the inequalities.
b.
x x x x + 2 3 4 5
c.
3x x + 5 < 1+ 2 2
d.
x+ 3 x 2 2x 4 2 3
e. (3x + 1)2 – (x + 2)(4x – 1) > 5(x – 1)2 + 6x f. (a + 3)x – 5 1
(a > –3)
g. (a4 + 4)x + 3 > 0 (a ) Algebra 8
Definition
quadratic inequality A quadratic inequality is an inequality that can be written in one of the forms ax2 + bx + c > 0,
ax2 + bx + c < 0,
ax2 + bx + c 0,
ax2 + bx + c 0,
for the real numbers a, b, and c, a 0.
We will never be equals.
We have seen how to solve linear inequalities such as 2(x + 3) + 3 5(x – 1). But how do we solve quadratic inequalities such as x2 – 5x –6? First we write the inequality in standard form, leaving only zero on the right side: x2 – 5x + 6 0.
In this example, we are looking for values of x that will make the quadratic on the left side greater than or equal to zero.
Note If and are used in the inequality, then remember that the zeros of the polynomial are included in the solution set. First, let us find the zeros of the polynomial x2 – 5x + 6: (x – 2)(x – 3) = 0 positivepositive=positive positivenegative=negative negativenegative=positive
x = 2 or x = 3. Then we construct a sign chart for each linear factor of the polynomial, and their product. x
¥
3
2
¥
x2
+
+
x3
+
+
+
(x 2)(x 3)
Since x2 – 5x + 6 0, we need to take the positive intervals. Therefore, the solution is the union of intervals (–, 2] and [3, –), i.e. x (–, 2] [3, ). Quadratic Inequalities
213
We can also construct the sign chart in one step. The zeros of the polynomial divide the real number line into three intervals, (–, 2], [2, 3] and [3, ). We know that the polynomial has constant sign in each of these three intervals. If we select a test number in each interval and evaluate the polynomial at that number, then the sign of the polynomial at this test number must be the sign for the whole interval. Let us try testing each interval in our problem. Choose a number from each interval, and substitute for x in the original inequality. For example, we could choose the numbers 1, 2.5, and 4. Test number
1
2.5
4
Value of the polynomial
2
–0.25
5
Sign of the polynomial
+
–
+
With this information, we can draw the sign chart. x x2
¥
3
2
+
5x + 6
¥
+
This is the same as the last line of the previous chart. We can also use the discriminant of a quadratic to help complete the sign chart. We have already seen that the discriminant of a quadratic equation ax2 + bx + c = 0 tells us it the equation has two real roots ( > 0), one double root ( = 0), or no real solutions ( < 0). If < 0, then the polynomial ax2 + bx + c always has the same sign as a. x
¥
¥
ax2 + bx + c
same sign as a
If = 0, then of the polynomial ax2 + bx + c has the same sign as a but we must consider the zero of the polynomial. x
x1=x2=
¥
ax2 + bx + c
b 2a
same sign as a
¥ same sign as a
If > 0, the polynomial ax2 + bx + c has the opposite sign to a between the zeros of the polynomial and the same sign as a in other intervals. x ax2 + bx + c
214
¥
x2
x1 same sign as a
opposite sign to a
¥ same sign as a
Algebra 8
EXAMPLE
7
Solution
Solve the inequality –x2 – 4 < –5x. –x2 – 4 < –5x –x2 + 5x – 4 < 0 (–x + 4)(x – 1) = 0 x = 4 or x = 1 x
¥
x2 + 5x 4
4
1
¥
+
Therefore, x (–, 1) (4, ). EXAMPLE
8
Solve the inequalities. a. 3x + 4 x2
Solution
c. 9x2 – 12x + 4 0
b. x(x + 2) > 35
d. x2
2x 1 <0 4
a. 3x + 4 x2 ; –x2 + 3x + 4 0 = 9 + 16 = 25 ; x1 = –1, x2 = 4 x
¥
x2 + 3x + 4
Plot the roots in the sign chart line in ascending order.
4
1
¥
+
So x [–1, 4]. b. 35 < x(x + 2) ; –x2 – 2x + 35 < 0 = 4 + 140 = 144 ; x1 = –7, x2 = 5 x
7
¥
x2 + 3x + 4
5
¥
+
So x (–, –7) (5, ). c. 9x2 – 12x + 4 0 = 144 – 144 = 0 ; x1 = x2 =
12 2 = 18 3 2 3
x ¥ 9x2 12x + 4
+
¥
+
2 So the only solution is x = . 3 Quadratic Inequalities
215
c. x2
2x 1 < 0 ; 4x2 – 2x + 1 < 0 4
= 4 – 16 = –12 < 0. x 4x2
¥
2x + 1
¥
+
+
+
+
+
4x2 – 2x + 1 is always positive, so there is no solution.
EXAMPLE
9
Consider the equation x2 – 2(m + 1)x + 1 = 0. For which values of m does the equation have a. no real root? b. one double root? c. two distinct real roots?
Solution
Let us check the sign of discriminant. = 4(m + 1)2 – 4 = 4m2 + 8m + 4 – 4 = 4m2 + 8m 4m2 + 8m = 0 m = –2 or m = 0 m ¥ 4m2 + 8m
0
2
+
¥
+
Now we can answer the question. a. For < 0, there is no real root, and m (–2, 0). b. For = 0, there is one double root, and m = –2 or m = 0. c. For > 0, there are two real roots, and m (–, –2) (0, ).
EXAMPLE
10
Solve the inequalities. a. (3 – x)(x3 – 2x2 – 8x)(x2 + 3) < 0
216
b.
(3x+ 2)(x 5) 0 x(x – 1)(x2 + x +1)
c.
(x2 x 6)(x 1)17 0 x8 (1 x2 )55 Algebra 8
Solution
a. First we find all the zeros of the polynomials, then we determine the sign for each polynomial and multiply the signs of each polynomial. (3 – x)(x3 – 2x2 – 8x)(x2 + 3) = (3 – x)x(x2 – 2x – 8)(x2 + 3) 3–x=0; x=3 x=0 x2 – 2x – 8 = 0 ; x = –2 or x = 4 x2 + 3 = 0 ; no real solution. x
(3
x)x(x2
¥
2
0
3
4
¥
3x
+
+
+
x
+
+
+
x2 2x 8
+
+
x2 + 3
+
+
+
+
+
+
+
2x
8)(x2
+ 3)
We need a value less than zero, so x (–, –2) (0, 3) (4, ).
positive = positive positive positive = negative negative negative = positive negative
b. First we find all the zeros of the polynomials. The equality part of the original inequality is satisfied for these zeros and they must be included in the final solution set. On the other hand, since division by zero is never allowed, the zeros of x4 – x must not be included in the solution set. (3x+ 2)(x 5) x(x 1)(x2 + x+ 1) 2 3x+ 2 = 0 ; x = 3 x5=0 ; x=5 x=0 x 1= 0 ; x = 1 x2 + x+ 1 = 0; no real solution. x
¥
2 3
0
1
5
¥
3x + 2
+
+
+
+
x5
+
x
+
+
+
x1
+
+
+x+1
+
+
+
+
+
(3x + 2)(x 5)
+
+
+
x2
x4 x
2 We need a value greater than or equal to zero, so x , (0 , 1) [5 , ). 3 Quadratic Inequalities
217
c.
(x2 x 6)(x 1)17 0 x8 (1 x2 )55
x2 – x – 6 = 0 ; x = –2 or x = 3 (x – 1)17 = 0 ; x – 1 = 0 ; x = 1 x8 = 0 ; x = 0 (double root) (1 – x2)55 = 0 ; 1 – x2 = 0 ; x = 1 or x = –1 (1 is also a double root) If the power is an odd number, you can ignore it when you calculate the zero. If the power is an even number, consider it just as 2 when you calculate the zero.
x x2 x 6
¥
2
1
0
1
3
¥
+
+
(x 1)17
+
+
x8
+
+
+
+
+
+
(1 x2)55
+
+
+
+
+
+
(x2 x 6)(x 1)17 x8(1 x2)55
We need a value less than or equal to zero, so x [–2, –1) [3, ). The signs of M N and M , N 0 are the same. N
Let us summarize the key steps to solving any inequality. 1. Write the polynomial inequality in standard form. 2. Find all zeros of the polynomial(s). 3. Determine the character of the roots. 4. Determine the sign of the coefficient of leading term of the polynomial(s). 5. Construct a sign chart. 6. In the sign chart, from right to left start with the sign of the coefficient of the leading term a. After each root change the sign, but if there is a double root do not change the sign. Let us solve Example 10c in another way.
EXAMPLE
11
Solution
Solve the inequality
(x2 x 6)(x 1)17 0. x8 (1 x2 )55
x2 – x – 6 = 0 ; x = –2 or x = 3 (x – 1)17 = 0 ; x – 1 = 0 ; x = 1 x8 = 0 ; x = 0 (double root) (1 – x2)55 = 0 ; 1 – x2 = 0 ; x = 1 or x = –1 (1 is also a double root)
218
Algebra 8
x (x2
x 6)(x
¥
1)17
2
1
+
x8(1 x2)55
0
1
+
3
+
¥
+
Therefore, x [–2, –1) [3, ). EXAMPLE
12
Solution
Solve the inequality (x2 – 2x – 8)(x2 + x – 12) 0. Find the zeros of the polynomials. x2 – 2x – 8 = 0 ; x = 4 or x = –2 x2 + x – 12 = 0 ; x = –4 or x = 3 x
¥
(x2 2x 8)(x2 + x 12)
4
+
2
3
4
+
¥
+
So, x [–4, –2] [3, 4]. EXAMPLE
13
Solution
Find the domain of the function y =
x+ 2 3x 12 x2
.
3x – 12x2 > 0 3x – 12x2 = 0 x = 0 or x =
1 4 x
3x
¥
1 4
0
12x2
¥
+
1 So x 0 , . 4 EXAMPLE
14
Solution
Solve the inequality (x – 1)2(x – 2)3(x – 3)4(x – 4)5 0. (x – 1)2 = 0 ; x – 1 = 0 ; x = 1 (double root) (x – 2)3 = 0 ; x – 2 = 0 ; x = 2 (x – 3)4 = 0 ; x – 3 = 0 ; x = 3 (double root) (x – 4)5 = 0 ; x – 4 = 0 ; x = 4 x
¥
(x 1)2(x 2)3(x 3)4(x 4)5
1
+
2
+
3
4
¥
+
So x (–, 2] {3} [4, ]. Quadratic Inequalities
219
EXAMPLE
15
Solution
x4 3x3 + 2x2 < 0. x3 5x2
Solve the inequality x4 3x3 + 2x2 <0 x3 5x2 x2 (x 1)(x 2) <0 x2 (x 5)
x = 0 (double root) x = 1, x = 2, x = 5 x
¥
x4 3x3 + 2x2
0
x3 5x2
1
2
+
5
¥
+
So x (–, 0) (0, 1) (2, 5).
EXAMPLE
16
Solution
x 1 x+ 1 < 2. x x 1
Solve the inequality
x 1 x+ 1 2 < 0 ; x 0 and x 1 x x 1 (x 1)2 x(x+1) 2 x( x 1) 0 x(x 1) 2 x2 x+ 1 <0 x(x 1) 2 x2 x+ 1 = 0 x = 1 or x =
1 2
x(x 1) = 0 x = 0, x = 1 x 2x2
x+1
x(x 1)
¥
1
1 2
0
+
1
+
¥
1 So x (– , –1) 0 , (1, ). 2 220
Algebra 8
EXAMPLE
17
Solution
Solve the inequality
1 2 1 2x . x+ 1 x2 x+ 1 x3 + 1
1 2 1 2x 0 ; x+ 1 x2 x+ 1 x3 + 1
x 1
x2 x+ 1 2(x+ 1) (1 2 x) 0 (x+ 1)(x2 x+ 1) x2 x 2 0 (x+ 1)(x2 x+ 1) x2 x 2 = 0 ; x = 1 or x = 2
x + 1 = 0 ; x = –1 (double root) x2 – x + 1 = 0 ; no real solution x x2
¥
1
x2
2
(x + 1)(x2 x + 1)
¥
+
So x (–, –1) (–1, 2].
EXAMPLE
18
Solution
Solve the inequality
2 x2 + 3x 2 . (1 x2 ) (x2 + 3x) . x 2
Since |2x2 + 3x – 2| and |x – 2| are non-negative, just check their roots. 2x2 + 3x – 2 = 0 ; x = –2, x = x–2=0; x=2
The absolute value of a number is never negative. |x| 0, x
0.
1 2
1 satisfy the inequality, so these values are in the solution set. However, 2 is not in 2 the solution set because it makes the denominator zero.
–2 and
1 – x2 = 0 ; x = –1, x = 1 x2 + 3x = 0 ; x = 0, x = –3 x |2x2
+ 3x 2|(1
x2)
(x2 + 3x)|x 2|
¥
3
So x (–, –3) [–1, 0) {–2, Quadratic Inequalities
1
+
0
1
+
¥
1 } [1, 2) (2, ). 2 221
EXAMPLE
19
Solution
Find the domain of the function y = 4 7
4
x2 6x 16 + 7 3x4 4x3 + x 1 if y is a real number. x2 12 x+ 11
3x4 4x3 + x 1 is a real number for all values of x because the index is an odd number.
x2 6 x 16 x2 6 x 16 is a real number if 0 x2 12 x+ 11 x2 12 x+11
Therefore, y is a real number if x2 6x 16 0. x2 12 x+ 11 x2 6 x 16 = 0 ; x = 2 , x = 8 x2 12 x+ 11 = 0 ; x = 1, x = 11 x x2
¥
6x 16
2
+
x2 12x + 11
1
8
+
¥
11
+
So x (–, –2] (1, 8] (11, ).
EXAMPLE
20
Solution
Solve the inequality 3
2
If ab < ac then b < c.
2
3 x 1 x 1
3 x 1 3x 1
<8
<2
x 3 3 x7
3
3 x 1
x 3
2 x1 < 8 3 x7 . 1
x 3 3 x 1 3 ; 2 x1 < (2 3 ) 3 x7
3x 3 3 x 7
3x 1 3(x 3) < 3(x 1) 3x 7 3x 1 3 x 9 <0 3x 3 3x 7 (3x 1)(3 x 7) (3 x 9)(3 x 3) <0 (3x 3)(3 x 7) 12 x 20 <0 (3x 3)(3 x 7) 12 x 20 = 0 ; x =
5 3
3x 3 = 0 ; x = 1 3x 7 = 0 ; x = 222
7 3 Algebra 8
x
¥
12x 20
5 3
1
(3x 3)(3x 7)
+
7 3
¥
+
5 7 So x ( , 1) , . 3 3
Check Yourself 3 Solve the inequalities. 1. x2 + 5x – 6 > 0
2. (x + 3)3(x – 1)2(x – 4) 0
2 x2 + 4) 4. (x 2)( <0 x2 4 Answers
2 1 5. 4x x 2
1. (–, –6) (1, )
2. (–, –3] {1} [4, ) 1 5. (–, 0] [ , ) 2
3.
2 3 4 + 2 x 2 + x 4 x2
x
4. (–2, –ñ2) (ñ2, 2)
3. (–2, 2) [6, )
‘Obvious’ is the most dangerous word in mathematics.
EXERCISES
6 .2
1. Determine the sign of each polynomial.
3. For which values of k does the equation
a. x2 – 5x + 4
b. 2x2 + x – 6
x2 + 2(1 – k)x + 1 = 0 have
c. 2x2 – 3x + 4
d. –16x2 + 8x – 1
a. no real solution?
e. –4x2 + 10x – 25
f. 12x2 + 4ñ3x + 1
c. two distinct real roots?
b. one double root?
4. For which values of b does the equation 3x2 + (b + 1)x + 1 = 0 have two distinct real roots?
2. Solve the inequalities. a. x2 < 9x – 20
b. 4x – 7x2 > 0
c. x(6x + 7) 0
d. 2x – 6 < 3x2
e. 3 x2 + 8 x
16 3
g. (2x – 1)2 > (x – 5)2 Quadratic Inequalities
f. x(x 1) + 1 <
5. For which values of a does the equation ax2 + (a + 1)x + 2 = 0 have no real root? 4 1 x+ 5 5
6. The product of a number and four plus the number is less than 15. Find all possible integer values of the number. 223
7. Solve the inequalities.
8. Solve the inequalities.
a. x(x – 1)2 > 0
a. 1
b. (2 – x)(3x + 1)(2x – 3) > 0 c. (3x – 2)(x – 3)3(x + 1)3(x + 2)4 < 0 d.
e.
(x 1)(3 x 2) >0 5 2x
x4 + x2 + 1 <0 x2 4x 5
g.
3x 2 <3 2x 3
2 x2 + 18 x 4 h. >2 x2 + 9x+ 8
j.
c.
(x+ 1)(x+ 2)( x+ 3) >0 (2 x 1)( x+ 4)(3 x)
f.
i.
b.
d.
1 3 k. x x2 0 x 2
m.
(x2 + 4x+ 4)(x 2)1001 <0 (x 2)2004 x+ 4
n. |6x2 – 2x + 1| 1 o.
x+ 2 <3 2x 3
p.
x2 3x+ 2 >1 x2 + 3x+ 2
q.
x2 3x 1 1 x2 + x+ 1
224
3 x (x2 + x+ 3) 2 x+ 2 x2 5x+ 2 x3 2
x 5x+ 6
a. y =
2x 1 2x 3 3 2 x +x x 3x2
1 3 > 2 3x 2 x 7x 4 3x2
(a2 + a+ 1)x2 + 4 >0 a 4 x a 4 x2
<0
2
9. Find the domain of each function.
x+ 1 3 1 x2 x2 2
l.
1 1 1 1 1 + + <0 x x2 x3 x4 x5
2 3
2
x 49
b. y =
1 x 4
c. y =
3x 6 x+ 2
d.
4
(x4 5x3 + 6 x2 )(1 x2 )
10. Solve the inequality
12 3 2 > 1. x + 2x x + 2x 2 2
4 3 4x2 8 11. Solve the inequality x + 3x + < 0. 2
x
Algebra 8
We saw in Chapter 1 that a set of simultaneons equations to solve is called a system of equations. A system that of equations includes more than one inequality is called an inequality system. To find the solution of a system, we solve each inequality separately and then find the intersection of the solutions. EXAMPLE
21
Solution
x2 7 x 8 > 0 Solve the inequality system . 2 x 4x+ 3 > 0
First find the zeros of each polynomial. (1) x2 – 7x – 8 > 0 x2 – 7x – 8 = 0 ; x = –1, x = 8 (2) x2 – 4x + 3 > 0 x2 – 4x + 3 = 0 ; x = 1, x = 3 We need to find values so that both polynomials are greater than zero. Let us check the chart. x
¥
1
1
3
8
¥
(1)
+
+
(2)
+
+
+
+
system
We can see that both polynomials are greater than zero when x (–, –1) (8, ). This is the intersection of the solutions. EXAMPLE
22
Solution
x2 + x 4 <1 Solve the inequality system . x x2 < 64
(1)
x2 + x 4 x2 + x 4 x <1; <0 ; x x x2 – 4 = 0 ; x = 2 ; x = 0
x2 4 <0 x
(2) x2 < 64 ; x2 – 64 < 0 x2 – 64 = 0 ; x = 8 x
¥
8
2
0
2
8
¥
(1)
+
+
+
(2)
+
+
system
This time, both polynomials need to be less than zero, so x (–8, –2) (0, 2). Quadratic Inequalities
225
EXAMPLE
23
Solution
x5 100 x3 Solve the inequality system . (x+ 9)(5 x x2 18) 0 x2 18 x+ 45
(1) x5 100x3 x5 – 100x3 0 ; x3(x2 – 100) = 0 x = 0, x = 10 (2)
(x+ 9)(5 x x2 18) 0 x2 18 x+ 45 x + 9 = 0 ; x = –9 5x – x2 – 18 = 0 ; no solution x2 – 18x + 45 = 0 ; x = 3, x = 15 x
¥
10
9
0
3
10
15
¥
(1)
+
+
+
+
(2)
+
+
+
+
system
So x [–10, –9] [10, 15). EXAMPLE
24
Solution
Solve the inequality 2x 1 0 x+ 2 > 0 2 2 x 1 < (x+ 2)
(1) 2x – 1 0
If
f ( x) g( x)
2x – 1 = 0 1 x= 2
f ( x) 0 then, g( x) 0 2 f ( x) g (x)
2 x 1 < x+ 2. (1) (2) (3)
(2) x + 2 > 0 x+2=0 x = –2
x
¥
(3) 2x – 1 < x2 + 4x + 4 x2 + 2x + 5 > 0 x2 + 2x + 5 = 0 since < 0, there is no real solution 1 2
2
¥
(1)
+
(2)
+
+
(3)
+
+
+
system
So x [ 226
1 , ). 2 Algebra 8
25
Solve the inequality
Solution
Case 1 __________________
Case 2 __________________
(1) x < 0 2 x + x 2 0 (2)
(1) x 0 2 x + x 2 > x2 (2)
(1) x < 0
(1) x 0
EXAMPLE
If
f ( x) g( x) then
x2 + x 2 > x.
2
(2) x + x – 2 = 0 ;
g( x) < 0 f ( x) 0
(2) x – 2 > 0 ;
x1 = –2, x2 = 1
x – 2 = 0, x = 2
or g( x) 0 f ( x) g2 (x)
x
Case 1
¥
2
0
1
2
¥
(1)
+
+
+
(2)
+
+
+
(1)
+
+
+
(2)
+
system
Case 2
system
So x (–, –2] (2, ).
EXAMPLE
26
Solution
Solve the inequality (x2 + x + 1)2 – 4(x2 + x + 1) + 3 < 0. For the inequality, we let t = x2 + x + 1. Then the original inequality becomes t2 – 4t + 3 < 0. First let us solve the inequality for t. t2 – 4t + 3 = 0 ; t = 1, t = 3 x
¥
t2 4t + 3
3
1
+
¥
+
So 1 < t < 3. Now solve for x. 1 < x2 + x + 1 < 3 0 < x2 + x < 2, which gives us the system of inequalities x2 + x > 0 (1) . x2 + x 2 < 0 (2) Quadratic Inequalities
227
(1) x2 + x > 0 x2 + x = 0 ; x = 0, x = –1 (2) x2 + x – 2 < 0 x2 + x – 2 = 0 ; x = –2, x = 1 x
¥
2
1
0
1
¥
(1)
+
+
+
+
(2)
+
+
system
So x (–2, –1) (0, 1).
27
Solve the inequality |x2 – 3x + 2| 2x – x2.
Solution
Case 1 ________________________________________
EXAMPLE
x2 3x+ 2 0 (1) x2 3x+ 2 2 x x2 (2)
(1) x2 – 3x + 2 0 x2 – 3x + 2 = 0 ; x = 1, x = 2 (2) x2 – 3x + 2 2x – x2 ; 2x2 – 5x + 2 0 1 2x2 – 5x + 2 = 0 ; x = , x = 2 2 Case 2 _______________________________________ x2 3x+ 2 < 0 (1) (x2 3x+ 2) 2 x x2 (2)
(1) x2 – 3x + 2 < 0 x2 – 3x + 2 = 0 ; x = 1, x = 2 (2) –(x2 – 3x + 2) 2x – x2 –x2 + 3x – 2 2x – x2 x–20 x–2=0; x=2 228
Algebra 8
x
Case 1
1 2
¥
1
2
¥
(1)
+
+
+
(2)
+
+
(1)
+
+
+
(2)
+
system
Case 2
system
1 So x , 1 (1, 2) {2}, i.e. 2 1 x , 2 . 2
Check Yourself 4 Solve the systems. x2 – 4x 0 1. x – 3 0
x2 4x+ 3 > 0 2. x2 (x 7)2 > 0
3.
x2 x > 1+ x
Answers 1. [4, ) 2. x 0, x 7, (–, 1) (3, ) 3. ( , ) 3
In the center of a square pond whose side measures 10 m grows a plant whose top reaches 1 m above the water level. If we pull the plant toward the bank, its top becomes even with the waters surface. What is the depth of the pond and the length of the plant?
Do not worry about your difficulties in mathematics. I can assure you that mine are still greater. Einstein Quadratic Inequalities
229
EXERCISES
6 .3
1. Solve the inequality systems.
4. Solve the inequalities.
x+ 2 > 0 a. 2 x 3 < 0
x 1 > 0 b. x2 2 x 3 < 0
x2 + 3x 10 > 0 c. 2 x2 + 11x 6 < 0
x 2 x+ 1 0 d. x2 1 x 3 > 0 2 x2 + 2 < 5x x2 x
x2 4x+ 3 < 0 e. 2 x 4 < 0
f.
x+ 2 x+ 1 > 1 g. x x 2 > 2
x 1 <0 3x h. x 1 x+ 1 < 0
a.
2 x+ 10 < 3x 5
b.
x+ 3 2 4x
c.
x2 x 12 < x
d.
17 15 x 2 x3 >0 x+ 3
e.
x+ 1 x 2 1
f.
x+2
5. Solve the inequalities. a.
x2 9 < x+ 2
b.
2 3x x2 2 x+ 3
c.
x2 x 2 x 1
d.
4 1 x < 2 x
e.
x2+ x3<4
2
2. Solve the inequality systems. x+ 3 3 x <2 a. x3 < 16 x 4 x2
x4
6. Solve the inequalities. b. 4x – 2 < x2 + 1 < 4x + 6
c. 5x 7 < 4 x + 2 3x < 4 x5 5 x x 25
b. |x2 – 3x – 15| < 2x2 – x
2+ x x 1
c.
6 x2 < x 1
b.
1+ x x 1
7. Solve the inequality
x+ 3 4 x 1 + x+ 8 6 x 1 > 1.
3. Solve the inequalities. a. |4 – 3x| 2 – x
a.
8. For which values of m are the following inequalities
satisfied for any real number x?
c. |2x2 + x + 11| > x2 – 5x + 6
a. 2x2 + 5x + m > x2 – x – 7
d. |x – 4| + |2x + 6| > 10
b. 2x2 + mx2 – 2x + m – 1 > (1 + m)x2 – 4x – m
e.
x3 x2 5x+ 6
2
f. |2x + 1| – |5x – 2| 1 g. 230
x2 x 12 x3
c. (m + 2)x2 – 3x + m – 2 < 0
2x
d. x2 – 6x + 4m2 > 7
9. Solve the inequality
(x2 + 6x + 14)2 – 9 (x2 + 6x + 15) + 9 < 0. Algebra 8
CHAPTER REVIEW TEST
6A
1. What is the solution of the inequality
5. What is the solution of the inequality
2 – 3(1 – x) < x + 1.8? A) x > 1.4
B) x < 0.4
D) x > –0.7
C) x > 0.7
E) x < 1.4
1 1 < ? x+ 2 x 1
A) x < 0 or x > 2
B) –2 < x < 1
C) x < –2 or x > 1
D) 0 < x < 2
E) –2 < x < 2
2. What is the solution of the inequality x2 + 5x – 14 0? A) [–2, 7]
B) [–7, 2]
D) [2, 7)
C) (–7, 2)
E) (–2, 7]
6. What is the greatest integer value of x which satisfies the inequality x2 – 13x + 36 < 0? A) 13
3. What is the solution of the inequality
B) 7
C) –8
D) 8
E) –13
x2 4 > 0? x2 1
A) (–, –2) (1, )
B) (–, –2) (2, )
C) (–, –1) (1, )
D) – {–1, 1}
E) (–, –2) (–1, 1) (2, )
7. What is the solution of the inequality 1 1 1 + + < 0? x x 2 x3
4. What is the solution of the inequality
A) (–, 0)
B) (0, )
C)
D)
E) {0}
(2 3x)(3 x 3) 0? 3x +1 1 2 A) , [1, ) 3 3
1 2 B) ; ; 1 3 3
2 C) (1, + ) 3
1 2 D) , (1, + ) 3 3
1 2 E) ; ; 1 3 3 Chapter Review Test 6A
8. How many integer values of x are there which satisfy the inequality |x + 2|
2 x +7 ?
A) 1
D) 5
B) 3
C) 4
E) 7 231
9. What is the solution of the inequality
13. What is the domain of the function
(1+ x2 )2001 (1 x)2002 0? ( x + 2)2003 (3 x)2005
A) x > –2
y=
B) x < –2
D) x > 3
C) x < 3
E) –2 < x < 3
x2
?
20 8 x x2
A) [–1, 7]
B) [–10, 2]
D) [5, 4)
C) (–10, 2)
E) (–2, 10]
14. What is the solution of the inequality system 10. The equation 2x(ax – 4) – x + 6 = 0 has no real 2
roots. What is the smallest possible integer value of a? A) –1
B) 2
C) 3
D) 4
E) 5
11. What is the solution of the inequality
B) (–2, –1)
D) (–5, –1) (1, )
A) (2, 3)
C) (1, 3)
E) (–1, 1) (5, )
D) [3, 2)
12. What is the sum of the integer values of x which x2 8x +7 < 0? ( x + 2)2
A) 32
C) 24
232
B) 28
D) 20
C) (–2, 3)
E) (–3, –2)
x < 2 ? ( x 3)( x 2) 0
A) (–2, 2) D) [3, 2)
satisfy the inequality
B) [2, 3]
15. What is the solution of the inequality system
( x 3)( x + 2) <1? x2 1
A) (–2, –1) (1, 3)
x2 5x +6 0 ? x2 + x 2 0
B) [2, 3]
C) (–2, 3)
E) (–3, –2)
16. What is the solution of the inequality 3 x x + 3 >1?
A) (–2, 2) E) 16
D) (–, 2)
B) (1, )
C) (–2, –3) E) (2, ) Algebra 8
6B
CHAPTER REVIEW TEST 1. What is the solution of the inequality
5. What is the solution of the inequality
x +1 x 1 > x? 2 3
A) x > 1
x2 > x +1?
B) x < –3 D) x < 1
C) x < 3
1 B) 1, 2
A) (–2, 4)
E) x > 6
C) (–, 2)
1 E) , 2
D) [2, )
2. What is the solution of the inequality (x – 6)(x + 3) 2 – 2x? A) [–5, 4]
6. What is the smallest possible integer value of x
B) [–4, 5]
D) [–5, 4)
C) (–4, 2)
E) (–4, 5]
3. What is the solution of the inequality x2 + x <1? 5x 3
which satisfies the inequality (x + 7)2(x – 4) 0? A) 3
B) 4
C) –8
D) 8
E) –13
7. What is the solution of the inequality (x2 + 8x – 9)(x2 – 4) 0?
3 A) , (1, 3) 5
3 B) , 1 (3, ) 5
C) (–, 1)
D) – {–1, 1}
A) (–, –2) [1, )
B) (–, –2] (2, )
C) (–, –1] [1, )
D) R – {–1, 1}
E) (–, –9] [–2, 1] [2, )
E) (–1, 1) (3, )
8. Which inequality is represented in this sign chart?
4. What is the solution of the inequality 1 5 + <1? 2 x 2+ x
-2
+
5
+
A) [–, –1) (1, 3)
B) [3, ]
A) x2 – 3x + 10 < 0
B) x2 > 3x – 10
C) (–, –2) (2, )
D) (1, 3)
C) x2 < 3x + 10
D) x2 – 2x + 5 0
E) (1, 3) Chapter Review Test 6B
E) x2 < 2x + 5 233
9. What is the solution of the inequality
13. What is the domain of the function
( x 1)3( x 2) 0? ( x 3)2
y = 4 x x2 +
B) [2, ]
A) (–, 1)
C) [1, 2] E) (2, )
D) [–1, 2)
What are the possible values of b? B) (–, –12]
D) (–, –12] [12, )
x2
A) (–2, 4]
?
B) [–1, 2]
D) [2, 4)
10. The equation 3x2 + bx + 12 = 0 has real roots. A) [–12, 12]
7
C) [12, )
E) (–1, 1) (5, )
E) (2, 4]
14. What is the solution of the inequality system
x5 >0 x+ 3 x+ 7 <0 x3
?
A) (2, 3)
B) [–7, 3]
D) [–3, 3)
11. What is the solution of the inequality 1 A) , 2
x 1? x +1
C) (–4, 2)
C) (–7, –3) E) (7, –2)
15. How many integer values of x are there which satisfy the inequality (x + 8)(x - 1)2(x – 5) < 0?
B) (–2, –1)
1 D) , 2
C) (1, 3)
A) 8
B) 9
C) 10
D) 11
E) 16
E) (–1, 1)
16. x1 and x2 are the roots of x2 – 3mx + m – 3 = 0.
12. What is the sum of the integer values of x which ( x 2)( x 4)x2 satisfy the inequality 0? ( x +1)2
A) 8 234
B) 9
C) 10
D) 11
1 1 + > 4 is given. x1 x2
What are the possible values of m? A) (–, )
E) 16
D) (3, 12)
B) [2, ]
C) (–, –3) E) (0, 12) Algebra 8
EXERCISES
1 .1
1. a. 6 b. 10 c. –11 d. 4x e. 5y f. 11a2 c. 2 d. 3x e.
y 2 3xy f. 3 2x
2. a. 3 b. 5 c. 6 d. 9 e. 4x f. 12xy g. aò15 h. 72xy
4. a. 2ñ2 b. 6ñ2 c. 9ñ3 d. 10ò10 e. 5ñ5 f. xyñx 5. a. 5ñ3 b. 7ñ5 c. 0 d. –2ñ6
e. 21ñ2 f. 2ñ5 g. 10ñ3 h. –13xñx i. 1 j.
4 3
6. a. ñ2 + 1 b. ñ3 – ñ2 c. ñ2 + ñ6 d. 2ñ2 + ñ3 e. ñ7 – 1
f. ñ3 + ñ2 g. 2 – ñ3 h. 2ñ2 i. ñ6 j. ñ2 k. ñ3 + 1
7. a. 6 b. 2 c. 4 d. 4
6 3 12
– 11 d. 11
11. a.
2 33 2 6
e. –1 – ñ2 f. 2ñ3 – 4 g. 2 + ñ3 h. 22 5 3 –1 c. 6ñ3 + 7ò11 d. 5 2
b.
EXERCISES 1
1. a. 212 b. 7 c. 5 7 d. xa 3
1
6
e. 215 f. 3 5 g. 3
c. 2 d.
5 4
e. ñ2 + 1 f. –2
l. 7 m. 9 n. 9 o. ñ6 + 1 p. 16 q. –1 r. 7. a. 4 b.
1 5
d. 2x e. 81 f. 13. ax+2y
a c. a d. ( )x 1 e. –2 f. 5 b 89 30
14. –
1 4
g. 1 h. 18 15. –2
28. 57122 – 57121
Answers to Exercises
j. 3 – ñ6 k. ñ5 + 1 l. 1
12. a. ñ2 b. 1 c. 9
–
1 3
2 c. ( )3 3
2. a. ña b. 3 b2
d. c xab
e. 4 a
18 7
j. 2 k. 3 l. 9 m.
1 5
n. 12 3 o.
16. 128
3. a. 6 3
b. 10 a
3
s. 2 6. a. 27 b. 55 c. 46 d. 24 e. 32x+1 f. 30x g. y – x h. 2100
8. a. 256 b. –729 c. –4 d. 64 e. a9 f. a
9. a. 2x+1 b. 0 c. 110 3x – 2
4 10. a. 6 b. 8 c. 5 d. 9 e. 4 f. { , 4} g. 3 h. 4 i. 3 j. –3 3
17. 2
18. x – 2
19. –4
20.
22 l. 0 m. ò14 22. 15 23. 0 3 19 29. 30. 2651 31. –121 8
f. 20 g. 2ñ7 h. 2ñ3 i. 2ñ2 j. 0 k. 27. 9
2 70 5 10 3
3 2 2 3 6
27 p. 24 313 q. 24 231 4. a. 1 5 39 17 5. a. 4 b. 26 c. 0 d. 54 e. 44 f. 17 g. h. i. 2 j. {–4, 4} k. 10 2 5
c. 100 d. ñ5 e. 3 f. 5 g. –3 h. 2x i. a2 6 a b. 6 24 33
i.
10. a. ñ3 b.
1 .2 2
1
12 3 – 8 23
9. a. ò15 + ò10
8. a. 2 b. 9 c. 24
b. 7 + ñ7 c. 4 – ñ6 – ñ2 d. 12 e. 6(ñ3 + ñ2) f. 4 g. –1 h. 8 i. 2 j. –2 k. 1 l. ñ3 m. ò13 c.
3. a. 5 b. 3
16 5
24. 3
11. 2
12. 268
21. a. 7 b. 8ñ3 c. 2.8 d. ñ5 e. 3 25. 3ñ2
26. 8 –
2 2
235
EXERCISES
2.
frequency (f)
1 2 1 3 3 5 3 3 1 2
3.
Percent
judo karate 10% 5%
35 30
wrestling 7%
25
football 20% basketball 12%
20 15
swiming 29%
10
volleyball 17%
Other
Expenses
5. a. 2705000 tons b. 2001 c. 1080000 tons
Other
EXERCISES
5 d. 5 e. 2 5 3 5
b. 0,
500 000 400 000 300 000 200 000 100 000
3 .1
1. a. 0 b. 0 c. 0 d. 0 2. a. 0, c.
600 000
2003
Pichard Whiting Hake
Hazelnut production in Turkey (tons)
2002
Blue fish
700 000
2001
d.
Scad Gray mullet
2000
Anchocy Horse mackerel
1999
4.
Books
0
Clothing
5 Food
1 2 3 4 5 6 7 8 9 10
Entertaiment
scores (x)
Rent
1.
2 .1
f.
5 8 3 1 b. 0, c. 0, d. 0, 2 3 7 9 4 5
3. a. 0, –2 4. 0,
5 2
5. –1, 2 6. a. 4 b. no real solution
7. a. –1,5 b. no real solution c. (–1 – ñ5), (–1 + ñ5) d. –2, 4 8. ñ3
5 3 1 2 2 3 1 c. 7 d. 5 e. –1, 2 f. –1, –2 g. 1 h. , i. , j. , 3 2 4 3 5 2 3
9. a. 0, 1
10. a. –a, –1 b. no real solution c.
5b ,1 2a
11. a. (2 + ñ5), (2 – ñ5), b. (ñ7 – 2), (–ñ7 – 2) c. (3 ò22) d. no real solution e. no real solution f. no real solution 3 1 g. 3 2ñ2 h. –1 7 12. a. 2 ñ2 b. 1, c. x d. 5 13 e. no real root f. 1 2 7 g. no real root 2 2 6 3 6 4 1 7 8 2 , 1 l. –1, 23 14. a. two real roots h. x i. , 4 j. , 1 k. 13. a. 1 33 b. 3 73 c. 1, 5 d. , 5 5 26 4 5 3 8 4 9 b. two real roots c. one double root d. one double root e. no real solution f. no real solution 15. a. a 40 9 7 7 7 9 1 b. a c. a 16. a. m b. m c. m 17. – , 1 18. 8 cm, 10 cm 40 2 2 2 40 4
19. 12 cm, 20 cm 236
20. 4, 6, 8, 10
21. 256 cm2, 441 cm2
22. 7, 49 Answers to Exercises
EXERCISES
72 40 4 5 11 6 10 5 18 8 24 c. e. , f. (3ñ2 – 2), 0 2. , b. , , d. , 49 21 3 6 10 5 3 3 7 7 7
1. a.
153 256
d. 10.
3 .2
5. m1 = –5, m2 = 4, x2
14 ,4 3
19. 7
11. 5
20. c = 2, 1,
7 5 12. , 3 3
1 5 2
EXERCISES
5 4
6. n1 = –4, n2 = 2, x2
13. –6, 2 21. –13
14. 1
15. k > 6
22. 1 , 1 13 2 2
2 3
7. 0
16. 6, 7
3.
2 5
4. a.
8. m = 2n 17. a b
17 21 b. 53 c. 9 4
9. k = 3
18. 0
23. 4 + 2ñ3 cm
3 .3
1. a. x2 – 1 = 0 b. 4x2 – 8x + 3 = 0 c. x2 – 4x = 0 d. x2 – 4x + 2 = 0 e. x2 – 2ñ3x + 1 = 0 f. p2q2x2 – (p2+q2)x+1=0 2. x2 + 2x + 5 = 0 3. x2 – 8x – 12 = 0 4. x2 + x + 27 = 0 5. a. x2 – 4 = 0 b. x2 + 8x + 7 = 0 c. 4x2 – 25 = 0 d. x2 – 2x + 1 = 0 e. x2 + 2mx – 8m2 + 27m – 18 = 0 f. mx2 – 2mx – 8m – 9 = 0
6. 5
4 13. 18(ñ2 + 1) 7 14. from A to B 60 km/h, from B to A 80 km/h 15. 160 km/h 16. 16 km/h 17. 7 women 18. 80 km/h 19. 24 hours, 48 hours 20. 20 km/h 21. height = 6 m, base = 12 m 22. 40 kg 25%, 20 kg 40% 23. 60 hours, 84 hours 24. 4 hours, 8 hours 25. 6 m, 8 m
7. –7, –6, –5 or 5, 6, 7 8. 9 9. 15 hours, 10 hours 10. 3 hours, 2 hours 11. 6 hours 12.
EXERCISES
3 .4
1. a. 2, 3 b. ñ3 c. ò10, 2ñ3 d. –5 ñ7 e. –5
1 3 5 f. 79 g. 4 h. 2 ñ3 i. 2 ñ2 2. a. b. (1 ò10), –1, 3 2 2
c. –2, –1, 0, 1 d. 1, 2 e. –6, 1
7 6 42 1 3. a. 0, , 1, 5 b. –10, (–4 ñ2), 2 c. 2 d. 5 e. f. 5, , 1 g. no solution 3 3 4
h. 1, 2, 4 i. 0 j. –3 ò15
4. a. 2 b.
11 – 21 c. –7 d. 1, 2 e. 12 f. 3 g. –1, 2 h. 0 i. –1, 7 j. no solution 2
8 5. a. 7, 8 b. 2 c. 0 d. 0, 2 e. 0 f. no solution g. – , 1 h. 2 3
g. 3 – ñ7, 2 h. 2 i. no solution d. –1, 4 Answers to Exercises
7. a. 0 b.
6. a. –1 b.
7 , 3 c. 2 d. –1, 5 e. –1, 3 f. no solution 3
2 1 3 3 c. –2, 0 d. 2 e. – , , 2 f. x (– , ] 8. a. 1 b. 0, 3 c. 3 2 2 4 2
1
237
EXERCISES
3 .5
5 7 1. a. (–6, –18), (18, 6) b. ( , – ) c. (–3, 6), (10, –7) d. (10, 15), (15, 10) e. (4, –1), (–4, 1) f. (3, 3), (–3, –3) 2 2 g. (–3, –4) h. (0, –5), (1, –4) 3 2. a. (–4, –4), (–6, –2) b. (0, 0), ( , 3) c. (0, 0) d. (1, 0) e. (2, 4), (4, 2), (ò22 – 4, –ò22 – 4), (–ò22 – 4, ò22 – 4) 2 7 7 7 ,– )(– , 2 2 2
f. (0, –2), (2, 0) g. (0, ñ2), (0, –ñ2) h. (2, 1), (–2, –1), (
EXERCISES 1. a. x = –1, y = 4
4 .1
b. x = 1, y = 0
2. a. {3}
b. 1
3. a. (–1, 1], (–3, 2) b. (2, 5), (–3, )
e. {3} f. 2
7 ) 2
c. {1, 3, 5, 7, 8, 9, 10}
d. {1, 2, 3, 4, 5, 7, 8, 9, 10}
4. a. A = {1, 2}, B = {a, b, c} b. A = {1, 2}, B = [2, 4]
5. a. {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)}, {(–1, –1), (–1, 1), (1, –1), (1, 1)}, {(1, –1), (1, 1), (2, –1), (2, 1), (3, –1), (3, 1)}, {(–1, 1), (–1, 2), (–1, 3), (1, 1), (1, 2), (1, 3)} b. {(December, December), (December, January), (December, February), (January, December), (January, January), (January, February), (February, December), (February, January), (February, February)}, {(2, 2), (2, 5), (2, 6), (5, 2), (5, 5), (5, 6), (6, 2), (6, 5), (6, 6)}, {(December, 2), (December, 5), (December, 6), (January, 2), (January, 5), (January, 6), (February, 2), (February, 5), (February, 6)}, {(2, December), (2, January), (2, February), (5, December), (5, January), (5, February), (6, December), (6, January), (6, February)} 6. a.
b.
B 6
1 2 3 · · · ·
1 2 3 5
· · · ·
b c d e
1
A
b.
· · · · ·
1 2 3 4 5
· · · · ·
4
0 1 2 3 4
A
7. a. {(0, 0), (1, 1), (2, 4), (3, 9)}
B 4
4 3 2
4
8. a.
c.
B
b. {(–2, –5/3), (–1, –1), (0, –1/3), (1, 1/3), (2, 1), (3, 5/3)}
2 -2
3
9. a.
A
b.
y 3 2 1
7
1 2
d.
y
y
1 1
x
10. a. {(December, winter), (September, fall), 1 (June, summer), (March, spring)}, {December, September, June, March}, {winter, fall, summer, spring} b. x = 2y – 3, [–43, 65], [–20, 34] c. x = y2 + 2, [2, 11], [–3, 3] 238
c.
y
x
1 -1
3 x
Answers to Exercises
x
11. a.
b.
y
c.
y
y
(0, 4) (2, 0)
x
(-2, 2)
12.
x
(0, 0)
1
(4, 2)
(2, 2)
(0, -3)
y
(0, 3)
(0, 0)
x
(4, -1)
x
-1
13. {(a, a), (a, b), (b, b), (b, c), (c, a), (c, b), (c, c), (c, d), (d, a), (d, d)} 14. 84
y
17.
5 4
15. use the fact that n(A) + n(B) includes n(A B) twice. 16. {(1, a, blue), (1, a, red), (1, b, blue), (1, b, red), (2, a, blue), (2, a, red), (2, b, blue), (2, b, red), (3, a, blue), (3, a, red), (3, b, blue), (3, b, red)}, use three coordinate axes that are perpendicular to each other in the space
1 -1
3
x
-5
4 .2
EXERCISES
x 2. a. f(x) = 5x + 2 b. f(x) = x2 – 2x c. f ( x) = + 3x3 2 3. a. multiply by 2, then subtract 4 b. add 1, then take the square root of all c. divide by three times itself increased by 1
1. a. function b. not c. not
+ 2 3 x2 1 4. a. –3, 33, 2x2 + 5x, –2x2 + 5x, 2x – 5ñx, 2x2 + 4hx – 5x + 2h2 – 5h b. 1 , 11 , 9 x2 1 , 6 x , , 2 18 9x + 2 x2 + 2 x4 + 2
6a + 3b 1 2 4a + 4ab + b2 + 2
6. a.
b.
y -2 -1
2 2 4 4 2 2 c. – , undefined, 4u3 +1 , 3 x +22 x + 2 , 6u +1 2 , 6u + 2 4u +1 2 3 8u + 4u x + 3x +5x + 3 u + 2 u u + 4u + 4u
x -1 -1
3 1 2
d.
y
y
3 2 1
3 2
1 2 -1
c.
y
x
7. a. 1
1
4
9
x
1/2
x
5.
8 3
7 4 d. 21; –3 ; – 7 b. 6, –1; –6 c. 2; 3 5
8. a. function b. not c. function d. function e. not f. not
-4
c. (–, 4) (4, ) d. (–, –3) (–3, 3) (3, ) e. [2, ) f. (–, –3) (–3, 0) (0, ) 5 10 10 1 3 3 5 g. [5, 6) (6, ) h. [5, 10] i. ,1 j. [1, 1) 1, k. 3, , 10. a. , b. [4, 1) ( 1, 3 ) 3 3 2 4 5 2 c. (–1, 1) (1, ) d. (–, –2) (4, ) e. [–1, 2] f. [1, 2] {0} g. (0, 1) (1, 3) h. [–5, –1) (1, 5] i. (–, –1] [11, ) 9. a.
b.
2 2 11. a. f ( x) = 3 x b. f ( x) = x 4x c. f ( x) = x 3 ( x + 2)(1 x) 2 x + 2x x5 b. –4, 4 c. 13. a. odd b. neither c. even d. neither e. even f. even g. odd h. odd i. neither j. odd
j. [–5, –2] [2, 3) (3, 5]
12. a. – 4 3 14. a. even b. odd c. even d. odd e. neither f. even Answers to Exercises
239
15. a.
(-4,6)
y
b.
(4,6)
c.
y
x
x (-2,-2)
(-4,6)
(-5,-6)
y
(-7,4)
(0,0)
x
y
(4,4)
(0,2) (-5,0) (-2,0)
(0,4)
(-2,-2)
x
(7,-4) (4,-6)
(5,0)
(-4,-4)
(0,-4)
(7,-4)
x
2
(0,-2)
(5,-6)
y
(-5,6)
(2,2)
(-7,4)
(2,0)
(-5,0)
(7,-4)
(-7,-4)
(2,-2)
y
(7,2) (-2,0)
(0,0)
16. a. (2,4)
(-7,2)
(7,4)
(-7,4)
y
(-2,4)
(0,4) (4,4)
(-4,4)
b.
(2,4)
x
y
(7,2)
(2,0) (5,0) (0,-2)
(-7,-2)
5
x -5
2
9 11 x
6
(-2,-4)
c.
(5,-6)
17. consider f(x2) – f(x1) where x1 < x2 on the given interval 18. 3 19. –2 20. a. , b. increasing on (–, –2] and [5, ), decreasing on [–2, 5] c. –6, 3, 6; 3 d. –4, 5, 8 e. (–, –6) (3, 6)
y
345
7
x
21. a. (-, –10) (–10, 10) (10, ), [–7, ) b. –5, 5 c. 2 d. undefined e. [–5, 0) (0, 5] f. (-, –10) (10, ) g. (–10, –6] {0} [6, 10) h. even i. increasing on [3, 6] and (10, ), decreasing on (–, –10) and [–6, –3], constant on (–10, –6], [-3, 0), (0, 3] and [6, 10) j. –7, k. [–3, 0) (0, 3] l. (–6, –3) (3, 6) 22. 0 23. d(t) = 15t 24. a. F(x) = 1.8x + 32 b. 1.8
10t if 0 t 90 225h2 if 0 h 2 25. a. V(t) = 10t b. V( h) = c. h(t ) = 15 1200 h 1500 if 2 h 3 2+ t 90 if 90 t 210 120 26. until his twenties the person gained weight and kept his weight constant until his forties, then he lost weight
because of a diet or an illness and after his mid-forties he started to gain weight again. 27. he left home at 8 a.m. and 1 during the day visited four places, probably for a sale, and at 6 p.m. left for home. 28. a. (–, 4] b. [3, ) c. 0, 6 1 d. (–, 1) (1, ) 29. 100 30. 0, 2, –1 31. –8 32. decreasing on , , increasing on 1 , 5 5 2 13 2x 3 33. –x – 3x 34. 35. 12 36. 9 3x
EXERCISES
4 .3
1. a. x2 + x, x2 + x + 2, x3 + x2 x 1, x5 +8 x4 +15 x3 , 240
1 with domain \ {–1, 1} x 1
x2 + 3x with domain \ {–5, 0} x +5
b.
x3 + 4 x2 +5 x, x3 + 2 x2 5 x,
c. x + 3+ x + 2, x + 3 x + 2, ( x + 3) x+ 2,
x+ 3 x+ 2
Answers to Exercises
d.
x 1+ x +1,
x 1 x +1,
2 x2 + 2 x , x4 + 2 x3 + 2 x2 + x x2 2 x +1
e. one-to-one 13. a.
8. x
2 2. a. 25 b. undefined c. –11 3. a. x, x 2 + x +1 , x +x 1 x +1 7 x +6 11 x 5 x + 3 10 x 3 b. x4 – 4x3 + 4x2, x2, x2 – 2x + 2, x + 2 c. , , , 2 x + 3 x + 3 4x +9 13 3 x
x4 41 c. f(x) = x3 – x + 4, g(x) = 2x2 – x 5. a. b. –1 5 11 9. a. not b. not c. one-to-one 10. a. one-to-one b. one-to-one c. not d. one-to-one
4. a. f(x) = ñx, g(x) = 2x – 4 c. 4 d. 9 6. 2 7. 2
x 1
x 1 x +1,
b. f ( x) = x7 , g( x) =
11. a. one-to-one b. one-to-one c. not
y (5, 4)
b. inverse function c. does not exist
12. a.
6x 4
x+ 4 x3
b.
d.
y (1, 4)
e.
x
(-2, 4) (3, 0)
x
(0, 0)
(4, -5)
f. inverse function does not exist
14. a. 5 b. 0 c. 3 d. –14 e. 7 15. a.
y (-5, 6)
(-3, -6)
(-3, -8)
5
x 9 +1 d. 3
x +5 4
y
(-3, 2) (-1, -2)
3
(-5, 6)
(7, 5)
(-2, 2)
x
c.
(7, -6)
x
(3, -5)
2x 1 2x 22 x 1 2x 1 6x 5 3x 7 b. c. 16. a. b. c. x d. e. 2x – 11 2x 3x 2 3x x2 2 x +1 2
3x 7 x 1 g. 17. a. 1 b. –1 c. 1 d. 17 18. a. 1 b. –2 19. a. find two different x-values that give the same x5 2x 4 16 y-value b. assume that for x1 x2, f(x1) f(x2) 20. –2 21. Compare the functions when the argument is x and –x
f.
22. Compare the functions when the argument is x and –x
23. a. x +9 3 b. x +11+ 4
24. Try to compose
each of the sides with g f
EXERCISES
5 .1
y
1. a.
b.
4
-1
0
3 2
-3 -4
2x2=3y 1
2. a. yes b. no c. yes d. no
Answers to Exercises
-
y=2x2
2 3
y
c.
1 x
y 4x2=0 2
y 0
-1
3
y=3x2
1 3
3y=x2
2y=3x2 y=3x2
y+4x2=0
-1 -
x
3. a.
1 1 7 25 b. c. – d. – 3 5 9 8
1 2
-2
1
x 2y+x2=0 y+2x2=0
241
4. a.
y -4 -3 -2 -1 0 1 2 3 4 1 x 2 -2 -
5. a. V(0, 0), x = 0, ymin = 0 b. V(0, 0), x = 0, ymax = 0 c. V(0, 2), x = 0, ymin = 2 d. V(0, –1), x = 0, ymax = –1 6 6 6 6 5 25 5 25 e. V( , f. V( , ), x = , ymin = ), x = , ymax = 7 5 7 5 4 8 4 8
9 2
2 5 2 5 g. V( , ), x = , ymin = 3 3 3 3
1 5 1 5 h. V( , ), x = , ymax = 2 4 2 4
3 75 3 75 i. V( , – ), x = , ymin = – j. V(–4, 0), x = –4, ymax = 0 2 4 2 4 5 41 k. V(–3, –5), x = –3, ymax = –5 l. V(1, 3), x = 1, ymin = 3 6. – 7. 1 8. – 9. a. (0, 0), (4, 0) 2 8 3 3 b. (3, 0), (0, 9) c. (0, 396) d. (–2ñ2, 0), (2ñ2, 0), (0, –8) e. (– , 0), (–1, 0), (0, 3) f. (0, –5) g. (– , 0), (0, 9) 2 4 h. (0, –4) 10. a. m (–2ñ2, 2ñ2) b. m = 2ñ2 c. m (–, -8
–2ñ2) (2ñ2, ) 11. 3 y
12. a.
y
b. ñ3 x
-ñ3
c.
7
y x
x
y
h.
-48
i.
1 2+ñ2 x
4
3 x
-3 -4
y
ñ5 x
-ñ5 -5
-2
x
x y
y
j.
13.
12
13
2 ñ2 -3
y
f.
1
-1
98
-7
y
y
e.
-4 1
-1
y
d.
5
-9
g.
-4
7
f(A)= [9, 7)
-1 1 4
-2 3
x
-3
4
4 5
x
x
-5 -8 -9
y
14.
15. a. ymax = 12, ymin = 3
3 -1 1
3
x
16. ymax = 64
17. ymin = –
9 4
1 18. a = – , b =1 2
19. y = 3(x – 5)2 20. –3(x – 1)2 + 2 2; –3(x – 1)2 0; 3(x – 1)2 0 21. m [0,
16 ] 9
-5
242
Answers to Exercises
y
22. a.
y=x2+4
y
b.
y=x2 y=x2 2
y=x2 4
-ñ2
y 16
-25
y=x2 2
g.
3
y=(x 4)2
-1
y
y=x2
4
y=(x+1)2
3)2
y
y
b.
c.
3
y
d.
1
3
1 2 3
x
-1
1
x
y
1
2
x
-1 2 3 4
x
-1
24. a. y = –(x – 1)2 + 3 b. y = f. y = –
1 (x – 3)(x + 3) 3
x
2
7
1
1
y=(x 2)2
7
y=x2
y=(x+1)2 2
23. a.
y=(x 2)2+3
x -2
y=(x
y
h. -1
-9 y=x2
y=x2
y=(x+5)2
x
x
4
x
y=x2
y
f.
-5
x
-2
-2
y=x2
y
d.
x
ñ2
x
e.
y
c.
1 1 x(x – 5) c. y = (x + 3)2 d. y = (x + 1)2 + 3 e. y = –x2 + 4x – 3 6 3
25. a. y = –3x2 + x + 4 b. y = –x2 + 4x – 4
c. y = (x + 1)2 – 4 d. y = –x2 – x – 4 27. –12 28. 33. 10 square units 34. m (–, –1) (0, )
18 5
29. 0 30.
1 2
26. a. y = x2 – 4 b. y = (x – 2)2 31.
35. 64 square units 36.
25 4
32.
9 square units 4
49 37. 2 38. (–3, 0) 39. 9 40. 1 20
41. 35 square units
Answers to Exercises
243
EXERCISES
6 .1
1. a. (–2, ) b. (–, 6] c. (–, –7) d. [4, ) e. [2, 4) f. (–1, 0] g. (–5, x
2. a.
¥
x
+
¥
x
6 f. x ; a+ 3
x
x
1. a. x2
¥
5x+4
x
d.
¥
1/4
c. (–, 0) (2, )
1/2 x
14/3
+ 0
b. x 0;
e. x > 2 ; 3
x
2/3 x
3 a4+4
–3 g. x > 4 ; a +4
4
1
2. a. (4, 5) b. (0,
x
¥
x 7 - + 2 3
x
6 .2
+
16x2+8x 1
x
c.
+
3
d. x 3;
6/(a+3)
EXERCISES
-5/3
1 3. a. x > – ; 2
+ 2
c. x > 2;
¥
3x 5
6/ñ3
ñ3x 6
x
b.
4x + 1
d.
-1/4
1 3 ) h. [0.5, ] 2 2
x
b. 2x2+x
+
3/2
-2
+
6
e.
¥
+
2x2
+ + + + + +
3x+4
x
f.
x 4x2+10x 25
x
c.
12x2+4ñ3x+1
4 7 4 4 f. ( , 1) g. (–, –4) (2, ) ) c. (– , – ] [0, ) d. e. x = – 7 6 3 5
4. (–, –2ñ3 – 1) (2ñ3 – 1, )
5. (3 – 2ñ2, 3 + 2ñ2)
¥
ñ3/6
+
+
3. a. (0, 2) b. 0, 2
6. {–6, –5, –4, –3, –2, –1, 0 1, 2}
1 3 2 2 5 7. a. (0, ), x 1 b. (– , – ) ( , 2) c. x –2, (– , –1) ( , 3) d. (– , ) (1, ) 3 2 3 3 2 1 3 7 e. (–4, –3) (–2, – 1) ( , 3) f. (–1, 5) g. (– , ) ( , ) h. (–8, –1) i. – {2} j. (–, –7] (–1, 0) (0, 1] (3, ) 2 2 3
k. (– ,
– 6 ] [–1, 0) [1, 2
6 4 1 11 ] l. (– , 1) ( , 2) m. (–, 2) – {–4, –2} n. [0, ] o. (– , 1) ( , ) 3 3 5 2
1 p. (–, –2) (–2, –1) (–1, 0) q. [– , 0] [1, ) 2
3 8. a. (0, 1) b. (0, 1) c. d. [ , 2) 2
9. a. – {7, –7}
b. (0, 16) (16, ) c. (–, –2) [2, ) d. [2, 3] [–1, 1] 10. a. (0, ñ3 – 1) (–4, –3) (–ñ3 – 1, –2) (1, 2) 11. x 0, (–2, 1) 244
Answers to Exercises
EXERCISES 1. a. (–2, c. (–8, –
6 .3
3 ) b. (1, 3) c. (–6, –5) d. (–1, 1) e. (1, 2) f. [1, 2) g. (2, 4) h. (0, 1) 2
13 ) (0, 5) 2
2. a. (0, 1) b. (–1, 1) (3, 5)
3 5 3. a. (– , 1] [ , ) b. (– , – ) (3, ) c. (–, –5) (–1, ) d. (–, –4) (0, ) 2 3
3 2 2 13 5 3 e. [ , 2) f. [ , ] g. (–, 3) 4. a. (3, ) b. [ , 4) c. [4, ) d. (–3, 1) e. [3, ) f. (9, ) 5. a. [3, ) b. [–2, – ] 2 7 3 5 10 2
c. [2, 3] d. [–15, 7. [1, 5) (10, )
TEST 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
B D A D B B C C C C
TEST 1. 2. 3. 4. 5. 6. 7. 8.
E B E A C D A D
13 – 5 417 ) e. [3, ) 64 2
TEST B B C B C D C D A B
6A 9. 10. 11. 12. 13. 14. 15. 16.
Answers to Exercises
– 5 –1 , 2
13 + 3 – 11 – 1 11+1 ] b. [–1, 3] c. [– 6, ) ( , 2 2 2
5 8. a. (2, ) b. (1, ) c. (– , – ) d. (–, –2) (2, ) 2
1A 11. 12. 13. 14. 15. 16. 17. 18. 19. 20.
6. a. [
1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
C D B C B D C B D D
TEST E B D D C B A B
1. 2. 3. 4. 5. 6. 7. 8.
D B A C E B E C
1B 11. 12. 13. 14. 15. 16. 17. 18. 19. 20.
TEST D C C B A C C D B C
1. 2. 3. 4. 5. 6. 7. 8.
C E C B E D C C
9. (–5, –1)
3A 9. 10. 11. 12. 13. 14. 15. 16.
TEST B B D A E C B D
1. 2. 3. 4. 5. 6. 7. 8.
B D E A D B A C
3B 9. 10. 11. 12. 13. 14. 15.
A E B D B B S
6B 9. 10. 11. 12. 13. 14. 15. 16.
C D A B E C D D 245
6]
246
Answers to Exercises
Answers to Exercises
247
248
Answers to Exercises
