MCR3U
1
Exam Review
Polynomials
A polynomial is an algebraic expression with w ith real coefficients and non-negative integer exponents. A polynomial with 1 term is called a monomial, 7 x . A polynomial with 2 terms is called a binomial, 3 x 2 9 . A polynomial with 3 terms is called a trinomial, 3 x 2 7 x 9 . The degree of the polynomial is determined by the value of the highest exponent of the variable in the polynomial. e.g. 3 x 2 7 x 9 , degree is 2. For polynomials with one variable, if the degree is 0, then it is called a constant. If the degree is 1, then it is called linear. If the degree is 2, then t hen it is called quadratic. If the degree is 3, then t hen it is called cubic. We can add and subtract polynomials by collecting like terms. e.g. Simplify S implify.. The negative in front of the brackets 5 x 4 x 2 2 x 4 2 x 3 3 x 2 5 applies to every term inside the brackets.
! 5 x 4 x 2 2 x 4 2 x 3 3 x 2 5
That is, you multiply each term by ±1.
! 5 x 4 x 4 2 x 3 x 2 3 x 2 2 5 ! 4 x 4 2 x 3 4 x 2 3 To multiply polynomials, multiply each term in the first polynomial by each term in the second. e.g. Expand and simplify.
x
2
4 x 2 2 x 3
! x 4 2 x 3 3 x 2 4 x 2 8 x 12 ! x 4 2 x 3 7 x 2 8 x 12 Factoring Polynomials
To expand means to write a product of polynomials po lynomials as a sum or a difference of terms. To factor means to write a sum or a difference of ter ms as a product of polynomials. Factoring is the inverse operation of expanding. Expanding 2 x 33 x 7 ! 6 x 2 5 x 21 Factoring Product of polynomials
Sum or difference of terms
MCR3U
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Exam Review
Types of factoring: Common Factors: factors that are common among each term. e.g. Factor, 3 3 2 2 2 Each term is divisible by 35 m n 21m n 56 m n
7
2
n.
! 7 m 2 n 5mn 2 3n 8 Factor by grouping: group terms to help in the factoring process. e.g. Factor, Group 4 x ± 4nx and ny ± y, factor each B : 1 6 x 9 x 2 4 y 2 A : 4 x ny 4nx y
! 4 x 4nx ny y ! 4 x n yn ! 4 x n y n ! 4 x y n
group
Recall n ±
! (1 3 x )2 4 y 2
1+6 x+9 x2 is a perfect square trinomial Difference of squares
! [(1 3 x ) 2 y ][(1 3 x ) 2 y ]
= ±(m ± n)
! (1 3 x 2 y )(1 3 x 2 y )
Common factor
Factoring ax b x c Find the product of ac. Find two numbers that multiply to ac and add to b. e.g. Factor, 2
: y 2 9 y 14
! y 2 7 y 2 y 14
B : 3 x 2 7 xy 6 y 2
Product = 14 = 2(7) Sum = 9 = 2 + 7
! 3 x 2 9 xy 2 xy 6 y 2
! y ( y 7) 2( y 7)
! 3 x( x 3 y ) 2 y ( x 3 y )
! ( y 2)( y 7)
! (3 x 2 y )( x 3 y )
Product = 3(±6) = ±18 = ±9(2) Sum = ± 7 = ±9 + 2 Decompose middle term ± 7 xy into ±9 xy + 2 xy. Factor by grouping.
Sometimes polynomials can be factored using special patterns . Perfect square trinomial e.g. Factor,
a 2 ab b ! ( a b)( a b) 2
2 A : 4 p 12 p 9
: 100 x 2 80 xy 16 y 2
! ( 2 p 3) 2
! 4( 25 x 2 20 xy 4 y 2 ) ! 4(5 x 2 y )(5 x 2 y ) b 2 ! ( a b)(a b) 9 x 2 4 y 2 ! (3 x 2 y )(3 x 2 y )
Difference of squares
e.g. Factor,
2
a
2
Things to think about when factoring: Is there a common factor? y y Can I factor by grouping? y Are there any special patterns? 2 y Check, can I factor x b x c ? 2 y Check, can I factor ax b x c ?
or a 2 2ab b 2 ! ( a b)(a b)
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Exam Review
Rational Expressions F
For polynomials F and G, a rational expression is formed when e.g.
G
,G
{
0.
3 x 7 21 x 14 x 9 2
Simplifying Rational Expressions
e.g. Simplify and state the restrictions. m2 9 ( m 3)(m 3) Factor the numerator and denominator. ! m 2 6m 9 (m 3)(m 3) Note the restrictions. m { 3
! !
( m 3)(m 3) Simplify.
( m 3)( m 3) m 3 m3
, m { 3
State the restrictions.
Multiplying and Dividing Rational Expressions e.g. Simplify and state the restrictions. 2 2 x 2 7 x x 2 3 x 2 x 9 x 4 x 3 v 2 A : 2 z 2 B : 2 x 1 x 14 x 49 x 5 x 4 x 5 x 4 x ( x 7) ( x 1)( x 2) ( x 3)( x 3) ( x 1)( x 3) Factor.
! ! !
( x 1)( x 1)
v
Note restrictions.
( x 7)( x 7)
( x 1)( x 2) x ( x 7) v ( x 1)( x 1) ( x 7)( x 7) x ( x 2) ( x 1)( x 7)
{ s1, 7
, x
!
Simplify.
!
State restrictions.
!
( x 4)( x 1) ( x 3)( x 3) ( x 4)( x 1) ( x 3)( x 3) ( x 4)( x 1) ( x 3)
!
( x 1)
, x
{
z
v v
( x 4)( x 1) ( x 4)( x 1) ( x 1)( x 3) ( x 4)( x 1) ( x 1)( x 3)
4, s 1, 3
Factor. Note restrictions. Invert and multiply. Note any n ew restrictions. Simplify.
State restrictions.
Adding and Subtracting Rational Expressions
e.g. Simplify and state the restrictions. 3 5 Factor. A : 2 Note restrictions. x 4 x 2 Simplify if possible. 3 5
! ! ! !
( x 2)( x 2) x 2 3 ( x 2)( x 2)
5( x 2) ( x 2)( x 2)
3 5 x 10
( x 2)( x 2)
2 x
! !
Add.
( x 2)( x 2) 5 x 7
Find LCD. Write all terms using LCD.
:
! , x
{ s2
State r estrictions.
2
xy
3 xy y 2
2 x ( x y )
2 y
3 y ( x y )
3 x
Factor. Note restrictions. Simplify if possible. Find LCD. Write all terms using LCD.
xy ( x y ) xy ( x y ) 2 y 3 x xy ( x y )
, x
{ 0, y, y { 0
Subtract. State restrictions.
Note that after addition or subtraction it may be possible to factor the numerator and simplify the expression further. Always reduce the answer to lowest terms.
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Exam Review
Radicals n
e.g.
a,
is called the radical sign, n is the index of the radical, and a is called the radicand.
3 is said to be a radical of order 2.
Like radicals:
5, 2 5,
3
3
8 is a radical of order 3.
Unlike radicals:
5
5,
3
5,
3
Same order, like radicands
Entire radicals:
8 , 16 ,
Different order
29
Mixed radicals: 4 2 , 2 3 , 5
Different radicands
7
A radical in simplest form meets the following conditions: For a radical of order n, the radicand has no factor that is the nth power of an integer. 8 Not simplest form
!
4v2
!
2 v2
The radicand contains no fractions. 3 2
!
2
!2
!
2
!
Simplest form
!
3 2
v
2 2
6 2
2
The radicand contains no factors with negative exponents. 1 1 a ! a
!
a
Simplest form
!
4
3
a
2
! !
3
2
3
Simplest form
a
a
!
22 2
v
a
6 6
1
The index of a radical must be as small as possible.
a a
2
Simplest form
Addition and Subtraction of Radicals
To add or subtract radicals, you add or subtract the co efficients of each radical. e.g. Simplify. 2 12 5 27
3
40
!2
4 v 3 5 9 v 3 3 4 v 10
! 22 !4
3
53 3 32
3 15 3 6 10
! 11
10
Express each radical in simplest form.
Collect like radicals. Add and subtract.
3 6 10
Multiplying Radicals av
b
!
ab , a
u 0, b u 0
e.g. Simplify.
2 2 3 2 3 3 ! 2 2 2 3 3 2 3 2 2 3 3 3 ! 23
62 6
! 2 18 3 ! 16
6
63
62 6
Use the distributive property to expand
Multiply coefficients together. Multiply radicands together.
Collect like terms. Express in simplest form.
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Exam Review
Conjugates
Opposite signs
a
b
c d and a
Same terms
b
c d are called conjugates.
Same terms
When conjugates are multiplied the result is a rational expression (no radicals). e.g. Find the product.
5 3 2 5 3 2 ! 5 3 2 2
2
! 5 9(2) ! 5 18 ! 13 Dividing Radicals
a b
!
a b
e.g. Simplify.
, a, b R , a u 0, b u 0
2 10 3 30
!
5
2 10 5
3 30 5
!2
10
!2
2 3 6
5
3
30 5
Prime Factorization
e.g.
180
Factor a number into its prime factors using the tree diagram method.
3
60 6 2
10 3
2
5
180 ! 2 2 3 2 5
Exponent Rules Rule Product Quotient
Power of a power
Description
Example
mn
va ! a m a z a n ! a mn a m n ! a mvn a
m
n
Power of a quotient
¨ a ¸ ! a n { ,b 0 © ¹ ª b º b n
Zero as an exponent
a0 ! 1
Negative exponents Rational Exponents
4 v 4 ! 47 54 z 52 ! 52 2
3 ! 3 2 4
n
a
m
!
1 a
m
an
!
n
a
m
5
92 !
! a n
8
¨ 3 ¸ ! 35 © ¹ ª 4 º 45 0 7 !1
,a { 0
m
5
m
1 9
2
4
27 3
!3
27 4
! 3 27
4
MCR3U e.g. Evaluate.
e.g. Simplify. 2
3
0
3
2 2
6
Exam Review
2
! 1 9 ! 102 ! !
1
¨ b 3 ¸ ©© 3 ¹¹ ! ª 2a º
Follow the order of operations. Evaluate brackets first.
!
102 1
!
100
b
3( 2 )
3 2
(2a ) b
6
2 a 2
2
2
b
a 4
! a
6
Power of a quotient.
3( 2 )
Power of a product.
6
6
b
6
Solving Exponential Equations
e.g. Solve for x. 2 9 x 8 ! 73 Add 8 to both sides. 9 x 2 ! 73 8 x 2
! 81
x 2
! 92
9 9
x 2 ! 2 x ! 2 2
Simplify.
x ! 4
Note LS and RS are powers of 9, so rewrite them as powers using the same base.
When the bases are the same, equate the exponents. Solve for x.
LS ! 9
x 2
8 RS ! 73
! 9 4 2 8 ! 81 8 ! 73 ! RS x ! 4 checks
Don¶t forget to check your solution!
Functions
A relation is a relationship between two sets. Relations can be described using: an equation an arrow diagram a graph g y ! 3 x 2 7 8 -1 0 7 in words 6 -3 ³output is three more than input´ 3 -5 a set of ordered pairs {(1, 2), (0, 3), (4, 8)}
a table x 1 2 3 4
2
-2
y 2 3 4 3
function notation 2 f ( x ) ! x 3 x The domain of a relation is the set of possible input values ( x values). The range is the set of possible output values ( y values). e.g. State the domain and range. A: {(1, 2), (0, 3), ( 4, 8)} Domain = {0, 1, 4} Range = {2, 3, 8}
B:
4
2
Looking at the graph we can see that y does not go below 0. Thus, Domain = R Range = { y | y u 0, y R}
C:
y
!
x 5
What value of x will make x ± 5 = 0? x = 5 The radicand cannot be less than zero, so Domain = { x | x u 5, x R } Range = { y | y
u
0, y R }
MCR3U
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Exam Review
A function is a special type of relation in which every element of the domain corresponds to exactly one element of the range. y ! x 7 and y ! x 2 15 are examples of functions. y ! s x is not a function because for every value of x there are two values of y. The vertical line test is used to determine if a graph of a relation is a function. If a vertical line can be passed along the entire length of the graph and it never touches more than one point at a time, then the relation is a function. e.g. A: B: The line passes through This passes more than one point, so this the vertical relation fails the vertical line test, so it is a line test. It is not a function. function. 4
4
2
2
Inverse Functions The inverse, f 1 , of a relation, f , maps each output of the original relation back onto the corresponding input value. The domain of the inverse is the range of the function, and the range of the inverse is the domain of the function. That is, if ( a, b) f , then (b, a ) f 1 . The graph of y ! f 1 ( x ) is the reflection of the graph y ! f ( x ) in the line y ! x .
e.g. Given f ( x) !
3x 1
Evaluate f ( 3) . f ( 3) ! f ( 3) ! f ( 3) !
3( 3) 1 5
9 1
5
. Evaluate 3 f ( 2) 1
« 3(2) 1» 1 - 5 ¼½ « 6 1» ! 3¬ 1 ¼ 5 ½
3 f (2) 1 ! 3¬
Replace all x¶s with ±3. Evaluate.
5 10
¨ 5 ¸ ª 5 º
! 3(1) 1 3 f (2) 1 ! 4
Determine f 1 ( x) .
x
! !
3 x 1 5 3 y 1
5 5 x ! 3 y 1
Rewrite f ( x ) as y ! Interchange x and y. Solve for y.
Evaluate f 1 ( 2) 3x 1
f 1 ( x ) !
5
f 1 ( 2) !
!
! 5 x 1 5 x 1 y ! 3 y
f 1 ( 2) !
3
@ f 1 ( x) !
You are not solving for f ( 2) .
! 3© ¹ 1
5 f ( 3) ! 2
y
You want to find the value of the expression 3 f ( 2) 1 .
5x 1 3
5 x 1 3 5( 2) 1 3 10 1 3 11 3
If you have not already determined
f 1 ( x ) do so. 1
Using f ( x ) , replace all x¶s with 2. Evaluate.
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Exam Review
e.g. Sketch the graph of the inverse of the given function y ! f ( x ) . 4
4
4
y ! f 1 ( x ) 2
2
2
Reflect the graph in the line y = x.
Draw the line y = x.
y ! f ( x)
-2
-2
-2
-4
-4
-4
The inverse of a function is not necessarily going to be a function. If you would like the inverse to also be a function, you may have to restrict the domain or range of the original function. For the example above, the inverse will only be a function if we restrict the domain to { x | x u 0, x R } or { x | x e 0, x R } . Transformations of Functions
To graph y
! af [k ( x p )] q
from the graph y
! f ( x)
consider:
a ± determines the vertical stretch. The graph y ! f ( x ) is stretched vertically by a factor of a. If a < 0 then the graph is reflected in the x-axis, as well. 1 k ± determines the horizontal stretch. The graph y ! f ( x ) is stretched horizontally by a factor of . If k k < 0 then the graph is also reflected in the y-axis. p ± determines the horizontal translation. If p > 0 the graph shifts to the right by p units. If p < 0 then the graph shifts left by p units.
q ± determines the vertical translation. If q > 0 the graph shifts up by q units. If q < 0 then the graph shifts down by q units. 4
When applying transformations to a graph the stretches and reflections should be applied before any translations. e.g. The graph of y ! f ( x ) is transformed into y ! 3 f ( 2 x 4) . Describe the transformations. First, factor inside the brackets to determine the values of k and p. y ! 3 f 2 x 2
e.g. Given the graph of y ! f ( x ) sketch the graph of y ! 2 f x 2 1
2
y
4
4
Stretch vertically b a factor of 2.
2
2
2
! f ( x)
4
Reflect in y-axis.
2
4
a ! 3, k ! 2, p ! 2 There is a vertical stretch of 3. 1 A horizontal stretch of . 2 The graph will be shifted 2 units to the right.
2
4
4 4
2
2
Shift to the right by 2.
Shift up by 1. 2
2
This is the graph of y
4
4
! 2 f x 2 1
MCR3U
Exam Review
9
4
a>0
Quadratic Functions
2
The graph of the quadratic function, f ( x ) ! a x 2 b x c , is a parabola. When a " 0 the parabola opens up. When a 0 the parabola opens down.
minimum -5
5
maximum -2
Vertex Form: f ( x) ! a( x h)
2
a<0
k
The vertex is ( h, k ) . The maximum or minimum value is k . The axis of symmetry is y = h. 2 Standard Form: f ( x) ! ax Factored Form: f ( x) ! a( x p)( x q) The zeroes are x ! p and x ! q . The y-intercept is c.
-4
b x c
Complete the square to change the standard form to vertex form. e.g. f ( x ) ! 2 x 2 12 x 7
f ( x ) ! 2 x f ( x ) ! 2 x
Factor the coefficient of x2 form the terms with x2 and x.
f ( x ) ! 2 x 2 6 x 7
Divide the coefficient of x by 2. Square this number. Add and subtract it.
2
6 x 32 32 7
2
6 x 3 2( 3 ) 7 2
f ( x ) ! 2( x 3) 2( 9) 7 2
f ( x ) ! 2( x 3) 25 2
Bring the last term inside the bracket outside the brackets.
2
Factor the perfect square trinomial inside the brackets.
Simplify.
Maximum and Minimum Values Vertex
form, maximum/minimum value is k .
Factored form: e.g. Determine the maximum or minimum value of f ( x) ! ( x 1)( x 7 ) . The zeroes of f ( x ) are equidistant from the axis of symmetry. The zeroes are x ! 1 and x ! 7 . 1 7 The axis of symmetry is x = 4. The axis of symmetry passes through the vertex. x ! 2 The x-coordinate of the vertex is 4. To find the y-coordinate of the vertex, x ! 4 evaluate f ( 4) . f (4) ! ( 4 1)(4 7) f (4) ! 3( 3) f (4) ! 9
The vertex is ( 4, 9 ) . Because a is positive ( a The minimum value is ±9.
! 1 ), the graph opens up.
Standard form: e.g. Determine the maximum or minimum value of f ( x) ! 2 x 2 10 x 10 without completing the square. g ( x) ! 2 x 2 10 x is a vertical translation of f ( x) ! 2 x 2 10 x 10 with y-intercept of 0. g ( x) ! 2 x( x 5) x ! 0, 5 are the zeroes. 2 Factor g ( x ) ! 2 x 10 x to determine 05 zeroes, then find the a xis of symmetry. Both ! 2.5 x ! x ! 2.5 is the x-coordinate of vertex. 2 f ( x ) and g ( x) will have the same x f ( 2.5) ! 2( 2.5)2 10( 2.5) 10 coordinates for the vertex. To find the ycoordinate for f(x) simply evaluate f(x) using The y-coordinate of vertex f ( 2.5) ! 22.5 the same x-coordinate. is 22.5. It is a maximum because the graph opens down.
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Exam Review
Zeroes To determine the number of zeroes of a quadratic function consider the form of the function. Vertex form: If a and k have opposite signs there are 2 zeroes (2 roots). If a and k have the same sign there are no zeroes (0 roots). If k = 0 there is one zero (1 root). Factored form: f ( x) ! a ( x p)( x q) 2 zeroes. The zeroes are x ! p and x ! q .
! a ( x p) 2 1 zero. The zero is x = p. Standard form: Check discriminant. D ! b 2 4ac If D 0 there are no zeroes. If D ! 0 there is 1 zero. f ( x )
If D " 0 there are 2 zeroes. To determine the zeroes of from the standard form use the quadratic formula . For , ax
2
b x c ! 0
use x !
bs
b
2
4ac
2a
to solve for x.
Reciprocal functions
The reciprocal function of a function, f , is defined as
1 f
. To help you graph y !
1 f ( x)
, you should
use the following: The vertical asymptotes of y ! As f ( x) increases, 1
1 f ( x)
1
will occur where f ( x) ! 0
f ( x)
decreases. As f ( x ) decreases,
f ( x)
increases.
1
0. f ( x ) f ( x) 1 The graph of y ! always passes through the points where f ( x) ! 1 or f ( x ) ! 1 . f ( x) You may find it helpful to sketch the graph of y ! f ( x ) first, before you graph the reciprocal.
For f ( x) " 0 ,
" 0 . For f ( x) 0 ,
1
4
e.g. Sketch the graph of y !
1
x 4 x Look at the function f ( x ) ! x 2 4 x . Factor it. f ( x) ! x( x 4) . The zeroes are x = 0, and x = 4. The vertical asymptotes will be at x = 0, and x = 4. 2
2 y ! x 4 x
.
You could sketch the graph of f ( x) ! x 2 4 x to see where the function increases and decreases, where f ( x) ! 1 or ±1. Use the information above to help you sketch the reciprocal.
2
5
-2
-4
4
2
y ! 5
-2
-4
Vertical asymptotes
1 x2 4 x
MCR3U
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Exam Review
Exponential Functions fx = 2x
6
fx = 2x-2+3
In general, the exponential function is defined by the equation, y ! a x
4
6
4
or f ( x ) ! a , a " 0, x R . Transformations apply to exponential functions the same way they do to all other functions. x
2
2
Exponential Growth and Decay Population growth and radioactive decay can be modelled using exponential functions. t
t
Growth: N (t ) ! N 0 ( 2) d
¨ 1 ¸ h Decay: N (t ) ! N 0 © ¹ ª 2 º
N 0 - initial amount
N 0 - initial amount
t ± time elapsed d ± doubling period N (t ) - amount at time t Compound Interest Calculating the future amount: Calculating
! P (1 i ) n
the present amount: P ! A(1 i) n
t ± time elapsed h ± half-life N (t ) - amount at time t
A
± future amount
P ± present (initial) amount
i ± interest rate per conversion period n ± number of conversion periods
Trigonometry
Given a right angle triangle we can use the following ratios Primary Trigonometric Ratios y x y sin U ! cos U ! tan U ! r r x Reciprocal Trigonometric Ratios 1 x 1 r 1 r cscU ! ! cot U ! ! sec U ! ! y sin U y tan U x cos U
r
y
U x
A
Trigonometry of Oblique Triangles Sine Law a b c
!
!
b
sin A sin B sin C Can be used when you know ASA, AAS, SSA
Cosine Law a 2 ! b 2 c 2 2bc cos A Can be used when you know SSS, SAS
C
c
a
B
MCR3U
C
When you know SSA it is considered the ambiguous case. Angle
Conditions
A 90Q
a b sin A a ! b sin A a " b sin A aeb
A " 90
Q
12
Exam Review # of Triangles 0
b
a b sinA
1 2
C A
0
B
a b
1
a"b
A
B
Trigonometric Identities
Pythagorean Identity: sin 2 U cos 2 U ! 1
Quotient Identity: tan U !
sin U cos U
e.g. Prove the identity. sin U 2 cos U 1 ! cos U 2
S ! sin
2
2
Work with each side separately. U 2 cos 2 U 1 Look for the quotient or Pythagorean identities. ! sin 2 U cos 2 U cos 2 U 1 You may need to factor, simplify or split terms up. When you are done, write a concluding statement. ! 1 cos 2 U 1 ! cos 2 U ! RS Since LS=RS then sin 2 U 2 cos 2 U 1 ! cos 2 U is true for all values of U . 2
Periodic Functions A periodic function has a repeat ing pattern. The cycle is the smallest complete repeating pattern. The axis of the curve is a horizontal line that is midway between the maximum and minimum values of the graph. The equation is max value min value y ! . 2 Trigonometric Functions The graphs of y ! sin U , y 1
The period is the length of the cycle. The amplitude is the magnitude of the vertical distance from the axis of the curve to the maximum or minimum value. The equation is max value min value a! 2
! cosU , and y ! tan U are shown below. y ! sin U y ! tan U y ! sin U 5
0.5
50
100
150
200
250
300
350
-0.5
Period = 360Û Amplitude = 1 Zeroes = 0Û, 180Û, 360Û«
50
100
150
200
250
300
-5
-1
y
1
! cosU y
0.5
50
-0.5
-1
100
150
200
250
300
350
! cosU
Period = 360Û Amplitude = 1 Zeroes = 90Û, 2 70Û«
y ! tan U Period = 180Û Zeroes = 0Û, 180Û, 360Û« Vertical asymptotes = 90Û, 270Û«
350
MCR3U
13
Exam Review
Transformations of Trigonometric Functions
Transformations apply to trig functions as they do to any other function. The graphs of y ! a sin k (U b) d and y y ! sin U and y ! cos U respectively.
! a cos k (U b) d
are transformations of the graphs
The value of a determines the vertical stretch, called the amplitude . It also tells whether the curve is reflected in the U -axis. The value of k determines the horizontal stretch. The graph is stretched by a factor of this value to determine the period of the transformation of y or y ! cos k U is
or y ! cos U .
Q
k
. We can use
180
Q
, k > 0. The period of y ! tan k U is , k > 0. k k The value of b determines the horizontal translation, known as the phase shift. The value of d determines the vertical translation. y ! d is the equation of the axis of the curve . e.g. e.g. 1 Q y ! cos 2U 1 y ! sin U 45 2 The period of y
! sin k U
360
! sin U
1
1
gx = cos 2x +1
2
1.5 0.5
gx = 0.5sinx+45
1
0.5
50
50
100
150
200
250
300
350
400
100
150
200
250
300
350
-0.5
-0.5
-1
fx = cos x
fx = sinx -1
400
