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R E V I E W, p a g e s 2 5 5 – 2 6 0
3.1
y 2 f(x)
y
1. Here is the graph of
y ⫽ f(x). Sketch the graph
of each function below.
Write the domain and
range of each
translation image.
y f(x )
x
4
2
0
4
2
y 3 f(x 4)
4
y f(x 3)
6
a) y ⫺ 2 ⫽ f(x)
Each point on the graph of
y ⴝ f(x) is translated 2 units up.
From the graph, the domain is:
x ç ⺢ ; the range is: y » ⴚ1
b) y ⫽ f(x ⫺ 3)
Each point on the graph of
y ⴝ f(x) is translated 3 units right.
From the graph, the domain is:
x ç ⺢ ; the range is: y » ⴚ3
c) y ⫹ 3 ⫽ f(x ⫹ 4)
Each point on the graph of y ⴝ f(x) is translated 4 units left and 3 units
down. From the graph, the domain is: x ç ⺢; the range is: y » ⴚ6
2. The graph of the function y = x3 - 2 is translated 3 units left and
4 units up. Write the equation of the translation image.
The equation of the translation image has the form y ⴚ k ⴝ (x ⴚ h)3 ⴚ 2,
with h ⴝ ⴚ3 and k ⴝ 4
So, the equation is: y ⴚ 4 ⴝ (x ⴙ 3)3 ⴚ 2, which simplifies to y ⴝ (x ⴙ 3)3 ⴙ 2
3.2
3. Here is the graph of y ⫽ g(x). On the same grid, sketch the graph of
y
each function below. Write the domain and range of each reflection
image.
a) y ⫽ ⫺g(x)
Reflect points on y ⴝ g(x) in
the x-axis:
(ⴚ2, ⴚ10) becomes (ⴚ2, 10)
(0, ⴚ2) becomes (0, 2)
(2, 6) becomes (2, ⴚ6)
Draw a smooth curve through
the points for the graph of
y ⴝ ⴚg(x).
The domain is: x ç ⺢
The range is: y ç ⺢
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b) y ⫽ g(⫺x)
Reflect points on y ⴝ g(x) in the
y-axis:
(ⴚ2, ⴚ10) becomes (2, ⴚ10)
(0, ⴚ2) stays as (0, ⴚ2)
(2, 6) becomes (ⴚ2, 6)
Draw a smooth curve through the
points for the graph of y ⴝ g(ⴚx).
The domain is: x ç ⺢
The range is: y ç ⺢
Chapter 3: Transforming Graphs of Functions—Review—Solutions
8
y g(x)
4
y g(x)
x
4
2
0
4
2
4
y g(x)
8
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4. The graph of f(x) = (x - 2)3 - 4 was reflected in the x-axis and its
y
image is shown. What is an equation of the image?
4
When the graph of y ⴝ f(x) is reflected in the x-axis, the equation of the
image is y ⴝ ⴚf(x). So, an equation of the image is:
f(x) ⴝ ⴚ[(x ⴚ 2)3 ⴚ 4]
f(x) ⴝ ⴚ(x ⴚ 2)3 ⴙ 4
2
x
0
2
3.3
5. Here is the graph of y ⫽ h(x). Sketch the graph of each function
6
y
below. Write the domain and range of each transformation image.
4
1
a) y ⫽ h(3x)
b) y ⫽ 2 h(x)
The graph of y ⴝ h(x) is
compressed horizontally by a
1
3
factor of .
For each point at the ends of
the line segments on y ⴝ h(x),
divide the x-coordinate by 3,
plot the new points then join
them for the graph of y ⴝ h(3x).
The domain is: ⴚ1 ◊ x ◊ 2
The range is: ⴚ2 ◊ y ◊ 3
The graph of y ⴝ h(x) is compressed
1
2
vertically by a factor of .
For each point at the ends of the
line segments on y ⴝ h(x), divide
the y-coordinate by 2, plot the
new points then join them for the
y h(x)
2
x
y 1 h(x)
2
0
2
y h(3x)
4
2
4
6
y 2h(3x)
1
2
graph of y ⴝ h(x).
The domain is: ⴚ3 ◊ x ◊ 6
The range is: ⴚ1 ◊ y ◊ 1.5
c) y ⫽ 2h(⫺3x)
The graph of y ⴝ h(x) is stretched vertically by a factor of 2, compressed
1
3
horizontally by a factor of , then reflected in the y-axis.
Use: (x, y) on y ⴝ h(x) corresponds to aⴚ , 2yb on y ⴝ 2h(ⴚ3x)
x
3
Point on y ⴝ h(x)
Point on y ⴝ 2h(ⴚ3x)
(ⴚ3, ⴚ2)
(1, ⴚ4)
(0, 3)
(0, 6)
(3, 2)
(ⴚ1, 4)
(6, 2)
(ⴚ2, 4)
Plot the points, then join them.
The domain is: ⴚ2 ◊ x ◊ 1
The range is: ⴚ4 ◊ y ◊ 6
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Chapter 3: Transforming Graphs of Functions—Review—Solutions
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6. The graph of y = g(x) is the image of the
graph of y = f(x) after a transformation.
Corresponding points are labelled.
Write an equation of the image graph
in terms of the function f.
y
8
A
y f(x)
4
A (1, 1)
8 4
B (1, 1)
The graph has not been translated, so an
equation of the image graph has the form:
y ⴝ af(bx)
4
4
x
8
B
y g(x)
8
A point (x, y) on y ⴝ f(x) corresponds to the point a , ayb on y ⴝ af(bx).
The image of A(1, 1) is a , 1ab , which is Aⴕ(ⴚ8, 4).
x
b
1
b
1
8
Compare coordinates: b ⴝ ⴚ and a ⴝ 4
An equation for y ⴝ g(x) is: y ⴝ 4f aⴚ xb
1
8
3.4
7. Here is the graph of y = f(x).
On the same grid, sketch the
graph of y - 4 = 3f 12(x - 5)2.
Write the domain and range of
the transformation image.
Compare: y ⴚ k ⴝ af (b(x ⴚ h))
to y ⴚ 4 ⴝ 3f(2(x ⴚ 5))
k ⴝ 4, a ⴝ 3, b ⴝ 2, and h ⴝ 5
A point (x, y) on the graph of
y ⴝ f(x) corresponds to the point
y
0
2
6
2
4
y f (x)
8
x
10
4
6
y 4 3f(2 (x 5))
8
x
a ⴙ 5, 3y ⴙ 4b on the graph of y ⴚ 4 ⴝ 3f(2(x ⴚ 5)).
2
Point on y ⴝ f(x)
Point on y ⴚ 4 ⴝ 3f(2(x ⴚ 5))
(0, ⴚ4)
(5, ⴚ8)
(1, ⴚ3)
(5.5, ⴚ5)
(4, ⴚ2)
(7, ⴚ2)
(9, ⴚ1)
(9.5, 1)
Plot the points, then join them.
From the graph of y ⴚ 4 ⴝ 3f(2(x ⴚ 5)), the domain is: x
and the range is: y » ⴚ 8
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8. The graph of y = g(x) is the image
4
of the graph of y = f(x) after a
combination of transformations.
Corresponding points are labelled.
Write an equation of the image graph
in terms of the function f.
y
2
y f(x)
0 A
2
B
2
2
4
x
4
B
A
y g(x )
The equation of the image graph can be
written as: y ⴚ k ⴝ af (b(x ⴚ h))
The horizontal distance between A and B is 2.
The vertical distance between A and B is 4.
The horizontal distance between Aⴕ and Bⴕ is 2.
The vertical distance between Aⴕ and Bⴕ is 2.
1
The graph of y ⴝ f(x) has been compressed vertically by a factor of and
2
1
2
reflected in the x-axis, so a ⴝ ⴚ .
There is no horizontal stretch or compression, so b ⴝ 1.
Since B(4, 0) lies on the x-axis, it will not move after the vertical
compression and reflection.
Determine the translation that would move B(4, 0) to Bⴕ(2, ⴚ3).
A translation of 2 units left and 3 units down is required, so h ⴝ ⴚ2 and
k ⴝ ⴚ3
1
An equation for the image graph is: y ⴙ 3 ⴝ ⴚ f(x ⴙ 2)
2
1
9. The point (2, 2) lies on the graph of y = 4x3. After a combination of
transformations, the equation of the image graph is
y + 6 = 5 a4 12(x - 3)23b. What are the coordinates of the point
that is the image of (2, 2)?
1
Compare: y ⴙ 6 ⴝ 5 a (2(x ⴚ 3)) 3b with y ⴚ k ⴝ af (b(x ⴚ h)) :
1
4
k ⴝ ⴚ6, a ⴝ 5, b ⴝ 2, and h ⴝ 3
1
4
A point (x, y) on the graph of y ⴝ x3 corresponds to the point
x
1
3
a ⴙ 3, 5y ⴚ 6b on the graph of y ⴙ 6 ⴝ 5 a (2(x ⴚ 3)) b .
2
4
Substitute x ⴝ 2 and y ⴝ 2 in the expression for the coordinates above.
2
a ⴙ 3, 5(2) ⴚ 6b ⴝ (4, 4)
2
The image of (2, 2) has coordinates (4, 4).
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3.5
10. Determine an equation of the inverse of each function, then sketch
graphs of the function and its inverse.
2
a) y = - 5 x + 3
b) y = (x - 3)2 + 7
y 2x 3
5
y (x 3)2 7
y
y
4
16
2
12
x
4
2
0
2
4
2
8
y 5x 15
2
4
4
0
Solve for y.
5x ⴝ ⴚ2y ⴙ 15
2y ⴝ ⴚ5x ⴙ 15
(y ⴚ 3)2 ⴝ x ⴚ 7
√
yⴚ3ⴝ— xⴚ7
√
yⴝ— xⴚ7ⴙ3
ⴚ5x ⴙ 15
2
2
5
The graph of y ⴝ ⴚ x ⴙ 3 is a line
2
5
with y-intercept 3 and slope ⴚ .
Reflect points on the graph of
2
5
y ⴝ ⴚ x ⴙ 3 in the line y ⴝ x.
Join the points for the graph
of y ⴝ
8
Write: x ⴝ (y ⴚ 3)2 ⴙ 7
Solve for y.
2
5
Write: x ⴝ ⴚ y ⴙ 3
yⴝ
4
√
y x 73
x
12 16
ⴚ 5x ⴙ 15
.
2
The graph of y ⴝ (x ⴚ 3)2 ⴙ 7
is the image of the graph of
y ⴝ x2 after a translation of 3
units right and 7 units up.
Reflect points on the graph
of y ⴝ (x ⴚ 3)2 ⴙ 7 in the
line y ⴝ x.
Join the points for the graph
√
of y ⴝ — x ⴚ 7 ⴙ 3.
11. Restrict the domain of the function y = f(x) so its inverse is a
function.
y
x f(y)
4
y f(x)
2
x
0
2
4
Sample response: Sketch the graph of the inverse by reflecting points in
the line y ⴝ x.
The inverse is a function if the domain of y ⴝ f(x) is restricted to x ◊ 3 or
x » 3.
50
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12. A graph was reflected in the line y = x. Its reflection image
y = g(x) is shown. Determine an equation of the original graph in
terms of x and y.
a)
b)
y
y g(x )
4
4
y
y g(x )
2
x
x
2
0
2
2
4
2
0
4
4
y 4x 5
4
y x 2 3
Use the line y ⴝ x to sketch
the graph of the inverse.
This line has y-intercept 5,
and slope ⴚ4, so its
equation is: y ⴝ ⴚ4x ⴙ 5
©P DO NOT COPY.
Use the line y ⴝ x to sketch the
graph of the inverse. This curve is a
parabola that has vertex (0, 3), and
is congruent to y ⴝ ⴚx2. So, its
equation is: y ⴝ ⴚx2 ⴙ 3
Chapter 3: Transforming Graphs of Functions—Review—Solutions
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