This document contains a practice test with multiple choice and written response questions about transformations of graphs of functions. Some key questions ask students to: 1) Identify which statement about transforming a graph is false. 2) Determine if reflecting a function's graph in the y-axis will produce its inverse. 3) Write equations of functions after transformations like translations, stretches, and reflections. 4) Sketch and describe transformations of a graph to satisfy a given equation. 5) Determine the domain and range of a transformed function. 6) Find the inverse of a function and restrict its domain to make the inverse a function.

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P R AC T I C E T E S T, p a g e s 2 6 1 – 2 6 4
1. Multiple Choice The graph of y + 2 = -3f a2(x + 5)b is the image
1
of the graph of y = f(x) after several transformations. Which
statement about how the graph of y = f(x) was transformed is
false?
A. The graph was reflected in the x-axis.
B. The graph was horizontally stretched by a factor of 2.
C. The graph was translated 2 units down.
D. The graph was reflected in the y-axis.
2. Multiple Choice Which statement about a function and its inverse
is not always true?
A. The inverse of a function is a function.
B. The domain of a function is the range of its inverse, and the
range of a function is the domain of its inverse.
C. The graph of the inverse of a function can be sketched by
reflecting the graph of the function in the line y = x.
D. Each point (x, y) on the graph of a function corresponds to the
point (y, x) on the graph of its inverse.
√
3. Write an equation of the function y = x - 3 after each
transformation below.
a) a translation of 2 units right and 5 units down
√
The equation of the image graph has the form y ؊ k ‫( ؍‬x ؊ h) ؊3,
where k ‫ 5؊ ؍‬and h ‫.2 ؍‬
√
So, an equation of the image graph is: y ؉ 5 ‫ ؍‬x ؊ 5
b) a reflection in the x-axis
√
The y-coordinates of points on the graph of y ‫ ؍‬x ؊ 3 change sign.
√
So, an equation of the image graph is: y ‫ ؊ ؍‬x ؊ 3
c) a reflection in the y-axis
√
The x-coordinates of points on the graph of y ‫ ؍‬x ؊ 3 change sign.
√
So, an equation of the image graph is: y ‫؊ ؍‬x ؊ 3
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d) a vertical stretch by a factor of 2 and a horizontal compression by
1
a factor of 3
√
The equation of the image graph has the form y ‫ ؍‬a bx ؊ 3, where
a ‫ 2 ؍‬and b ‫.3 ؍‬
√
So, an equation of the image graph is: y ‫3 2 ؍‬x ؊ 3
4. Here is the graph of y = f(x). y y ϭ f(x)
On the same grid, use 4
transformations to sketch the
2
graph of y - 5 = -3f(2x).
x
Describe the transformations.
Ϫ4 Ϫ2 0 2 4
Compare: y ؊ k ‫ ؍‬af (b(x ؊ h)) to Ϫ2 y Ϫ 5 ϭ Ϫ3f(2x )
y ؊ 5 ‫3؊ ؍‬f(2x)
Ϫ4
k ‫ ,5 ؍‬a ‫ ,3؊ ؍‬b ‫ ,2 ؍‬and h ‫0 ؍‬
A point (x, y) on y ‫ ؍‬f(x) corresponds Ϫ6
to the point a ؉ h, ay ؉ kb on
x
b
y ؊ k ‫ ؍‬af (b(x ؊ h)).
Substitute the values above.
A point on y ؊ 5 ‫3؊ ؍‬f(2x) has coordinates a , ؊3y ؉ 5b .
x
2
Transform some points on the lines.
Point on Point on
y ‫ ؍‬f(x) y ؊ 5 ‫3؊ ؍‬f(2x)
(؊4, 4) (؊2, ؊7)
(0, 0) (0, 5)
(4, 4) (2, ؊7)
Draw 2 lines through the points for the graph of y ؊ 5 ‫3؊ ؍‬f(2x).
1
The graph of y ‫ ؍‬f(x) was compressed horizontally by a factor of ,
2
stretched vertically by a factor of 3, reflected in the x-axis, then
translated 5 units up.
©P DO NOT COPY. Chapter 3: Transforming Graphs of Functions—Practice Test—Solutions 53

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√
5. Here is the graph of y = x.
√1
a) Use this graph to sketch a graph of y - 1 = -2 2(x - 4).
y √
yϭ x
2
x
0 2 4 6 8 10
Ϫ2
Ί1
y Ϫ 1 ϭ Ϫ2 2 (x Ϫ 4)
Compare: y ؊ k ‫ ؍‬af (b(x ؊ h)) to
√1
y ؊ 1 ‫2؊ ؍‬ (x ؊ 4)
2
1
k ‫ ,1 ؍‬a ‫ ,2؊ ؍‬b ‫ , ؍‬and h ‫4 ؍‬
2
A point (x, y) on y ‫ ؍‬f(x) corresponds to the point a ؉ h, ay ؉ kb
x
b
on y ؊ k ‫ ؍‬af (b(x ؊ h)).
Substitute the values above.
√1
A point on y ؊ 1 ‫2؊ ؍‬ (x ؊ 4) has coordinates: (2x ؉ 4, ؊2y ؉ 1)
2
√
Transform some points on y ‫ ؍‬x.
Point on
√
Point on √1
y‫ ؍‬x y ؊ 1 ‫2؊ ؍‬ (x ؊ 4)
2
(0, 0) (4, 1)
(1, 1) (6, ؊1)
(4, 2) (12, ؊3)
√1
Join the points with a smooth curve for the graph of
y ؊ 1 ‫2؊ ؍‬ (x ؊ 4).
2
b) Write the domain and range of the function in part a.
From the graph, the domain is x » 4; and the range is y ◊ 1.
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6. a) Determine an equation of the inverse of y = (x - 4)2 + 5.
Interchange x and y in the equation.
x ‫( ؍‬y ؊ 4)2 ؉ 5 Solve for y.
(y ؊ 4)2 ‫ ؍‬x ؊ 5
√
y؊4‫ —؍‬x؊5
√
y‫ —؍‬x؊5؉4
b) Sketch the graph of the inverse. Is the inverse a function? Explain.
y
y ϭ (x Ϫ 4)2 ϩ5
8
√
y ϭ x Ϫ5ϩ4
6
4
2 √
y ϭϪ x Ϫ5ϩ4
x
0 2 4 6 8
Graph y ‫( ؍‬x ؊ 4)2 ؉ 5.
This is the graph of y ‫ ؍‬x2 after a translation of 4 units right and
5 units up.
Interchange the coordinates of points on this graph, then plot the new
√
points to get the graph of y ‫ — ؍‬x ؊ 5 ؉ 4.
No, the inverse is not a function because its graph does not pass the
vertical line test.
c) If your answer to part b is yes, explain how you know. If your
answer to part b is no, determine a possible restriction on the
domain of y = (x - 4)2 + 5 so its inverse is a function, then
write the equation of the inverse function.
The domain can be restricted by considering each part of the graph of
y ‫( ؍‬x ؊ 4)2 ؉ 5 to the right and left of the vertex.
So, one restriction is x » 4 and the equation of the inverse function is:
√
y‫ ؍‬x؊5؉4
Another restriction is x ◊ 4 and the equation of the inverse function
√
is: y ‫ ؊ ؍‬x ؊ 5 ؉ 4
©P DO NOT COPY. Chapter 3: Transforming Graphs of Functions—Practice Test—Solutions 55