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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
1. Multiple Choice The graph of y + 2 = -3f a2(x + 5)b is the image
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 ⴚ k ⴝ (x ⴚ h) ⴚ3,
where k ⴝ ⴚ5 and h ⴝ 2.
√
So, an equation of the image graph is: y ⴙ 5 ⴝ x ⴚ 5
b) a reflection in the x-axis
√
The y-coordinates of points on the graph of y ⴝ x ⴚ 3 change sign.
√
So, an equation of the image graph is: y ⴝ ⴚ x ⴚ 3
c) a reflection in the y-axis
√
The x-coordinates of points on the graph of y ⴝ x ⴚ 3 change sign.
√
So, an equation of the image graph is: y ⴝ ⴚx ⴚ 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 ⴝ a bx ⴚ 3, where
a ⴝ 2 and b ⴝ 3.
√
So, an equation of the image graph is: y ⴝ 2 3x ⴚ 3
4. Here is the graph of y = f(x).
y
On the same grid, use
transformations to sketch the
graph of y - 5 = -3f(2x).
Describe the transformations.
y f(x)
4
2
x
4
2
Compare: y ⴚ k ⴝ af (b(x ⴚ h)) to
y ⴚ 5 ⴝ ⴚ3f(2x)
k ⴝ 5, a ⴝ ⴚ3, b ⴝ 2, and h ⴝ 0
A point (x, y) on y ⴝ f(x) corresponds
0
2
2
4
y 5 3f(2x )
4
6
x
to the point a ⴙ h, ay ⴙ kb on
b
y ⴚ k ⴝ af (b(x ⴚ h)).
Substitute the values above.
x
A point on y ⴚ 5 ⴝ ⴚ3f(2x) has coordinates a , ⴚ3y ⴙ 5b .
2
Transform some points on the lines.
Point on
y ⴝ f(x)
Point on
y ⴚ 5 ⴝ ⴚ3f(2x)
(ⴚ4, 4)
(ⴚ2, ⴚ7)
(0, 0)
(0, 5)
(4, 4)
(2, ⴚ7)
Draw 2 lines through the points for the graph of y ⴚ 5 ⴝ ⴚ3f(2x).
1
2
The graph of y ⴝ f(x) was compressed horizontally by a factor of ,
stretched vertically by a factor of 3, reflected in the x-axis, then
translated 5 units up.
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√
x.
5. Here is the graph of y =
√1
a) Use this graph to sketch a graph of y - 1 = -2 2(x - 4).
y
y
2
√
x
x
0
2
2
4
6
8
10
1
y 1 2 2 (x 4)
Compare: y ⴚ k ⴝ af (b(x ⴚ h)) to
√1
y ⴚ 1 ⴝ ⴚ2
2
(x ⴚ 4)
1
2
k ⴝ 1, a ⴝ ⴚ2, b ⴝ , and h ⴝ 4
A point (x, y) on y ⴝ f(x) corresponds to the point a ⴙ h, ay ⴙ kb
x
b
on y ⴚ k ⴝ af (b(x ⴚ h)).
√1
Substitute the values above.
A point on y ⴚ 1 ⴝ ⴚ2
(x ⴚ 4) has coordinates: (2x ⴙ 4, ⴚ2y ⴙ 1)
√
Transform some points on y ⴝ x.
Point on
√
yⴝ x
Point on
(0, 0)
(4, 1)
(1, 1)
(6, ⴚ1)
(4, 2)
(12, ⴚ3)
2
√1
y ⴚ 1 ⴝ ⴚ2
√1
2
(x ⴚ 4)
Join the points with a smooth curve for the graph of
y ⴚ 1 ⴝ ⴚ2
2
(x ⴚ 4).
b) Write the domain and range of the function in part a.
From the graph, the domain is x
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» 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 ⴝ (y ⴚ 4)2 ⴙ 5
Solve for y.
(y ⴚ 4)2 ⴝ x ⴚ 5
√
yⴚ4ⴝ— xⴚ5
√
yⴝ— xⴚ5ⴙ4
b) Sketch the graph of the inverse. Is the inverse a function? Explain.
y
y (x 4)2 5
8
y 6
√
x 54
4
2
√
y x 54
x
0
2
4
6
8
Graph y ⴝ (x ⴚ 4)2 ⴙ 5.
This is the graph of y ⴝ 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 ⴝ — x ⴚ 5 ⴙ 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 ⴝ (x ⴚ 4)2 ⴙ 5 to the right and left of the vertex.
So, one restriction is x » 4 and the equation of the inverse function is:
√
yⴝ xⴚ5ⴙ4
Another restriction is x ◊ 4 and the equation of the inverse function
√
is: y ⴝ ⴚ x ⴚ 5 ⴙ 4
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