9-3 Transformations of Quadratic Functions
Describe how the graph of each function is
2
related to the graph of f (x) = x .
2
1. g(x) = x − 11
2
The graph of f (x) = x + c represents a translation up
or down of the parent graph. Since c = –11, the
translation is down. So, the graph is shifted down 11
units from the parent function.
x
SOLUTION: 2
SOLUTION: 2. h(x) =
2
3. h(x) = −x + 8
The function can be written f (x) = ax + c, where a
= –1 and c = 8. Since a is –1, the graph is reflected
across the x-axis. Since c = 8 the graph is 2
translated up 8 units. So, the graph of y = –x +8 is
2
the graph of y = x reflected across the x-axis and
translated up 8 units.
2
2
4. g(x) = x + 6
SOLUTION: SOLUTION: 2
2
The graph of f (x) = ax stretches or compresses the
parent graph vertically. Since a =
=
2
, the graph of y
The graph of f (x) = x + c represents a translation up
or down of the parent graph. Since c = 6, the
translation is up. So, the graph is shifted up 6 units
from the parent function.
2
x is the graph of y = x vertically compressed.
5. g(x) = −4x
SOLUTION: 2
3. h(x) = −x + 8
2
The coefficient of the x -term is negative, so the
graph is reflected across the x-axis. The graph of f
SOLUTION: 2
2
The function can be written f (x) = ax + c, where a
= –1 and c = 8. Since a is –1, the graph is reflected
across the x-axis. Since c = 8 the graph is 2
translated up 8 units. So, the graph of y = –x +8 is
2
the graph of y = x reflected across the x-axis and
translated up 8 units.
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2
(x) = ax stretches or compresses the parent graph
vertically. Since a = 4, the graph is vertically
2
stretched. So, the graph of y = −4x is the graph of y
2
= x reflected across the x-axis and vertically
stretched.
Page 1
9-3 Transformations of Quadratic Functions
5. g(x) = −4x
2
7. MULTIPLE CHOICE Which is an equation for
the function shown in the graph?
SOLUTION: 2
The coefficient of the x -term is negative, so the
graph is reflected across the x-axis. The graph of f
2
(x) = ax stretches or compresses the parent graph
vertically. Since a = 4, the graph is vertically
2
stretched. So, the graph of y = −4x is the graph of y
2
= x reflected across the x-axis and vertically
stretched.
A g(x) =
2
x +2
B g(x) = −5x2 − 2
C g(x) =
2
x −2
D g(x) = − x2 − 2
SOLUTION: The function is a parabola which can be written f (x)
2
2
6. h(x) = −x − 2
SOLUTION: 2
The function can be written f (x) = ax + c, where a
= –1 and c = –2. Since a is –1, the graph is reflected
across the x-axis. Since c = –2, the graph is
2
translated down 2 units. So, the graph of y = −x − 2
2
is the graph of y = x reflected across the x-axis and
translated down 2 units.
= ax + c. Since the graph opens upward, the leading
coefficient must be positive, so eliminate choices B
and D. The parabola is translated down 2 units, so c
= –2. This eliminates choice A. The only equation for
the function shown in the graph is g(x) =
2
x − 2.
So, the correct choice is C.
Describe how the graph of each function is
related to the graph of f (x) = x 2.
8. g(x) = −10 + x
2
SOLUTION: 2
The graph of f (x) = x + c represents a translation up
or down of the parent graph. Since c = –10, the
translation is down. So, the graph is shifted down 10
units from the parent function.
7. MULTIPLE CHOICE Which is an equation for
the function shown in the graph?
9. h(x) = −7 − x
2
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A g(x) =
x +2
2
B g(x) = −5x − 2
2
SOLUTION: Page 2
2
The function can be written f (x) = ax + c, where a
= –1 and c = –7. Since a is –1, the graph is reflected
9-3 Transformations of Quadratic Functions
9. h(x) = −7 − x
2
11. h(x) = 6 +
x
2
SOLUTION: 2
The function can be written f (x) = ax + c, where a
= –1 and c = –7. Since a is –1, the graph is reflected
across the x-axis. Since c = –7 the graph is 2
translated down 7 units. So, the graph of y = −7 − x
2
is the graph of y = x reflected across the x-axis and
translated down 7 units.
SOLUTION: 2
The function can be written f (x) = ax + c, where a
=
and c = 6. Since 0 < a < 1 and c is positive, the
graph of y = 6 +
2
2
x is the graph of y = x vertically
compressed and shifted up 6 units.
2
10. g(x) = 2x + 8
12. g(x) = −5 −
SOLUTION: x
2
2
The function can be written f (x) = ax + c, where a
= 2 and c = 8. Since a > 1 and c is positive, the graph
2
2
of y =2x + 8 is the graph of y = x vertically
stretched and shifted up 8 units.
SOLUTION: 2
The function can be written f (x) = ax + c, where a
=
and c = –5. Since a is negative, the graph is
reflected across the x-axis Since |a| >1, the graph is
vertically stretched. Since c < 0, the graph is shifted
down 5 units. So, the graph of y = −5 −
2
x is the
2
graph of y = x reflected across the x-axis, vertically
stretched, and shifted down 5 units.
11. h(x) = 6 +
x
2
SOLUTION: 2
The function can be written f (x) = ax + c, where a
=
and c = 6. Since 0 < a < 1 and c is positive, the
graph of y = 6 +
2
2
13. h(x) = 3 +
x
2
x is the graph of y = x vertically
compressed and shifted up 6 units.
SOLUTION: 2
The function can be written f (x) = ax + c, where a
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=
and c = 3. Since a > 1 and c is positive, the
graph of y = 3 +
Page 3
2
2
x is the graph of y = x vertically
9-3 Transformations of Quadratic Functions
13. h(x) = 3 +
x
2
15. h(x) = 1.35x + 2.6
2
SOLUTION: SOLUTION: 2
2
The function can be written f (x) = ax + c, where a
=
and c = 3. Since a > 1 and c is positive, the
graph of y = 3 +
2
The function can be written f (x) = ax + c, where a
= 1.35 and c = 2.6. Since a > 1 and c is positive, the
2
2
graph of y = 1.35x + 2.6 is the graph of y = x
vertically stretched and shifted up 2.6 units.
2
x is the graph of y = x vertically
stretched and shifted up 3 units.
16. g(x) =
2
x +
SOLUTION: 2
14. g(x) = 0.25x − 1.1
2
The function can be written f (x) = ax + c, where a
SOLUTION: 2
The function can be written f (x) = ax + c, where a
= 0.25 and c = –1.1. Since 0 < a < 1 and c is
2
negative, the graph of y = 0.25x − 1.1 is the graph
2
of y = x vertically compressed and shifted down 1.1
units.
and c =
. Since 0 < a < 1 and c is positive,
2
the graph of y = x +
is the graph of y = x
vertically compressed and shifted up
2
unit.
2
17. h(x) = 1.01x − 6.5
2
15. h(x) = 1.35x + 2.6
SOLUTION: SOLUTION: 2
The function can be written f (x) = ax + c, where a
= 1.35 and c = 2.6. Since a > 1 and c is positive, the
2
2
graph of y = 1.35x + 2.6 is the graph of y = x
vertically stretched and shifted up 2.6 units.
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=
2
The function can be written f (x) = ax + c, where a
= 1.01 and c = –6.5. Since a > 1 and c is negative,
2
2
the graph of y = 1.01x − 6.5 is the graph of y = x
vertically stretched and shifted down 6.5 units.
Page 4
9-3 Transformations of Quadratic Functions
D
2
17. h(x) = 1.01x − 6.5
SOLUTION: 2
The function can be written f (x) = ax + c, where a
= 1.01 and c = –6.5. Since a > 1 and c is negative,
2
2
the graph of y = 1.01x − 6.5 is the graph of y = x
vertically stretched and shifted down 6.5 units.
E
F
Match each equation to its graph.
A
18. y =
B
2
x −4
SOLUTION: 2
The function can be written f (x) = ax + c, where a
=
and c = –4. Since 0 < a < 1 and c is negative, the
C
2
2
x − 4 is the graph of y = x vertically
graph of y =
compressed and shifted down 4 units. So, the correct
choice is C.
2
19. y = − x − 4
SOLUTION: 2
The function can be written f (x) = –ax + c, where a
D
=
and c = –4. The negative sign in front of the a
indicates that the function is a reflection of the graph
2
of y = x about the x-axis. Since 0 < a < 1 and c is
2
negative, the graph of y = – x − 4 is a reflection of
2
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E
the graph of y = x vertically compressed and shifted
down 4 units. So, the correct choice is A.
Page 5
20. y =
2
x +4
graph of y =
2
2
x − 4 is the graph of y = x vertically
compressed and shifted down 4 units. So, the correct
9-3 Transformations of Quadratic Functions
choice is C.
2
the function is a reflection of the graph of y = x
about the x-axis. Since c is positive, the graph of y =
2
2
−x + 2 is a reflection of the graph of y = x shifted
up 2 units. So, the correct choice is D.
2
23. y = 3x + 2
2
19. y = − x − 4
SOLUTION: 2
SOLUTION: 2
The function can be written f (x) = –ax + c, where a
and c = –4. The negative sign in front of the a
=
indicates that the function is a reflection of the graph
The function can be written f (x) = ax + c, where a
= 3 and c = 2. Since a > 1 and c is positive, the graph
2
2
of y = 3x + 2 is the graph of y = x vertically
stretched, and shifted up 2 units. So, the correct
choice is E.
2
of y = x about the x-axis. Since 0 < a < 1 and c is
2
negative, the graph of y = – x − 4 is a reflection of
2
the graph of y = x vertically compressed and shifted
down 4 units. So, the correct choice is A.
20. y =
2
x +4
24. SQUIRRELS A squirrel 12 feet above the ground
2
drops an acorn from a tree. The function h = −16t +
12 models the height of the acorn above the ground
in feet after t seconds. Graph the function and
compare this graph to the graph of its parent
function.
SOLUTION: 2
SOLUTION: 2
The function can be written f (x) = ax + c, where a
The function can be written f (x) = –at + c, where a
= 16 and c = 12. The negative sign in front of the a
indicates that the function is a reflection of the graph
2
=
and c = 4. Since 0 < a < 1 and c is positive, the
graph of y =
2
x + 4 is the graph of y = x
2
vertically compressed and shifted up 4 units. So, the
correct choice is B.
of h = t about the t-axis. Since a > 1 and c is
2
positive, the graph of h = −16t + 12 is a reflection of
2
the graph of h = t vertically stretched, and shifted up
12 units.
2
21. y = −3x − 2
SOLUTION: 2
The function can be written f (x) = –ax + c, where a
= 3 and c = –2. The negative sign in front of the a
indicates that the function is a reflection of the graph
2
of y = x about the x-axis. Since a > 1 and c is
2
negative, the graph of y = −3x − 2 is a reflection of
2
the graph of y = x vertically stretched, and shifted
down 2 units. So, the correct choice is F.
CCSS REGULARITY List the functions in
order from the most stretched vertically to the
least stretched vertically graph.
2
2
25. g(x) = 2x , h(x) =
22. y = −x + 2
SOLUTION: x
2
SOLUTION: 2
The function can be written f (x) = –x + c, where c
2
= 2. The negative sign in front of the x indicates that
2
the function is a reflection of the graph of y = x
about the x-axis. Since c is positive, the graph of y =
2
2
−x + 2 is a reflection of the graph of y = x shifted
up 2 units. So, the correct choice is D.
2
2
The functions can be written as f (x) =ax . For g(x) =
2
2x , a = 2 so the graph is stretched vertically. For h
(x) =
2
x ,a =
so the graph is compressed vertically. Therefore, the functions in order from the
most stretched vertically to the least compressed
graph is g(x), h(x).
23. y = 3x + 2
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Page 6
SOLUTION: 2
2
The function can be written f (x) = ax + c, where a
26. g(x) = −3x , h(x) =
x
2
vertically; and since 6 > 4, h(x) is more stretched
2
than g(x) . For f (x) = 0.3x , a = 0.3 so the graph is
compressed vertically. Therefore, the functions in
order from the most stretched vertically to the least
compressed graph is h(x), g(x), and f (x).
9-3 Transformations of Quadratic Functions
CCSS REGULARITY List the functions in
order from the most stretched vertically to the
least stretched vertically graph.
2
25. g(x) = 2x , h(x) =
x
2
28. g(x) = −x , h(x) =
2
x , f (x) = −4.5x
2
SOLUTION: 2
2
The functions can be written as f (x) =ax . For g(x) =
2
−x , a = 1 so the graph is neither stretched nor
SOLUTION: 2
The functions can be written as f (x) =ax . For g(x) =
compressed. For h(x) =
2
x ,a =
so the graph is 2
2x , a = 2 so the graph is stretched vertically. For h
(x) =
2
x ,a =
so the graph is compressed graph is also stretched vertically. Since 4.5 >
vertically. Therefore, the functions in order from the
most stretched vertically to the least compressed
graph is g(x), h(x).
2
26. g(x) = −3x , h(x) =
x
2
SOLUTION: 2
The functions can be written as f (x) =ax . For g(x) =
2
−3x , a = 3 so the graph is stretched vertically. For h
(x) =
2
x ,a =
so the graph is compressed 2
2
2
SOLUTION: The functions can be written as f (x) =ax . For g(x) =
2
−4x , a = 4 so the graph is stretched vertically. For h
2
(x) = 6x , a = 6 so the graph is also stretched
vertically; and since 6 > 4, h(x) is more stretched
2
than g(x) . For f (x) = 0.3x , a = 0.3 so the graph is
compressed vertically. Therefore, the functions in
order from the most stretched vertically to the least
compressed graph is h(x), g(x), and f (x).
2
is more stretched than h(x). Therefore, the functions
in order from the most stretched vertically to the
least compressed graph is f (x), h(x), and g(x).
29. ROCKS A rock falls from a cliff 300 feet above the
ground. Another rock falls from a cliff 700 feet
above the ground.
a. Write functions that model the height h of each
rock after t seconds.
b. If the rocks fall at the same time, how much
sooner will the first rock reach the ground?
a. The formula h = -16t2 + h 0 can be used to
approximate the number of seconds t it takes for the
rock to reach height h from an initial height of h 0 in
feet.
For the first rock, the initial height is 300 feet.
2
2
28. g(x) = −x , h(x) =
, f (x)
SOLUTION: vertically. Therefore, the functions in order from the
most stretched vertically to the least compressed
graph is g(x), h(x).
27. g(x) = −4x , h(x) = 6x , f (x) = 0.3x
2
stretched vertically. For f (x) = −4.5x , a = 4.5, so the
2
x , f (x) = −4.5x
Therefore, the function for the first rock is h = −16t
+ 300.
For the second rock, the initial height is 700.
Therefore, the function for the second rock is h =
−16t2 + 700.
b. In order to figure out how long it takes each rock
to reach the ground, set the height equal to 0 for both
equations.
2
SOLUTION: 2
The functions can be written as f (x) =ax . For g(x) =
2
−x , a = 1 so the graph is neither stretched nor
compressed. For h(x) =
2
x ,a =
It takes the first rock about 4.3 seconds to reach the
ground.
so the graph is 2
stretched vertically. For f (x) = −4.5x , a = 4.5, so the
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graph is also stretched vertically. Since 4.5 >
, f (x)
is more stretched than h(x). Therefore, the functions
Page 7
It takes the second rock about 6.6 seconds to reach
the ground.
Since 6.6 - 4.3 = 2.3, the first rock will reach the
graph is also stretched vertically. Since 4.5 >
, f (x)
is more stretched than h(x). Therefore, the functions
9-3 Transformations
of Quadratic
Functionsto the
in order from the most
stretched vertically
least compressed graph is f (x), h(x), and g(x).
29. ROCKS A rock falls from a cliff 300 feet above the
ground. Another rock falls from a cliff 700 feet
above the ground.
a. Write functions that model the height h of each
rock after t seconds.
b. If the rocks fall at the same time, how much
sooner will the first rock reach the ground?
SOLUTION: a. The formula h = -16t2 + h 0 can be used to
approximate the number of seconds t it takes for the
rock to reach height h from an initial height of h 0 in
feet.
For the first rock, the initial height is 300 feet.
2
Therefore, the function for the first rock is h = −16t
+ 300.
For the second rock, the initial height is 700.
Therefore, the function for the second rock is h =
−16t2 + 700.
It takes the second rock about 6.6 seconds to reach
the ground.
Since 6.6 - 4.3 = 2.3, the first rock will reach the
ground about 2.3 seconds sooner than the second
rock.
30. SPRINKLERS The path of water from a sprinkler
can be modeled by quadratic functions. The following
functions model paths for three different sprinklers.
2
Sprinkler A: y = −0.35x + 3.5
2
Sprinkler B: y = −0.21x + 1.7
2
Sprinkler C: y = −0.08x + 2.4
a. Which sprinkler will send water the farthest?
Explain.
b. Which sprinkler will send water the highest?
Explain.
c. Which sprinkler will produce the narrowest path?
Explain.
SOLUTION: All three of the quadratic functions are of the form f
2
(x) = ax + c. Use a graphing calculator to compare
the graphs of the three functions.
b. In order to figure out how long it takes each rock
to reach the ground, set the height equal to 0 for both
equations.
It takes the first rock about 4.3 seconds to reach the
ground.
It takes the second rock about 6.6 seconds to reach
the ground.
Since 6.6 - 4.3 = 2.3, the first rock will reach the
ground about 2.3 seconds sooner than the second
rock.
30. SPRINKLERS The path of water from a sprinkler
can be modeled by quadratic functions. The following
functions model paths for three different sprinklers.
2
Sprinkler A: y = −0.35x + 3.5
2
Sprinkler B: y = −0.21x + 1.7
2
a. The value of a in the function for Sprinkler C is
the smallest of the three. Therefore, the parabola for
Sprinkler C is the most compressed vertically, so it
will send water the farthest.
b. The value of c in the function for Sprinkler A is
the largest of the three. Therefore, the parabola for
Sprinkler A is translated up the most, so it will send
water the highest.
c. The value of a in the function for Sprinkler A is
the largest of the three. Therefore, the parabola for
Sprinkler A is the most stretched vertically, so it will
expand the least, and therefore have the narrowest
path.
31. GOLF The path of a drive can be modeled by a
quadratic function where g(x) is the vertical distance
in yards of the ball from the ground and x is the
horizontal distance in yards.
Sprinkler C: y = −0.08x + 2.4
a. Which sprinkler will send water the farthest?
Explain.
b. Which sprinkler will send water the highest?
Explain.
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c. Which sprinkler will produce the narrowest path?
Explain.
Page 8
c. The value of a in the function for Sprinkler A is
the largest of the three. Therefore, the parabola for
Sprinkler A is the most stretched vertically, so it will
9-3 Transformations
of Quadratic
Functions
expand the least, and
therefore have
the narrowest
path.
31. GOLF The path of a drive can be modeled by a
quadratic function where g(x) is the vertical distance
in yards of the ball from the ground and x is the
horizontal distance in yards.
b. If the red tee is 30 yards closer to the hole, then g
(x) will be translated 30 yards to the right to obtain h
(x). So, for h(x) h = 200 + 30 or 230.
2
Therefore, the function is h(x) = 0.0005(x – 230) +
20.
Describe the transformations to obtain the
graph of g(x) from the graph of f (x).
2
32. f (x) = x + 3
2
g(x) = x − 2
SOLUTION: Graph f (x) and g(x) on the same coordinate plane.
a. How can you obtain g(x) from the graph of f (x) =
2
x .
b. A second golfer hits a ball from the red tee, which
is 30 yards closer to the hole. What function h(x) can
be used to describe the second golfer’s shot?
SOLUTION: a. For g(x) = 0.0005(x – 200)2 + 20, a = –0.0005, h
= 200, and k = 20. Since |–0.0005| < 1 and –0.0005 <
0, there is a reflection across the x-axis and the
graph will be compressed vertically (or be
wider). Since h = 200 and 200 > 0, there is a
translation 200 yards to the right. Since k = 20 and 20
> 0, there is a translation up 20 yards.Thus, the graph
of g(x) is the graph of f (x) translated 200 yards right,
compressed vertically, reflected in the x-axis, and
translated up 20 yards.
b. If the red tee is 30 yards closer to the hole, then g
(x) will be translated 30 yards to the right to obtain h
(x). So, for h(x) h = 200 + 30 or 230.
In order to obtain the graph of g(x) from the graph of
f(x), you need to subtract 5 from the y-value of each
point. Therefore, you need to translate the graph of f
(x) down 5 units.
2
33. f (x) = x − 4
2
g(x) = x + 7
SOLUTION: Graph f (x) and g(x) on the same coordinate plane.
2
Therefore, the function is h(x) = 0.0005(x – 230) +
20.
Describe the transformations to obtain the
graph of g (x) from the graph of f (x).
2
32. f (x) = x + 3
2
g(x) = x − 2
SOLUTION: Graph f (x) and g(x) on the same coordinate plane.
In order to obtain the graph of g(x) from the graph of
f(x), you need to add 11 to the y-value of each point.
Therefore, you need to translate the graph of f (x) up
11 units.
34. f (x) = −6x
2
g(x) = −3x
2
SOLUTION: Graph f (x) and g(x) on the same coordinate plane.
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In order to obtain the graph of g(x) from the graph of
Page 9
In order to obtain the graph of g(x) from the graph of
f(x), you need to add 11 to the y-value of each point.
Therefore, you need to translate the graph of f (x) up
11 units.
9-3 Transformations
of Quadratic Functions
34. f (x) = −6x
2
g(x) = −3x
2
SOLUTION: Graph f (x) and g(x) on the same coordinate plane.
In order to obtain the graph of g(x) from the graph
off (x), you need to take the y -value of each
point. Therefore, g(x) is a dilation of the graph of f
(x) that is compressed vertically.
35. COMBINING FUNCTIONS An engineer created
a self-refueling generator that burns fuel according to
2
the function g(t) = –t + 10t + 200, where t
represents the time in hours and g(t) represents the
number of gallons remaining.
a. How long will it take for the generator to run out
of fuel? b. The engine self-refuels at a rate of 40 gallons per
hour. Write a linear function h(t) to represent the
refueling of the generator. c. Find T(t) = g(t) + h(t). What does this new
function represent?
d. Will the generator run out of fuel? If so, when?
SOLUTION: a. The generator will run out of fuel when g(t) =
0.Solving we get:
Since time cannot be negative, t = 20 h.
So, the engine will run out of fuel after 20 hours.
b. If the engine refuels 40 gallons every hour, then the linear function describing this would be h(t) =
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40t. c. Find the sum of the two functions.
Since time cannot be negative, t = 20 h.
So, the engine will run out of fuel after 20 hours.
b. If the engine refuels 40 gallons every hour, then the linear function describing this would be h(t) =
40t. c. Find the sum of the two functions.
This new function T(t) represents the amount of fuel
in the tank while the engine is operating and
refueling.
d. The generator will now run out of fuel when T(t) =
0.
Use the quadratic formula to determine the solution
for t.
The time must be positive, so t ≈ 53.72.
Therefore, the generator will run out of fuel after
about 53.72 hours or 53 hours 43 minutes.
36. REASONING Are the following statements
sometimes, always, or never true? Explain.
2
a. The graph of y = x + k has its vertex at the
origin.
b. The graphs of y = ax2 and its reflection over the
x-axis are the same width.
2
c. The graph of y = x + k, where k ≥ 0, and the graph of a quadratic with vertex at (0, –3) have the
same maximum or minimum point.
SOLUTION: Page 10
a. The vertex for y = x2 + k is (0, k). This will equal
(0, 0) only when k = 0. For any other value, the
graph will be translated up or down. down. Since k ≥ 0, the points cannot be the same. 2
Therefore, the graph of y = x + k and the graph of
a quadratic with a vertex at (0, –3) will never have
the same maximum or minimum point.
Therefore, the generator will run out of fuel after
about 53.72 hours or 53 hours 43 minutes.
9-3 Transformations
of Quadratic Functions
2
36. REASONING Are the following statements
sometimes, always, or never true? Explain.
2
a. The graph of y = x + k has its vertex at the
origin.
b. The graphs of y = ax2 and its reflection over the
x-axis are the same width.
2
c. The graph of y = x + k, where k ≥ 0, and the graph of a quadratic with vertex at (0, –3) have the
same maximum or minimum point.
37. CHALLENGE Write a function of the form y = ax
+ c with a graph that passes through the points (−2,
3) and (4, 15).
SOLUTION: 2
Substitute the points into the function y = ax + c.
SOLUTION: a. The vertex for y = x2 + k is (0, k). This will equal
(0, 0) only when k = 0. For any other value, the
graph will be translated up or down. 2
Therefore, the graph of y = x + k will sometimes
have its vertex at the origin.
b. The reflection of y = ax2 over the x-axis will have
Next set both equations equal to each other to solve
for a.
2
the equation y = –ax . Since |a| = |–a|, both graphs
are dilated by a factor of a and will have the same
2
width. Therefore, the graph of y = ax and its
reflection over the x-axis will always have the same
width.
2
c. The vertex of the graph of y = x + k, if k ≥ 0 is (0, k) and it will be a minimum point since the graph
will open up. A quadratic with a vertex of (0, –3) will
have a minimum point at (0, –3) if the graph opens up
and a maximum point at (0, –3) if the graph opens
down. Since k ≥ 0, the points cannot be the same. 2
Therefore, the graph of y = x + k and the graph of
a quadratic with a vertex at (0, –3) will never have
the same maximum or minimum point.
2
37. CHALLENGE Write a function of the form y = ax
+ c with a graph that passes through the points (−2,
3) and (4, 15).
The value of a = 1, so the value of c is 15 – 16(1) or
2
–1. Therefore, the function is f (x) = x − 1.
38. CCSS ARGUMENTS Determine whether all
quadratic functions that are reflected across the yaxis produce the same graph. Explain your answer.
SOLUTION: Sample answer: Not all reflections over the y-axis
produce the same graph. If the vertex of the original
graph is not on the y-axis, the graph will not have the
y-axis as its axis of symmetry and its reflection
across the y-axis will be a different parabola.
39. OPEN ENDED Write a quadratic function that
opens downward and is wider than the parent graph.
SOLUTION: 2
SOLUTION: 2
Substitute the points into the function y = ax + c.
If the parent graph is f (x) = ax , then it opens
upward. In order to open downward, the value of a
2
must be negative. So, f (x) = – x is a quadratic
function that opens downward.
In order to be wider than the parent graph, the value
of |a| must be between 0 and 1.. Therefore, if a =
, the graph will be wider.
So, f (x) =
is an example of a quadratic
function that opens downward and is wider than the
parent graph.
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Next set both equations equal to each other to solve
for a.
Page 11
40. WRITING IN MATH Describe how the values of
2
2
representations for the functions y = ax , y = x + k,
Sample answer: Not all reflections over the y-axis
produce the same graph. If the vertex of the original
graph is not on the y-axis, the graph will not have the
y-axis as its axis ofofsymmetry
andFunctions
its reflection
9-3 Transformations
Quadratic
across the y-axis will be a different parabola.
So, f (x) =
is an example of a quadratic
function that opens downward and is wider than the
parent graph.
39. OPEN ENDED Write a quadratic function that
opens downward and is wider than the parent graph.
40. WRITING IN MATH Describe how the values of
2
2
representations for the functions y = ax , y = x + k,
SOLUTION: , the graph will be wider.
SOLUTION: 2
If the parent graph is f (x) = ax , then it opens
upward. In order to open downward, the value of a
2
must be negative. So, f (x) = – x is a quadratic
function that opens downward.
In order to be wider than the parent graph, the value
of |a| must be between 0 and 1.. Therefore, if a =
, the graph will be wider.
So, f (x) =
is an example of a quadratic
function that opens downward and is wider than the
parent graph.
2
For y = ax , the a stretches the parent graph verticall
|a| < 1. When a is negative, the graph is reflected ov
multiplied by a factor of a. 2
For y = x + k, the parent graph is translated up if k i
in the table will all have the constant k added to them
2
For y = ax + k, the graph will either be stretched ve
value of a and then will be translated up or down dep
graph is reflected over the x-axis. The y-values in the
constant k added to them.
Consider the following example. 40. WRITING IN MATH Describe how the values of
2
y =x
2
2
representations for the functions y = ax , y = x + k,
SOLUTION: 2
For y = ax , the a stretches the parent graph verticall
|a| < 1. When a is negative, the graph is reflected ov
multiplied by a factor of a. 2
For y = x + k, the parent graph is translated up if k i
in the table will all have the constant k added to them
2
For y = ax + k, the graph will either be stretched ve
value of a and then will be translated up or down dep
graph is reflected over the x-axis. The y-values in the
constant k added to them.
x
y
–4
16
–2
4
0
0
2
4
4
16
2
6
4
18
2
y =x +2
Consider the following example. y =x
x
y
–4
16
–2
4
2
0
0
2
4
4
16
x
y
–4
18
–2
6
0
0
2
y =x +2
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41. SHORT RESPONSE A tutor charges a flat fee of
$55 and $30 for each hour of work. Write a function
that represents the total charge C, in terms of the
number of hours h worked.
SOLUTION: Let h = the numbers of hours worked. Then thePage
total12
charge is the flat fee, $55, plus $30 per hour for each
hour worked, or 30h. So C = 55 + 30h.
x
0
2
4
–4
–2
y
18
0 Functions
6
18
9-3 Transformations
of6 Quadratic
41. SHORT RESPONSE A tutor charges a flat fee of
$55 and $30 for each hour of work. Write a function
that represents the total charge C, in terms of the
number of hours h worked.
SOLUTION: Let h = the numbers of hours worked. Then the total
charge is the flat fee, $55, plus $30 per hour for each
hour worked, or 30h. So C = 55 + 30h.
Therefore, choice C can be eliminated.
Since a = 2, this function is a dilation of the graph of
2
y = x that is stretched vertically. A parabola that is
stretched vertically will become narrower. So, the
correct choice is D.
43. Candace is 5 feet tall. If 1 inch is about 2.54
centimeters, how tall is Candace to the nearest
centimeter?
F 123 cm
G 26 cm
H 13 cm
J 152 cm
SOLUTION: 2
42. Which best describes the graph of y = 2x ?
A a line with a y-intercept of (0, 2) and an xintercept at the origin
B a parabola with a minimum point at (0, 0) and that
2
is wider than the graph of y = x C a parabola with a maximum point at (0, 0) and
2
that is narrower than the graph of y = x D a parabola with a minimum point at (0, 0) and that
2
is narrower than the graph of y = x SOLUTION: 2
An equation in the form y = ax is a quadratic
equation whose graph is a parabola. So, choice A
cannot be correct as a line would have a linear
equation in the form y = mx + b.
2
For y = 2x , a = 2. Since the value of a > 0, this
parabola opens up and thus has a minimum at (0, 0).
Therefore, choice C can be eliminated.
Since a = 2, this function is a dilation of the graph of
2
y = x that is stretched vertically. A parabola that is
stretched vertically will become narrower. So, the
correct choice is D.
43. Candace is 5 feet tall. If 1 inch is about 2.54
centimeters, how tall is Candace to the nearest
centimeter?
F 123 cm
G 26 cm
H 13 cm
J 152 cm
SOLUTION: Candace is about 152 centimeters. So, the correct
choice is J.
44. While in England, Imani spent 49.60 British pounds
on a pair of jeans. If this is equivalent to $100 in U.S.
currency, how many British pounds would Imani
have spent on a sweater that cost $60?
A 8.26 pounds
B 29.76 pounds
C 2976 pounds
D 19.84 pounds
SOLUTION: Imani would have spent about 29.76 British pounds
on the sweater. So, the correct choice is B.
Solve each equation by graphing.
2
45. x + 6 = 0
SOLUTION: 2
Graph the related function f (x) = x + 6.
Candace is about 152 centimeters. So, the correct
choice is J.
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44. While in England, Imani spent 49.60 British pounds
on a pair of jeans. If this is equivalent to $100 in U.S.
currency, how many British pounds would Imani
Page 13
There are no x-intercepts of the graph. So, there are
Imani would have of
spent
about 29.76
British pounds
9-3 Transformations
Quadratic
Functions
on the sweater. So, the correct choice is B.
Solve each equation by graphing.
2
45. x + 6 = 0
There are no x-intercepts of the graph. So, there are
no real solutions to this equation. The solution is ∅.
2
46. x − 10x = −24
SOLUTION: Rewrite the equation in standard form.
SOLUTION: 2
Graph the related function f (x) = x + 6.
2
Graph the related function f (x) = x − 10x + 24.
There are no x-intercepts of the graph. So, there are
no real solutions to this equation. The solution is ∅.
The x-intercepts of the graph appear to be at 4 and 6,
so the solutions are 4 and 6.
Check:
2
46. x − 10x = −24
SOLUTION: Rewrite the equation in standard form.
2
Graph the related function f (x) = x − 10x + 24.
2
47. x + 5x + 4 = 0
SOLUTION: 2
Graph the related function f (x) = x + 5x + 4.
The x-intercepts of the graph appear to be at 4 and 6,
so the solutions are 4 and 6.
Check:
The x-intercepts of the graph appear to be at –1 and
–4, so the solutions are –1 and –4.
Check:
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2
47. x + 5x + 4 = 0
Page 14
9-3 Transformations of Quadratic Functions
2
2
47. x + 5x + 4 = 0
48. 2x − x = 3
SOLUTION: 2
Graph the related function f (x) = x + 5x + 4.
SOLUTION: Rewrite the equation in standard form.
2
Graph the related function f (x) = 2x − x – 3.
The x-intercepts of the graph appear to be at –1 and
–4, so the solutions are –1 and –4.
Check:
The x-intercepts of the graph appear to be at –1 and
1.5, so the solutions are –1 and 1.5.
Check:
2
48. 2x − x = 3
SOLUTION: Rewrite the equation in standard form.
2
49. 2x − x = 15
2
Graph the related function f (x) = 2x − x – 3.
SOLUTION: Rewrite the equation in standard form.
2
Graph the related function f (x) = 2x − x – 15.
The x-intercepts of the graph appear to be at –1 and
1.5, so the solutions are –1 and 1.5.
Check:
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The x-intercepts of the graph appear to be at –2.5
Page 15
and 3, so the solutions are –2.5 and 3.
Check:
9-3 Transformations of Quadratic Functions
2
2
49. 2x − x = 15
50. 12x = −11x + 15
SOLUTION: Rewrite the equation in standard form.
SOLUTION: Rewrite the equation in standard form.
2
2
Graph the related function f (x) = 2x − x – 15.
Graph the related function f (x) = 12x + 11x + 15.
The x-intercepts of the graph appear to be at –2.5
and 3, so the solutions are –2.5 and 3.
Check:
The x-intercepts are located between –2 and –1 and
table using an increment of 0.1 for the x-values locat
between 0 and 1.
x
y
–1.9
7.42
–1.8
4.08
–1.7
0.98
–1.6
–1.88
–1.5
–4.5
–1.4
–6.8
x
y
0.1
–2.59
0.2
–2.16
0.3
–1.71
0.4
–1.24
0.5
–0.75
0.6
–0.2
For the first table, the function value that is closest to
changes is 0.98. So the first root is approximately –1.
function value that is closest to zero when the sign ch
second root is approximately 0.6. Thus, the roots are
0.6.
2
50. 12x = −11x + 15
Find the vertex, the equation of the axis of
symmetry, and the y-intercept of each graph.
SOLUTION: Rewrite the equation in standard form.
2
Graph the related function f (x) = 12x + 11x + 15.
51. The x-intercepts are located between –2 and –1 and
table using an increment of 0.1 for the x-values locat
between 0 and 1.
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SOLUTION: Find the vertex.
Because the parabola opens up, the vertex is located
at the minimum point of the parabola. It is located at
(0, 4).
Find the axis of symmetry.
The axis of symmetry is the line that goes through
Page 16
the vertex and divides the parabola into congruent
halves. It is located at x = 0.
halves. It is located at x = 2.
Find the y-intercept.
The y-intercept is the point where the graph
intersects the y-axis. It is located at (0, 6), so the yintercept is 6.
For the first table, the function value that is closest to
changes is 0.98. So the first root is approximately –1.
function value that is closest to zero when the sign ch
9-3 Transformations
of Quadratic Functions
second root is approximately 0.6. Thus, the roots are
0.6.
Find the vertex, the equation of the axis of
symmetry, and the y-intercept of each graph.
53. 51. SOLUTION: Find the vertex.
Because the parabola opens up, the vertex is located
at the minimum point of the parabola. It is located at
(0, 4).
Find the axis of symmetry.
The axis of symmetry is the line that goes through
the vertex and divides the parabola into congruent
halves. It is located at x = 0.
Find the y-intercept.
The y-intercept is the point where the graph
intersects the y-axis. It is located at (0, 4), so the yintercept is –4.
SOLUTION: Find the vertex.
Because the parabola opens down, the vertex is
located at the maximum point of the parabola. It is
located at (–2, 6).
Find the axis of symmetry.
The axis of symmetry is the line that goes through
the vertex and divides the parabola into congruent
halves. It is located at x = –2.
Find the y-intercept.
The y-intercept is the point where the graph
intersects the y-axis. It is located at (0, 2), so the yintercept is 2.
54. CLASS TRIP Mr. Wong’s American History class
will take taxis from their hotel in Washington, D.C.,
to the Lincoln Memorial. The fare is $2.75 for the
first mile and $1.25 for each additional mile. If the
distance is m miles and t taxis are needed, write an
expression for the cost to transport the group.
SOLUTION: Number of Miles
52. SOLUTION: Find the vertex.
Because the parabola opens up, the vertex is located
at the minimum point of the parabola. It is located at
(2, 2).
Find the axis of symmetry.
The axis of symmetry is the line that goes through
the vertex and divides the parabola into congruent
halves. It is located at x = 2.
Find the y-intercept.
The y-intercept is the point where the graph
intersects the y-axis. It is located at (0, 6), so the yintercept is 6.
Cost of Each
Cost of
Taxi
Each Taxi
1
2.75
2.75
2
2.75 + 1.25(1)
4.00
3
2.75 + 1.25(2)
5.25
4
2.75 + 1.25(3)
6.50
m
2.75 + 1.25(m –
C
1)
So the equation for the cost of one taxi is C = 2.75 +
1.25 (m −1).
The cost for one taxi is C = 2.75 + 1.25 (m −1).
Therefore, if t taxis are needed, the cost to transport
the group is t • C or 2.75t + 1.25t(m −1).
Solve each inequality. Graph the solution on a
number line.
55. −3t + 6 ≤ −3
SOLUTION: eSolutions Manual - Powered by Cognero
53. Page 17
The solution set is {t|t ≥ 3}.
1.25 (m −1).
The cost for one taxi is C = 2.75 + 1.25 (m −1).
Therefore, if t taxisofare
needed, the
cost to transport
9-3 Transformations
Quadratic
Functions
the group is t • C or 2.75t + 1.25t(m −1).
Solve each inequality. Graph the solution on a
number line.
55. −3t + 6 ≤ −3
The solution set is {d| d > −125}.
Determine whether each trinomial is a perfect
square trinomial. If so, factor it.
2
58. 16x − 24x + 9
SOLUTION: SOLUTION: The first term is a perfect square.
2
2
16x = (4x)
The last term is a perfect square.
The solution set is {t|t ≥ 3}.
9 = (3)
The middle term is equal to 2ab.
24x = 2(4x)(3)
2
So, 16x − 24x + 9 is a perfect square trinomial.
.
2
56. 59 > −5 − 8f
SOLUTION: 2
59. 9x + 6x + 1
The solution set is {f |f > −8}.
SOLUTION: The first term is a perfect square.
2
2
9x = (3x)
The last term is a perfect square.
2
57. −2 −
< 23
SOLUTION: 1 = (1)
The middle term is equal to 2ab.
6x = 2(3x)(1)
2
So, 9x + 6x + 1 is a perfect square trinomial.
2
60. 25x − 60x + 36
The solution set is {d| d > −125}.
SOLUTION: The first term is a perfect square.
2
2
25x = (5x)
The last term is a perfect square.
Determine whether each trinomial is a perfect
square trinomial. If so, factor it.
36 = (6)
The middle term is equal to 2ab.
60x = 2(5x)(6)
2
So, 25x − 60x + 36 is a perfect square trinomial.
2
58. 16x − 24x + 9
SOLUTION: The first term is a perfect square.
2
2
16x = (4x)
The last term is a perfect square.
2
9 = (3)
The middle term is equal to 2ab.
24x = 2(4x)(3)
2
So, 16x − 24x + 9 is a perfect square trinomial.
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2
2
2
61. x − 8x + 81
SOLUTION: The first term is a perfect square.
2
2
x = (x)
The last term is a perfect square.
2
81 = (9)
The middle term is not equal to 2ab.
8x ≠ 2(x)(9)
Page 18
60x = 2(5x)(6)
2
So, 25x − 60x + 36 is a perfect square trinomial.
9-3 Transformations of Quadratic Functions
2
61. x − 8x + 81
SOLUTION: The first term is a perfect square.
2
2
x = (x)
The last term is a perfect square.
2
81 = (9)
The middle term is not equal to 2ab.
8x ≠ 2(x)(9)
2
So, x − 8x + 81 is not a perfect square trinomial.
2
62. 36x − 84x + 49
SOLUTION: The first term is a perfect square.
2
2
36x = (6x)
The last term is a perfect square.
2
49 = (7)
The middle term is equal to 2ab.
84x = 2(6x)(7)
2
So, 36x − 84x + 49 is a perfect square trinomial.
2
63. 4x − 3x + 9
SOLUTION: The first term is a perfect square.
2
2
4x = (2x)
The last term is a perfect square.
2
9 = (3)
The middle term is not equal to 2ab.
3x ≠ 2(2x)(3)
2
So, 4x − 3x + 9 is not a perfect square trinomial.
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