# Document 156171

```9-3 Transformations of Quadratic Functions
2
Describe how the graph of each function is related to the graph of f (x) = x .
2
1. g(x) = x − 11
SOLUTION: 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.
2. h(x) =
x
2
SOLUTION: 2
The graph of f (x) = ax stretches or compresses the parent graph vertically. Since a =
, the graph of y =
2
x is
2
the graph of y = x vertically compressed.
2
3. h(x) = −x + 8
SOLUTION: 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
2
2
x-axis. Since c = 8 the graph is translated up 8 units. So, the graph of y = –x +8 is the graph of y = x reflected
across the x-axis and translated up 8 units.
2
4. g(x) = x + 6
Page 1
2
3. h(x) = −x + 8
SOLUTION: 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
2
2
x-axis. Since c = 8 the graph is translated up 8 units. So, the graph of y = –x +8 is the graph of y = x reflected
across the x-axis and translated up 8 units.
2
4. g(x) = x + 6
SOLUTION: 2
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.
5. g(x) = −4x
2
SOLUTION: 2
2
The coefficient of the x -term is negative, so the graph is reflected across the x-axis. The graph of f (x) = ax
stretches or compresses the parent graph vertically. Since a = 4, the graph is vertically stretched. So, the graph of y
2
2
= −4x is the graph of y = x reflected across the x-axis and vertically stretched.
2
6. h(x) = −x − 2
SOLUTION: eSolutions
Page 2
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
2
x-axis. Since c = –2, the graph is translated down 2 units. So, the graph of y = −x − 2 is the graph of y = 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
2
x-axis. Since c = –2, the graph is translated down 2 units. So, the graph of y = −x − 2 is the graph of y = x
reflected across the x-axis and translated down 2 units.
2
7. MULTIPLE CHOICE Which is an equation for the function shown in the graph?
A g(x) =
2
x +2
B g(x) = −5x2 − 2
C g(x) =
2
x −2
D g(x) = − x2 − 2
SOLUTION: 2
The function is a parabola which can be written f (x) = 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.
Page 3
The function is a parabola which can be written f (x) = 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) =
is C.
2
x − 2. So, the correct choice
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.
9. h(x) = −7 − 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
2
the x-axis. Since c = –7 the graph is translated down 7 units. So, the graph of y = −7 − x is the graph of y = x
reflected across the x-axis and translated down 7 units.
2
2
10. g(x) = 2x + 8
SOLUTION: 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 of y =2x
2
+ 8 is the graph of y = x vertically stretched and shifted up 8 units.
11. h(x) = 6 +
x
2
2
SOLUTION: Page 4
11. h(x) = 6 +
x
2
SOLUTION: 2
The function can be written f (x) = ax + c, where a =
2
=6+
and c = 6. Since 0 < a < 1 and c is positive, the graph of y
2
x is the graph of y = x vertically compressed and shifted up 6 units.
12. g(x) = −5 −
x
2
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
2
x is the graph of y = x reflected across the x-axis, vertically stretched, and shifted down 5
units.
13. h(x) = 3 +
x
2
SOLUTION: 2
The function can be written f (x) = ax + c, where a =
+
2
and c = 3. Since a > 1 and c is positive, the graph of y = 3
2
x is the graph of y = x vertically stretched and shifted up 3 units.
Page 5
13. h(x) = 3 +
x
2
SOLUTION: 2
The function can be written f (x) = ax + c, where a =
+
2
and c = 3. Since a > 1 and c is positive, the graph of y = 3
2
x is the graph of y = x vertically stretched and shifted up 3 units.
2
14. g(x) = 0.25x − 1.1
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 negative, the
2
2
graph of y = 0.25x − 1.1 is the graph of y = x vertically compressed and shifted down 1.1 units.
2
15. h(x) = 1.35x + 2.6
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 graph of y
2
2
= 1.35x + 2.6 is the graph of y = x vertically stretched and shifted up 2.6 units.
2
eSolutions
by Cognero
16. g(x)Manual
= x- Powered
+
SOLUTION: Page 6
2
16. g(x) =
x +
SOLUTION: 2
The function can be written f (x) = ax + c, where a =
y=
2
x +
2
and c =
. Since 0 < a < 1 and c is positive, the graph of
is the graph of y = x vertically compressed and shifted up
unit.
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, the graph of
2
2
y = 1.01x − 6.5 is the graph of y = x vertically stretched and shifted down 6.5 units.
Match each equation to its graph.
A
B
C
Page 7
B
C
D
E
F
18. y =
2
x −4
SOLUTION: 2
The function can be written f (x) = ax + c, where a =
y=
2
and c = –4. Since 0 < a < 1 and c is negative, the graph of
2
x − 4 is the graph of y = x vertically compressed and shifted down 4 units. So, the correct choice is C.
2
19. y = − x − 4
Page 8
2
The function can be written f (x) = ax + c, where a =
and c = –4. Since 0 < a < 1 and c is negative, the graph of
2
2
9-3 Transformations
y = x − 4 is theof
graph
of y = xFunctions
vertically 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 =
and c = –4. The negative sign in front of the a indicates
2
that the function is a reflection of the graph of y = x about the x-axis. Since 0 < a < 1 and c is negative, the graph of
2
2
y = – x − 4 is a reflection of the graph of y = x vertically compressed and shifted down 4 units. So, the correct
choice is A.
2
20. y =
x +4
SOLUTION: 2
The function can be written f (x) = ax + c, where a =
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 vertically compressed and shifted up 4 units. So, the correct choice is B.
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
2
that the function is a reflection of the graph of y = x about the x-axis. Since a > 1 and c is negative, the graph of y =
2
2
−3x − 2 is a reflection of the graph of y = x vertically stretched, and shifted down 2 units. So, the correct choice is
F.
2
22. y = −x + 2
SOLUTION: 2
2
The function can be written f (x) = –x + c, where c = 2. The negative sign in front of the x indicates that the
2
2
function is a reflection of the graph of y = x about the x-axis. Since c is positive, the graph of y = −x + 2 is a
2
reflection of the graph of y = x shifted up 2 units. So, the correct choice is D.
2
23. y = 3x + 2
SOLUTION: 2
The function can be written f (x) = ax + c, where a = 3 and c = 2. Since a > 1 and c is positive, the graph of y = 3x
2
+ 2 is the graph of y = x vertically stretched, and shifted up 2 units. So, the correct choice is E.
2
2
24. SQUIRRELS A squirrel 12 feet above the ground 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
The function can be written f (x) = –at + c, where a = 16 and c = 12. The negative sign in front of the a indicates
2
that the function is a reflection of the graph of h = t about the t-axis. Since a > 1 and c is positive, the graph of h =
eSolutions Manual
Page 9
2
2
−16t + 12 is a reflection of the graph of h = t vertically stretched, and shifted up 12 units.
23. y = 3x + 2
SOLUTION: 2
The function can be
f (x) =Functions
ax + c, where a = 3 and c = 2. Since a > 1 and c is positive, the graph of y = 3x
9-3 Transformations
of written
2
+ 2 is the graph of y = x vertically stretched, and shifted up 2 units. So, the correct choice is E.
2
2
24. SQUIRRELS A squirrel 12 feet above the ground 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
The function can be written f (x) = –at + c, where a = 16 and c = 12. The negative sign in front of the a indicates
2
that the function is a reflection of the graph of h = t about the t-axis. Since a > 1 and c is positive, the graph of h =
2
2
−16t + 12 is a reflection of the graph of h = t vertically stretched, and shifted up 12 units.
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
SOLUTION: 2
2
2
The functions can be written as f (x) =ax . For g(x) = 2x , a = 2 so the graph is stretched vertically. For h(x) =
x ,
so the graph is compressed vertically. Therefore, the functions in order from the most stretched vertically to a=
the least compressed graph is g(x), h(x).
2
26. g(x) = −3x , h(x) =
x
2
SOLUTION: 2
2
The functions can be written as f (x) =ax . For g(x) = −3x , a = 3 so the graph is stretched vertically. For h(x) =
2
so the graph is compressed vertically. Therefore, the functions in order from the most stretched x ,a =
vertically to the least compressed graph is g(x), h(x).
2
2
27. g(x) = −4x , h(x) = 6x , f (x) = 0.3x
2
SOLUTION: 2
2
2
The functions can be written as f (x) =ax . For g(x) = −4x , a = 4 so the graph is stretched vertically. For h(x) = 6x ,
2
a = 6 so the graph is also stretched vertically; and since 6 > 4, h(x) is more stretched 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
28. g(x) = −x , h(x) =
2
x , f (x) = −4.5x
2
Page 10
SOLUTION: 2
2
2
2
2
The functions can be written as f (x) =ax . For g(x) = −4x , a = 4 so the graph is stretched vertically. For h(x) = 6x ,
2
a = 6 so the graph is also stretched vertically; and since 6 > 4, h(x) is more stretched than g(x) . For f (x) = 0.3x , a
9-3 Transformations
= 0.3 so the graph of
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
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) = −x , a = 1 so the graph is neither stretched nor compressed.
For h(x) =
2
x ,a =
2
so the graph is stretched vertically. For f (x) = −4.5x , a = 4.5, so the graph is also stretched
vertically. Since 4.5 >
, f (x) 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?
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.
2
For the first rock, the initial height is 300 feet. 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.
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
Sprinkler C: y = −0.08x + 2.4
eSolutions
Manualsprinkler
- Powered will
by Cognero
a. Which
send water the farthest? Explain.
b. Which sprinkler will send water the highest? Explain.
c. Which sprinkler will produce the narrowest path? Explain.
Page 11
It takes the secondofrock
seconds to reach the ground.
9-3 Transformations
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: 2
All three of the quadratic functions are of the form f (x) = ax + c. Use a graphing calculator to compare the graphs
of the three functions.
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.
2
a. How can you obtain g(x) from the graph of f (x) = 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.
eSolutions
Page 12
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.
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
9-3 Transformations
ofthe
function Functions
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.
2
a. How can you obtain g(x) from the graph of f (x) = 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.
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 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.
Page 13
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.
9-3 Transformations
2
33. f (x) = x − 4
2
g(x) = x + 7
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.
In order to obtain the graph of g(x) from the graph of f (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 the
2
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:
Page 14
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?
9-3 Transformations
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) = 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
eSolutions
Manual
- Powered
c. The
graph
of y =by
x Cognero
+ k, where
or minimum point.
SOLUTION: Page 15
k ≥ 0, and the graph of a quadratic with vertex at (0, –3) have the same maximum
Therefore, the generator will run out of fuel after about 53.72 hours or 53 hours 43 minutes.
9-3 Transformations
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: 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 the equation y = –ax2. Since |a| = |–a|, both graphs are dilated
2
by a factor of a and will have the same width. Therefore, the graph of y = ax and its reflection over the x-axis will
always have the same width.
c. The vertex of the graph of y = x2 + 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
2
at (0, –3) if the graph opens down. Since k ≥ 0, the points cannot be the same. 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).
SOLUTION: 2
Substitute the points into the function y = ax + c.
Next set both equations equal to each other to solve for a.
2
The value of a = 1, so the value of c is 15 – 16(1) or –1. Therefore, the function is f (x) = x − 1.
38. CCSS ARGUMENTS Determine whether all quadratic functions that are reflected across the y-axis produce the
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
Page 16
different parabola.
39. OPEN ENDED Write a quadratic function that opens downward and is wider than the parent graph.
2
The value of a = 1, so the value of c is 15 – 16(1) or –1. Therefore, the function is f (x) = x − 1.
38. CCSS ARGUMENTS Determine whether all quadratic functions that are reflected across the y-axis produce the
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
If the parent graph is f (x) = ax , then it opens upward. In order to open downward, the value of a must be negative.
2
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.
40. WRITING IN MATH Describe how the values of a and k affect the graphical and tabular representations for the
2
2
2
functions y = ax , y = x + k, and y = ax + k.
SOLUTION: 2
For y = ax , the a stretches the parent graph vertically if |a| > 1 or compresses the parent graph if 0 < |a| < 1. When
a is negative, the graph is reflected over the x-axis. The y-values in the table will all be multiplied by a factor of a. 2
For y = x + k, the parent graph is translated up if k is positive and down if k is negative. The y-values in the table
will all have the constant k added to them or subtracted from them. 2
For y = ax + k, the graph will either be stretched vertically or compressed vertically based upon the value of a and
then will be translated up or down depending on the value of k. When a is negative, the graph is reflected over the xaxis. The y-values in the table will be multiplied by a factor of a and the constant k added to them.
Consider the following example. y =x
x
y
–4
16
–2
4
2
y = 3x
0
0
2
y =x +2
2
4
4
16
x
y
–4
48
–2
12
2
0
0
2
12
4
48
2
y = 3x + 2
Page 17
2
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.
40. WRITING IN MATH Describe how the values of a and k affect the graphical and tabular representations for the
2
2
2
functions y = ax , y = x + k, and y = ax + k.
SOLUTION: 2
For y = ax , the a stretches the parent graph vertically if |a| > 1 or compresses the parent graph if 0 < |a| < 1. When
a is negative, the graph is reflected over the x-axis. The y-values in the table will all be multiplied by a factor of a. 2
For y = x + k, the parent graph is translated up if k is positive and down if k is negative. The y-values in the table
will all have the constant k added to them or subtracted from them. 2
For y = ax + k, the graph will either be stretched vertically or compressed vertically based upon the value of a and
then will be translated up or down depending on the value of k. When a is negative, the graph is reflected over the xaxis. The y-values in the table will be multiplied by a factor of a and the constant k added to them.
Consider the following example. y =x
x
y
–4
16
–2
4
2
y = 3x
0
0
2
4
4
16
x
y
–4
48
–2
12
–2
6
2
12
4
48
2
14
4
50
2
y =x +2
–4
18
0
0
2
x
y
2
0
0
y = 3x + 2
2
6
4
18
x
y
–4
50
–2
14
0
0
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.
2
42. Which best describes the graph of y = 2x ?
A a line with a y-intercept of (0, 2) and an x-intercept at the origin
B a parabola with a minimum point at (0, 0) and that is wider than the graph of y = x2 eSolutions Manual - Powered by Cognero
Page 18
2
represents the total charge C, in terms of the number of hours h worked.
SOLUTION: Let h = the numbers
hours worked.
Then the total charge is the flat fee, \$55, plus \$30 per hour for each hour
9-3 Transformations
of of
Functions
worked, or 30h. So C = 55 + 30h.
2
42. Which best describes the graph of y = 2x ?
A a line with a y-intercept of (0, 2) and an x-intercept at the origin
B a parabola with a minimum point at (0, 0) and that is wider than the graph of y = x2 2
C a parabola with a maximum point at (0, 0) and that is narrower than the graph of y = x D a parabola with a minimum point at (0, 0) and that is narrower than the graph of y = x2 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.
2
Since a = 2, this function is a dilation of the graph of 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
eSolutions
SOLUTION: Page 19
2
Graph the related function f (x) = x + 6.
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.
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.
2
Graph the related function f (x) = x − 10x + 24.
The x-intercepts of the graph appear to be at 4 and 6, so the solutions are 4 and 6.
Check:
2
eSolutions
by Cognero
5x + 4- Powered
=0
47. x +Manual
Page 20
SOLUTION: 2
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 –1 and –4, so the solutions are –1 and –4.
Check:
2
48. 2x − x = 3
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 1.5, so the solutions are –1 and 1.5.
Check:
Page 21
2
48. 2x − x = 3
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 1.5, so the solutions are –1 and 1.5.
Check:
2
49. 2x − x = 15
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 –2.5 and 3, so the solutions are –2.5 and 3.
Check:
Page 22
2
49. 2x − x = 15
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 –2.5 and 3, so the solutions are –2.5 and 3.
Check:
2
50. 12x = −11x + 15
SOLUTION: Rewrite the equation in standard form.
2
Graph the related function f (x) = 12x + 11x + 15.
The x-intercepts are located between –2 and –1 and between 0 and 1. Make a table using an increment of 0.1 for
the x-values located between –2 and –1 and between 0 and 1.
Page 23
x
y
–1.9
7.42
–1.8
4.08
–1.7
0.98
–1.6
–1.88
–1.5
–4.5
–1.4
–6.88
–1.3
–9.02
–1.2
–10.9
–1.1
–
2
50. 12x = −11x + 15
SOLUTION: Rewrite the equation in standard form.
2
Graph the related function f (x) = 12x + 11x + 15.
The x-intercepts are located between –2 and –1 and between 0 and 1. Make a table using an increment of 0.1 for
the x-values located between –2 and –1 and 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
–6.88
–1.3
–9.02
–1.2
–10.9
–1.1
–
12.6
x
y
0.1
–2.59
0.2
–2.16
0.3
–1.71
0.4
–1.24
0.5
–0.75
0.6
–0.24
0.7
0.29
0.8
0.84
0.9
1.41
For the first table, the function value that is closest to zero when the sign changes is 0.98. So the first root is
approximately –1.7. For the second table, the function value that is closest to zero when the sign changes is –0.24.
So the second root is approximately 0.6. Thus, the roots are approximately –1.7 and 0.6.
Find the vertex, the equation of the axis of symmetry, and the y-intercept of each graph.
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 y-intercept is –4.
Page 24
.
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 y-intercept is –4.
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 y-intercept is 6.
53. 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 y-intercept 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
Cost of Each Taxi
Cost of 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 –1)
C
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
Page 25
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.
9-3 Transformations
The y-intercept is the point where the graph intersects the y-axis. It is located at (0, 2), so the y-intercept 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
Cost of Each Taxi
Cost of 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 –1)
C
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: The solution set is {t|t ≥ 3}.
.
56. 59 > −5 − 8f
SOLUTION: The solution set is {f |f > −8}.
57. −2 −
< 23
SOLUTION: The solution set is {d| d > −125}.
Determine whether each trinomial is a perfect square trinomial. If so, factor it.
Page 26
The solution set is {f |f > −8}.
57. −2 −
< 23
SOLUTION: 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: 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.
2
59. 9x + 6x + 1
SOLUTION: The first term is a perfect square.
2
2
9x = (3x)
The last term is a perfect square.
2
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
SOLUTION: The first term is a perfect square.
2
2
25x = (5x)
The last term is a perfect square.
2
36 = (6)
The middle term is equal to 2ab.
60x = 2(5x)(6)
2
Page 27
6x = 2(3x)(1)
2
So, 9x + 6x + 1 is a perfect square trinomial.
2
60. 25x − 60x + 36
SOLUTION: The first term is a perfect square.
2
2
25x = (5x)
The last term is a perfect square.
2
36 = (6)
The middle term is equal to 2ab.
60x = 2(5x)(6)
2
So, 25x − 60x + 36 is a perfect square trinomial.
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.
Page 28
The last term is a perfect square.
2
9 = (3)
The middle term is not equal to 2ab.
3x ≠ 2(2x)(3)
9-3 Transformations
2
So, 4x − 3x + 9 is not a perfect square trinomial.
Page 29
```