Download Graph each polynomial equation by making a table of values. 1. f (x

5-4 Analyzing Graphs of Polynomial Functions
Graph each polynomial equation by making a
table of values.
4
3
2
1. f (x) = 2x – 5x + x – 2x + 4
SOLUTION: Make a table of values.
4
3
2
2. f (x) = –2x + 4x + 2x + x – 3
SOLUTION: Make a table of values.
Plot the points on the coordinate plane and connect
them by a smooth curve.
Plot the points on the coordinate plane and connect
them by a smooth curve.
4
3
2
3. f (x) = 3x – 4x – 2x + x – 4
4
3
2
2. f (x) = –2x + 4x + 2x + x – 3
SOLUTION: Make a table of values.
SOLUTION: Make a table of values.
Plot the points on the coordinate plane and connect
them by a smooth curve.
Plot the points on the coordinate plane and connect
them by a smooth curve.
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Page 1
5-4 Analyzing Graphs of Polynomial Functions
4
3
2
3. f (x) = 3x – 4x – 2x + x – 4
4
3
2
4. f (x) = –4x + 5x + 2x + 3x + 1
SOLUTION: Make a table of values.
SOLUTION: Make a table of values.
Plot the points on the coordinate plane and connect
them by a smooth curve.
Plot the points on the coordinate plane and connect
them by a smooth curve.
4
3
2
4. f (x) = –4x + 5x + 2x + 3x + 1
SOLUTION: Make a table of values.
Determine the consecutive integer values of x
between which each real zero of each function
located. Then draw the graph.
3
2
5. f (x) = x – 2x + 5
SOLUTION: Plot the points on the coordinate plane and connect
them by a smooth curve.
The change in sign indicates that there is a zero
between –2 and –1.
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Page 2
5-4 Analyzing Graphs of Polynomial Functions
Determine the consecutive integer values of x
between which each real zero of each function
located. Then draw the graph.
3
4
3
2
6. f (x) = –x + x + 2x + x + 1
SOLUTION: 2
5. f (x) = x – 2x + 5
SOLUTION: The changes in sign indicate that there are zeros
between 2 and 3 and at –1.
The change in sign indicates that there is a zero
between –2 and –1.
4
3
2
7. f (x) = –3x + 5x + 4x + 4x – 8
4
3
2
SOLUTION: 6. f (x) = –x + x + 2x + x + 1
SOLUTION: The changes in sign indicate that there are zeros
between 2 and 3 and at –1.
The changes in sign indicate that there are zeros
between 0 and 1 and between 2 and 3.
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Page 3
5-4 Analyzing Graphs of Polynomial Functions
4
3
2
7. f (x) = –3x + 5x + 4x + 4x – 8
4
3
2
8. f (x) = 2x – x – 3x + 2x – 4
SOLUTION: SOLUTION: The changes in sign indicate that there are zeros
between 0 and 1 and between 2 and 3.
The changes in sign indicate that there are zeros
between –2 and –1 and between 1 and 2.
4
3
2
8. f (x) = 2x – x – 3x + 2x – 4
SOLUTION: Graph each polynomial function. Estimate the xcoordinates at which the relative maxima and
relative minima occur. State the domain and
range for each function.
3
2
9. f (x) = x + x – 6x –3
SOLUTION: The changes in sign indicate that there are zeros
between –2 and –1 and between 1 and 2.
Graph the polynomial.
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Page 4
5-4 Analyzing Graphs of Polynomial Functions
Graph each polynomial function. Estimate the xcoordinates at which the relative maxima and
relative minima occur. State the domain and
range for each function.
3
Relative maxima at
;
Relative minima at
Domain: {All real numbers}
Range: {All real numbers}
3
2
10. f (x) = 3x – 6x – 2x + 2
SOLUTION: 2
9. f (x) = x + x – 6x –3
SOLUTION: Graph:
Graph the polynomial.
Relative maxima at
;
Relative minima at
Domain: {All real numbers}
Range: {All real numbers}
Relative maxima at
;
Relative minima at
Domain: {All real numbers}
Range: {All real numbers}
3
3
2
11. f (x) = –x + 4x – 2x – 1
SOLUTION: 2
10. f (x) = 3x – 6x – 2x + 2
SOLUTION: Graph:
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Graph:
Page 5
Relative maxima at
;
Relative minima at
5-4 Analyzing
Graphs
of Polynomial Functions
Domain: {All
real numbers}
Range: {All real numbers}
3
2
11. f (x) = –x + 4x – 2x – 1
Relative maxima at
;
Relative minima at
Domain: {All real numbers}
Range: {All real numbers}
3
2
12. f (x) = –x + 2x – 3x + 4
SOLUTION: SOLUTION: Graph:
Graph:
Relative maxima at
;
Relative minima at
Domain: {All real numbers}
Range: {All real numbers}
No Relative maxima: No Relative minima:
Domain: {All real numbers}
Range: {All real numbers}
3
2
12. f (x) = –x + 2x – 3x + 4
SOLUTION: Graph:
13. CCSS SENSE-MAKING Annual compact disc
sales can be modeled by the quartic function
4
3
2
f (x) = 0.48x – 9.6x + 53x – 49x + 599, where x is
the number of years after 1995 and f (x) is annual
sales in millions.
a. Graph the function for
b. Describe the turning points of the graph, its end
behavior, and the intervals on which the graph is
increasing or decreasing.
c. Continue the graph for x = 11 and x = 12. What
trends in compact disc sales does the graph suggest?
d. Is it reasonable that the trend will continue
indefinitely? Explain.
SOLUTION: a.
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Page 6
trends in compact disc sales does the graph suggest?
d. Is it reasonable that the trend will continue
indefinitely? Explain.
5-4 Analyzing Graphs of Polynomial Functions
SOLUTION: a.
d. Sample answer: No; with so many other forms of
media on the market today, CD sales will not
increase dramatically. In fact, the sales will probably
decrease. The function appears to be accurate only
until about 2005.
Complete each of the following.
a. Graph each function by making a table of
values.
b. Determine the consecutive integer values of
x between which each real zero is located.
c. Estimate the x-coordinates at which the
relative maxima and minima occur.
3
2
14. f (x) = x + 3x
SOLUTION: a.
b. Sample answer: Relative maximum at x = 5 and
relative minimum at x ≈ 9.5. f (x) →∞ as x →∞ and f
(x) →∞ as x → – ∞. The graph increases when x <
5 and x > 9.5 and decreases when 5 < x < 9.5.
c.
Sample answer: This suggests a dramatic increase in
sales.
d. Sample answer: No; with so many other forms of
media on the market today, CD sales will not
increase dramatically. In fact, the sales will probably
decrease. The function appears to be accurate only
until about 2005.
Complete each of the following.
a. Graph each function by making a table of
values.
b. Determine the consecutive integer values of
x between which each real zero is located.
c. Estimate the x-coordinates at which the
relative maxima and minima occur.
3
2
14. f (x) = x + 3x
b. The changes in sign indicate that there are zeros
at –3 and at 0.
c. Relative maxima at x = –2; Relative minima at x =
0.
3
2
15. f (x) = –x + 2x – 4
SOLUTION: a.
SOLUTION: a.
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b. The changes in sign indicate that there are zeros
at –3 and at 0.
c. Relative maxima at x = –2; Relative minima at x =
5-4 Analyzing
Graphs of Polynomial Functions
0.
3
2
15. f (x) = –x + 2x – 4
SOLUTION: a.
b. The changes in sign indicate that there are zeros
between –2 and –1.
c. Relative maxima at x = 1; Relative minima at x =
0.
3
2
16. f (x) = x + 4x – 5x
SOLUTION: a.
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b. The changes in sign indicate that there are zeros
between –2 and –1.
c. Relative maxima at x = 1; Relative minima at x =
0.
3
2
16. f (x) = x + 4x – 5x
SOLUTION: a.
b. The changes in sign indicate that there are zeros
at –5, 0 and 1.
c. Relative maxima at x = –3; Relative minima
between x = 0 and x = 1.
3
2
17. f (x) = x – 5x + 3x + 1
SOLUTION: a.
Page 8
b. The changes in sign indicate that there are zeros
at –5, 0 and 1.
c. Relative Graphs
5-4 Analyzing
Functions
maxima of
at Polynomial
x = –3; Relative
minima
between x = 0 and x = 1.
3
2
b. The changes in sign indicate that there are zeros
at 1, between –1 and 0, between x = 4 and x = 5.
c. Relative maxima at
; Relative minima at x
3.
3
2
18. f (x) = –2x + 12x – 8x
17. f (x) = x – 5x + 3x + 1
SOLUTION: a.
SOLUTION: a.
b. The changes in sign indicate that there are zeros
at 1, between –1 and 0, between x = 4 and x = 5.
b. The changes in sign indicate that there are zeros
at 0, between x = 0 and x = 1, and between x = 5 and
x = 6.
c. Relative maxima: near x = 4; Relative minima
between x = 0 and x = 1.
c. Relative maxima at
; Relative minima at x
3.
3
2
19. f (x) = 2x – 4x – 3x + 4
3
2
18. f (x) = –2x + 12x – 8x
SOLUTION: a.
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SOLUTION: a.
Page 9
b. The changes in sign indicate that there are zeros
at 0, between x = 0 and x = 1, and between x = 5 and
x = 6.
c. Relative Graphs
5-4 Analyzing
Polynomial
Functions
maxima:ofnear
x = 4; Relative
minima
between x = 0 and x = 1.
3
2
19. f (x) = 2x – 4x – 3x + 4
The changes in sign indicate that there are zeros
between x = –2 and x = –1, between x = 0 and x = 1,
and between x = 2 and x = 3.
c. Relative maxima near x = -0.3. Relative minima
near x = 1.6.
4
20. f (x) = x + 2x – 1
SOLUTION: a.
SOLUTION: a.
b.
The changes in sign indicate that there are zeros
between x = –2 and x = –1, between x = 0 and x = 1,
and between x = 2 and x = 3.
c. Relative maxima near x = -0.3. Relative minima
near x = 1.6.
b. The changes in sign indicate that there are zeros
between x = –2 and x = –1 and between x = 0 and x
= 1.
c. No relative maxima; Relative minima: near x = –1
4
20. f (x) = x + 2x – 1
SOLUTION: a.
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4
2
21. f (x) = x + 8x – 12
SOLUTION: a.
Page 10
b. The changes in sign indicate that there are zeros
between x = –2 and x = –1 and between x = 0 and x
= 1.
5-4 Analyzing
Graphs of Polynomial Functions
c. No relative maxima; Relative minima: near x = –1
4
2
21. f (x) = x + 8x – 12
SOLUTION: a.
b. The changes in sign indicate that there are zeros
between x = –2 and x = –1 and between x = 1 and x
= 2.
c. minima: near x = 0.
22. FINANCIAL LITERACY The average annual
price of gasoline can be modeled by the cubic
3
2
function f (x) = 0.0007x – 0.014x + 0.08x + 0.96,
where x is the number of years after 1987 and f (x) is
the price in dollars.
a. Graph the function for
.
b. Describe the turning points of the graph and its
end behavior.
c. What trends in gasoline prices does the graph
suggest?
d. Is it reasonable that the trend will continue
indefinitely? Explain.
SOLUTION: a.
b. The changes in sign indicate that there are zeros
between x = –2 and x = –1 and between x = 1 and x
= 2.
c. minima: near x = 0.
22. FINANCIAL LITERACY The average annual
price of gasoline can be modeled by the cubic
3
2
function f (x) = 0.0007x – 0.014x + 0.08x + 0.96,
where x is the number of years after 1987 and f (x) is
the price in dollars.
a. Graph the function for
.
b. Describe the turning points of the graph and its
end behavior.
c. What trends in gasoline prices does the graph
suggest?
d. Is it reasonable that the trend will continue
indefinitely? Explain.
SOLUTION: a.
b. Sample answer: The graph has a relative minimum
at x = 10 and then increases as x increases.
c. The graph suggests a fairly steep continuous
increase and gas prices at $5 per gallon by 2012,
which could be possible.
d. Sample answer: While it is possible for gasoline
prices to continue to soar at this rate, it is likely that
alternate forms of transportation and fuel will slow
down this rapid increase.
Use a graphing calculator to estimate the xcoordinates at which the maxima and minima of
each function occur. Round to the nearest
hundredth.
3
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2
23. f (x) = x + 3x – 6x – 6
SOLUTION: Graph the function.
Page 11
which could be possible.
d. Sample answer: While it is possible for gasoline
prices to continue to soar at this rate, it is likely that
5-4 Analyzing
Graphs
of Polynomial
alternate forms
of transportation
andFunctions
fuel will slow
down this rapid increase.
Use a graphing calculator to estimate the xcoordinates at which the maxima and minima of
each function occur. Round to the nearest
hundredth.
3
From the graph, the function has no relative maxima
and relative minima.
4
3
2
25. f (x) = –2x + 5x – 4x + 3x – 7
SOLUTION: Graph the function.
2
23. f (x) = x + 3x – 6x – 6
SOLUTION: Graph the function.
From the graph, the relative maxima is at x = 1.34
and there is no relative minima for the function.
5
3
2
26. f (x) = x – 4x + 3x – 8x – 6
From the graph, the relative maxima is at x = –2.73
and relative minima is at x = 0.73.
3
SOLUTION: Graph the function.
2
24. f (x) = –2x + 4x – 5x + 8
SOLUTION: Graph the function.
From the graph, the relative maxima is at x = –1.87
and relative minima is at x = 1. 52.
From the graph, the function has no relative maxima
and relative minima.
4
3
2
25. f (x) = –2x + 5x – 4x + 3x – 7
SOLUTION: Graph the function.
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Sketch the graph of polynomial functions with
the following characteristics.
27. an odd function with zeros at –5, –3, 0, 2 and 4
SOLUTION: An odd-degree function has an odd number of real
zeros and the end behavior is in opposite directions.
So draw a graph that crosses the x-axis at -5, -3, 0, 2,
and 4. Page 12
5-4 Analyzing
Graphs
Polynomial
From the graph,
the of
relative
maximaFunctions
is at x = –1.87
and relative minima is at x = 1. 52.
Sketch the graph of polynomial functions with
the following characteristics.
27. an odd function with zeros at –5, –3, 0, 2 and 4
SOLUTION: An odd-degree function has an odd number of real
zeros and the end behavior is in opposite directions.
So draw a graph that crosses the x-axis at -5, -3, 0, 2,
and 4. 28. an even function with zeros at –2, 1, 3, and 5
SOLUTION: An even-degree function has an even number of real
zeros and the end behavior is in the same direction.
So draw a graph that crosses the x-axis at -2, 1, 3,
and 5.
29. a 4-degree function with a zero at –3, maximum at x
= 2, and minimum at x = –1
SOLUTION: A 4-degree function has 4 zeros so the graph will
cross the x-axis 4 times. Draw a graph with a
maximum at 2, a minimum at –1. Since this is an
even degree function, the end behavior is in the same
direction. eSolutions Manual - Powered by Cognero
29. a 4-degree function with a zero at –3, maximum at x
= 2, and minimum at x = –1
SOLUTION: A 4-degree function has 4 zeros so the graph will
cross the x-axis 4 times. Draw a graph with a
maximum at 2, a minimum at –1. Since this is an
even degree function, the end behavior is in the same
direction. 30. a 5-degree function with zeros at –4, –1, and 3,
maximum at x = –2
SOLUTION: A 5-degree function has 5 zeros and end behavior in
opposite directions. Draw a graph with zeros at –4, –
1, and 3, and a maximum at x = –2.
31. an odd function with zeros at –1, 2, and 5 and a
negative leading coefficient
SOLUTION: An odd function with a negative leading coefficient
has end behavior in opposite directions and an odd
number of zeros. Draw a graph that has zeros at –1,
2, and 5.
Page 13
33. DIVING The deflection d of a 10-foot-long d diving
board can be calculated using the function d(x) =
5-4 Analyzing Graphs of Polynomial Functions
2
3
0.015x – 0.0005x , where x is the distance between
the diver and the stationary end of the board in feet.
31. an odd function with zeros at –1, 2, and 5 and a
negative leading coefficient
SOLUTION: An odd function with a negative leading coefficient
has end behavior in opposite directions and an odd
number of zeros. Draw a graph that has zeros at –1,
2, and 5.
a. Make a table of values of the function for
.
b. Graph the function.
c. What does the end behavior of the graph suggest
as x increases?
d. Will this trend continue indefinitely? Explain your
reasoning.
SOLUTION: a.
32. an even function with a minimum at x = 3 and a
positive leading coefficient
SOLUTION: An even function with a a positive leading coefficient
has end behavior in the same direction. Draw a
graph with a minimum at x = 3 with an even number
of zeros.
b. Graph of the function.
33. DIVING The deflection d of a 10-foot-long d diving
board can be calculated using the function d(x) =
2
3
0.015x – 0.0005x , where x is the distance between
the diver and the stationary end of the board in feet.
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a. Make a table of values of the function for
.
c. d(x) increases as x increases.
d. No, the value of x cannot be greater than 10Page
as 14
the length of the diving board is 10 ft.
Complete each of the following.
b. Zeros are at: x = –3.5, –1, 0 and 3.
c. Since the graph has 3 turning points, the smallest
possible degree of the polynomial function is (3 + 1)
or 4.
d. Domain: {all real numbers}; Range:
5-4 Analyzing Graphs of Polynomial Functions
c. d(x) increases as x increases.
d. No, the value of x cannot be greater than 10 as
the length of the diving board is 10 ft.
Complete each of the following.
a. Estimate the x-coordinate of every turning
point and determine if those coordinates are
relative maxima or relative minima.
b. Estimate the x-coordinate of every zero.
c. Determine the smallest possible degree of
the function.
d. Determine the domain and range of the
function.
36. SOLUTION: a. Relative maxima: x = – 2 and x = 2.5; Relative
minima: x = 1
b. The zeros are at: x = –3.5, and x = – 0.5.
c. Since the graph has 3 turning points, the smallest
possible degree of the polynomial function is (3 + 1)
or 4.
d. Domain: {all real numbers}; Range:
34. SOLUTION: a. Relative maxima: x = – 3.5 and x = –1; Relative
minima: x = –2.5 and x = 2
b. The zeros are at: x = –1.75, –0.25 and 3.5.
c. Since the graph has 4 turning points, the smallest
possible degree of the polynomial function is (4 + 1)
or 5.
d. Domain: {all real numbers}; Range: {all real
numbers};
37. SOLUTION: a. Relative maxima: x = – 1and x = –2.5; Relative
minima: x = –3.5, x = –2, and x = 1
b. The zeros are at: x = –3.75, –3.25, –2, –1.75, –
0.25, and 2.9.
c. Since the graph has 5 turning points, the smallest
possible degree of the polynomial function is (5 + 1)
or 6.
d. Domain: {all real numbers}; Range:
35. SOLUTION: a. Relative maxima: x = – 0.5; Relative minima: x = –
2.5 and x = 1.5
b. Zeros are at: x = –3.5, –1, 0 and 3.
c. Since the graph has 3 turning points, the smallest
possible degree of the polynomial function is (3 + 1)
or 4.
d. Domain: {all real numbers}; Range:
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38. SOLUTION: a. Relative maxima: x = – 3.5, x = – 1.75 and x = 1;
Relative minima: x = –2.5, x = –1 and x = 2
Page 15
b. The zeros are at: x = –4, –3, 0, 1.5, and 2.75.
c. Since the graph has 6 turning points, the smallest
possible degree of the polynomial function is (6 + 1)
0.25, and 2.9.
c. Since the graph has 5 turning points, the smallest
possible degree of the polynomial function is (5 + 1)
or 6.
5-4 Analyzing
Graphs of Polynomial Functions
d. Domain: {all real numbers}; Range:
c. Since the graph has 2 turning points, the smallest
possible degree of the polynomial function is (2 + 1)
or 3.
d. Domain: {all real numbers}; Range: {all real
numbers};
40. CCSS REASONING The number of subscribers
using pagers in the United States can be modeled by 4
3
2
f (x) = 0.015x – 0.44x + 3.46x – 2.7x + 9.68
where x is the number of years after 1990 and f (x) is
the number of subscribers in millions.
38. SOLUTION: a. Relative maxima: x = – 3.5, x = – 1.75 and x = 1;
Relative minima: x = –2.5, x = –1 and x = 2
b. The zeros are at: x = –4, –3, 0, 1.5, and 2.75.
c. Since the graph has 6 turning points, the smallest
possible degree of the polynomial function is (6 + 1)
or 7.
d. Domain: {all real numbers}; Range: {all real
numbers};
a. Graph the function.
b. Describe the end behavior of the graph.
c. What does the end behavior suggest about the
number of pager subscribers?
d. Will this trend continue indefinitely? Explain your
reasoning.
SOLUTION: a. Use a graphing calculator to graph the function.
39. SOLUTION: a. Relative maxima: x = – 2; Relative minima: x = 1
b. The zeros are at: x = –3, –0.5 and 2.
c. Since the graph has 2 turning points, the smallest
possible degree of the polynomial function is (2 + 1)
or 3.
d. Domain: {all real numbers}; Range: {all real
numbers};
40. CCSS REASONING The number of subscribers
using pagers in the United States can be modeled by 4
3
2
f (x) = 0.015x – 0.44x + 3.46x – 2.7x + 9.68
where x is the number of years after 1990 and f (x) is
the number of subscribers in millions.
a. Graph the function.
b. Describe the end behavior of the graph.
c. What does the end behavior suggest about the
number of pager subscribers?
d. Will this trend continue indefinitely? Explain your
eSolutions Manual - Powered by Cognero
reasoning.
SOLUTION: b. As x increases, f (x) increases.
c. Sample answer: The graph suggests that the
number of pager subscribers will increase
dramatically and continue to increase.
d. Sample answer: The graph is unreasonable for
since pager use is currently decreasing rapidly
and pagers have been replaced by more efficient
products.
41. PRICING Jin’s vending machines currently sell an
average of 3500 beverages per week at a rate of
$0.75 per can. She is considering increasing the
price. Her weekly earnings can be represented by
2
f (x) = –5x + 100x + 2625
where x is the number of $0.05 increases. Graph the
function and determine the most profitable price for
Jin.
SOLUTION: Page 16
Graph the function
.
dramatically and continue to increase.
d. Sample answer: The graph is unreasonable for
since pager use is currently decreasing rapidly
and pagersGraphs
have been
by more
efficient
5-4 Analyzing
of replaced
Polynomial
Functions
products.
41. PRICING Jin’s vending machines currently sell an
average of 3500 beverages per week at a rate of
$0.75 per can. She is considering increasing the
price. Her weekly earnings can be represented by
2
f (x) = –5x + 100x + 2625
where x is the number of $0.05 increases. Graph the
function and determine the most profitable price for
Jin.
From the graph, the graph attains it maximum at x =
10. The most profitable price for Jin is $0.75 + 0.05
(10) or $1.25.
For each function,
a. determine the zeros, x- and y-intercepts, and
turning points,
b. determine the axis of symmetry, and
c. determine the intervals for which it is
increasing, decreasing, or constant.
4
2
42. f (x) = x – 8x + 16
SOLUTION: Graph the function.
SOLUTION: Graph the function
.
From the graph, the graph attains it maximum at x =
10. The most profitable price for Jin is $0.75 + 0.05
(10) or $1.25.
For each function,
a. determine the zeros, x- and y-intercepts, and
turning points,
b. determine the axis of symmetry, and
c. determine the intervals for which it is
increasing, decreasing, or constant.
4
2
42. f (x) = x – 8x + 16
a. The zeros of the function are: x = 2, –2;
x-intercepts: x = 2, –2;
y-intercept: y = 16
Turning points: x = –2, 0, 2
b. The axis of symmetry is x = 0.
c. The function is increasing in the intervals
and and decreasing in 5
and 3
43. f (x) = x – 3x + 2x – 4
SOLUTION: Graph the function.
SOLUTION: Graph the function.
a.
x-intercepts: x = 2, –2;
y-intercept: y = 16
Turning points: x = –2, 0, 2
The
zeros
of the function
eSolutions
Manual
- Powered
by Cogneroare:
x = 2, –2;
a. The zero of the function is:
.
x-intercept:
;
y-intercept: y = –4;
Page 17
Turning points:
;
b. The graph is not symmetric about any line. So,
b. The axis of symmetry is x = 0.
c. The function is increasing in the intervals
and and decreasing in and 5-4 Analyzing Graphs of Polynomial Functions
there is no axis of symmetry.
c. Increasing:
and Decreasing:
5
3
4
and ;
3
44. f (x) = –2x + 4x – 5x
43. f (x) = x – 3x + 2x – 4
SOLUTION: Graph the function.
SOLUTION: Graph the function.
a. The zeros of the function are:
x-intercepts:
;
y-intercept: y = 0;
Turning point:
;
b. No axis of symmetry.
c. Increasing:
Decreasing:
a. The zero of the function is:
.
x-intercept:
;
y-intercept: y = –4;
Turning points:
;
b. The graph is not symmetric about any line. So,
there is no axis of symmetry.
c. Increasing:
and and Decreasing:
4
;
.
45. 3
44. f (x) = –2x + 4x – 5x
SOLUTION: Graph the function.
SOLUTION: a. Graph the function.
a. The zeros of the function are:
x-intercepts:
;
y-intercept: y = 0;
Turning point:
;
b. No axis of symmetry.
c. Increasing:
eSolutions Manual - Powered by Cognero
Decreasing:
.
Zeros: No;
x-intercepts: No;
y-intercept: y = 5;
Turning points: No;
b. No axis of symmetry.
c. Increasing: x > 0
Decreasing:
Constant:
Page 18
Turning point:
;
b. No axis of symmetry.
c. Increasing:
5-4 Analyzing
Graphs of Polynomial Functions
Decreasing:
you make from this new view of the graph?
SOLUTION: a. degree: 4; leading coefficient: 1;
End behavior:
b.
45. SOLUTION: a. Graph the function.
The changes in sign indicate that there are zeros
between x = 1 and x = 2 and between x = 2 and x =
3. So, there are two zeros.
c.
Zeros: No;
x-intercepts: No;
y-intercept: y = 5;
Turning points: No;
b. No axis of symmetry.
c. Increasing: x > 0
Decreasing:
Constant:
46. MULTIPLE REPRESENTATIONS Consider the
following function.
4
3
2
f (x) = x – 8.65x + 27.34x – 37.2285x + 18.27
d.
a. ANALYTICAL What are the degree, leading
coefficient, and end behavior?
b. TABULAR Make a table of integer values f (x) if
How many zeros does the function appear to have from the table?
c. GRAPHICAL Graph the function by using a
graphing calculator.
d. GRAPHICAL Change the viewing window to [0,
4] scl: 1 by [–0.4, 0.4] scl: 0.2. What conclusions can
you make from this new view of the graph?
SOLUTION: a. degree: 4; leading coefficient: 1;
End behavior:
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eSolutions
Sample answer: Sometimes it is necessary to have a
more accurate viewing window or to change the
interval values of the table function in order to assess
the graph more accurately.
47. REASONING Explain why the leading coefficient
and the degree are the only determining factors in the
Page 19
end behavior of a polynomial function.
SOLUTION: Sample answer: Sometimes it is necessary to have a
more accurate viewing window or to change the
interval values
of theoftable
functionFunctions
in order to assess
5-4 Analyzing
Graphs
Polynomial
the graph more accurately.
47. REASONING Explain why the leading coefficient
and the degree are the only determining factors in the
end behavior of a polynomial function.
SOLUTION: As the x-values approach large positive or negative
numbers, the term with the largest degree becomes
more and more dominant in determining the value of
f(x).
48. REASONING The table below shows the values of
g(x), a cubic function. Could there be a zero between
x = 2 and x = 3? Explain your reasoning.
50. CCSS ARGUMENTS Determine whether the
following statement is sometimes, always, or never
true. Explain your reasoning.
For any continuous polynomial function, the ycoordinate of a turning point is also either a
relative maximum or relative minimum.
SOLUTION: Sample answer: Always; the definition of a turning
point of a graph is a point in which the graph stops
increasing and begins to decrease, causing a
maximum or stops decreasing and begins to increase,
causing a minimum.
SOLUTION: Sample answer: No; the cubic function is of degree 3
and cannot have any more than three zeros. Those
zeros are located between –2 and –1, 0 and 1, and 1
and 2.
49. OPEN ENDED Sketch the graph of an odd
polynomial function with 6 turning points and 2 double
roots.
SOLUTION: Sample answer:
Odd polynomial functions have opposite end
behavior. Graph a polynomial function with 6 turning
points. Two of these are double roots.
51. REASONING A function is said to be even if for
every x in the domain of f , f (x) = f (–x). Is every
even-degree polynomial function also an even
function? Explain.
SOLUTION: 2
Sample answer: No; f (x) = x + x is an even degree,
but
.
52. REASONING A function is said to be odd if for
every x in the domain, –f (x) = f (–x). Is every odddegree polynomial function also an odd function?
Explain.
SOLUTION: 3
2
Sample answer: No; f (x) = x + 2x is an odd degree,
but
.
53. WRITING IN MATH How can you use the
characteristics of a polynomial function to sketch its
graph?
SOLUTION: Sample answer: From the degree, you can determine
whether the graph is even or odd and the maximum
number of zeros and turning points for the graph.
You can create a table of values to help you find the
approximate locations of turning points and zeros.
The leading coefficient can be used to determine the
end behavior of the graph, and, along with the
degree, build the shape of the graph.
50. CCSS ARGUMENTS Determine whether the
following statement is sometimes, always, or never
true. Explain your reasoning.
For any continuous polynomial function, the ycoordinate of a turning point is also either a
relative maximum or relative minimum.
SOLUTION: eSolutions
Manual - Powered by Cognero
Sample answer: Always; the definition of a turning
point of a graph is a point in which the graph stops
54. Which of the following is the factorization of 2x – 15
2
+x ?
A. (x – 3)(x – 5)
B. (x – 3)(x + 5)
C. (x + 3)(x – 5)
D. (x + 3)(x + 5)
SOLUTION: First rewrite the equation with the terms in
descending order by degree. Then factor. Page 20
You can create a table of values to help you find the
approximate locations of turning points and zeros.
The leading coefficient can be used to determine the
end behavior
of the of
graph,
and, along
with the
5-4 Analyzing
Graphs
Polynomial
Functions
degree, build the shape of the graph.
54. Which of the following is the factorization of 2x – 15
2
+x ?
A. (x – 3)(x – 5)
B. (x – 3)(x + 5)
C. (x + 3)(x – 5)
D. (x + 3)(x + 5)
2
56. Which polynomial represents (4x + 5x – 3)(2x – 7)?
F 8x3 – 18x2 – 41x – 21
3
2
G 8x + 18x + 29x – 21
H 8x3 – 18x2 – 41x + 21
3
2
J 8x + 18x – 29x + 21
SOLUTION: First rewrite the equation with the terms in
descending order by degree. Then factor. SOLUTION: The correct choice is H.
57. SAT/ACT The figure shows the graph of a
polynomial function f (x). Which of the following
could be the degree of f (x)?
The correct choice is B.
55. SHORT RESPONSE In the figure below, if x = 35
and z = 50, what is the value of y?
A2
B3
C4
D5
E6
SOLUTION: The vertically opposite angles are congruent.
Therefore, the interior angles of the triangle are
By the Triangle Angle-Sum Theorem
SOLUTION: The graph intersects x-axis at four points. Therefore,
the degree of the polynomial is 4. The correct choice
is C.
For each graph,
a. describe the end behavior,
b. determine whether it represents an odddegree or an even-degree function, and
c. state the number of real zeros.
2
56. Which polynomial represents (4x + 5x – 3)(2x – 7)?
F 8x3 – 18x2 – 41x – 21
3
2
G 8x + 18x + 29x – 21
H 8x3 – 18x2 – 41x + 21
3
2
J 8x + 18x – 29x + 21
SOLUTION: eSolutions Manual - Powered by Cognero
The correct choice is H.
57. SAT/ACT The figure shows the graph of a
58. Page 21
SOLUTION: a.
b. Since the end behavior is in the same direction, it
a.
b. Since the end behavior is in the same direction, it
is an even-degree function.
c. The graph intersects the x-axis at six points, so
there are six real zeros.
SOLUTION: The graph intersects x-axis at four points. Therefore,
the degree Graphs
of the polynomial
is 4. The
correct choice
5-4 Analyzing
of Polynomial
Functions
is C.
For each graph,
a. describe the end behavior,
b. determine whether it represents an odddegree or an even-degree function, and
c. state the number of real zeros.
60. 58. SOLUTION: a.
b. Since the end behavior is in the same direction, it
is an even-degree function.
c. The graph intersects the x-axis at two points, so
there are two real zeros.
SOLUTION: a.
b. Since the end behavior is in the same direction,
it is an even-degree function.
c. The graph intersects the x-axis at four points, so
there are four real zeros.
Simplify.
3
2
61. (x + 2x – 5x – 6) ÷ (x + 1)
SOLUTION: Use synthetic division method.
59. SOLUTION: a.
b. Since the end behavior is in the same direction, it
is an even-degree function.
c. The graph intersects the x-axis at six points, so
there are six real zeros.
3
2
62. (4y + 18y + 5y – 12) ÷ (y + 4)
SOLUTION: Use synthetic division method.
3
2
63. (2a – a – 4a) ÷ (a – 1)
SOLUTION: 60. SOLUTION: a.
eSolutions
Manual - Powered by Cognero
b. Since
the end behavior is in the same direction,
it is an even-degree function.
c. The graph intersects the x-axis at four points, so
Page 22
66. y – 5y – 8y + 40
SOLUTION: 5-4 Analyzing
Graphs of Polynomial Functions
3
2
63. (2a – a – 4a) ÷ (a – 1)
SOLUTION: 2
67. a + 6a – 16
SOLUTION: 2
68. b – 4b – 21
SOLUTION: 64. CHEMISTRY Tanisha needs 200 milliliters of a
48% concentration acid solution. She has 60% and
40% concentration solutions in her lab. How many
milliliters of 40% acid solution should be mixed with
60% acid solution to make the required amount of
48% acid solution?
SOLUTION: Let x be the amount of 60% acid solution.
Therefore,
2
69. 6x – 5x – 4
SOLUTION: 2
70. 4x – 7x – 15
SOLUTION: 60% solution : 80 mL
40% solution : 200 – 80 = 120 mL
Factor.
2
65. x + 6x + 3x + 18
SOLUTION: 2
66. y – 5y – 8y + 40
SOLUTION: 2
67. a + 6a – 16
SOLUTION: eSolutions Manual - Powered by Cognero
2
68. b – 4b – 21
SOLUTION: Page 23