2-2 Polynomial Functions
Graph each function.
6. f (x) = 32x5 – 16
SOLUTION: 5
5
The graph of f (x) = 32x − 16 is the graph of y = x stretched vertically by a factor of 32 and translated 16 units
down.
ANSWER: Describe the end behavior of the graph of each polynomial function using limits. Explain your reasoning
using the leading term test.
12. f (x) = −5x7 + 6x4 + 8
SOLUTION: The degree is 7, and the leading coefficient is −5. Because the degree is odd and the leading coefficient is negative,
ANSWER: The degree is 7, and the leading coefficient is −5. Because the degree is odd and the leading coefficient is negative,
18. f (x) = x(x + 1)(x – 3)
SOLUTION: The degree is 3, and the leading coefficient is 1. Because the degree is odd and the leading coefficient is positive,
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ANSWER: The degree is 3, and the leading coefficient is 1. Because the degree is odd and the leading coefficient is positive,
ANSWER: The degree is 7, and the leading coefficient is −5. Because the degree is odd and the leading coefficient is negative,
2-2 Polynomial Functions
18. f (x) = x(x + 1)(x – 3)
SOLUTION: The degree is 3, and the leading coefficient is 1. Because the degree is odd and the leading coefficient is positive,
ANSWER: The degree is 3, and the leading coefficient is 1. Because the degree is odd and the leading coefficient is positive,
State the number of possible real zeros and turning points of each function. Then determine all of the
real zeros by factoring.
24. f (x) = x6 – 8x5 + 12x4
SOLUTION: The degree of f (x) is 6, so it will have at most six real zeros and five turning points.
So, the zeros are 0, 2, and 6.
ANSWER: 6 real zeros and 5 turning points; 0, 6, and 2
30. f (x) = 6x5 – 150x3
SOLUTION: The degree of f (x) is 5, so it will have at most five real zeros and four turning points.
So, the zeros are 0, 5, and −5.
ANSWER: 5 real zeros and 4 turning points; 0, 5, and –5
For each function, (a) apply the leading-term test, (b) determine the zeros and state the multiplicity of
any repeated zeros, (c) find a few additional points, and then (d) graph the function.
36. f (x) = 2x(x + 5)2(x – 3)
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a.
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5 real zeros and 4 turning points; 0, 5, and –5
For each function, (a) apply the leading-term test, (b) determine the zeros and state the multiplicity of
repeated zeros,
(c) find a few additional points, and then (d) graph the function.
2-2 any
Polynomial
Functions
2
36. f (x) = 2x(x + 5) (x – 3)
SOLUTION: a.
The degree is 4, and the leading coefficient is 2. Because the degree is even and the leading coefficient is positive,
b. The zeros are 0, −5, and 3. The zero −5 has multiplicity 2 since (x + 5) is a factor of the polynomial twice.
c. Sample answer: Evaluate the function for a few x-values in its domain.
Choose x-values that fall in the intervals determined by the zeros of the function. Interval
x
−6
−1
1
4
(–∞, –5)
(–5, 0)
(0, 3)
(3, ∞)
f(x)
108
128
−144
648
d. Evaluate the function for several x-values in its domain.
x
−5
−4
−3
−2
0
2
3
f(x)
0
56
144
180
0
−196
0
Use these points to construct a graph.
ANSWER: a. The degree is 4, and the leading coefficient is 2. Because the degree is even and the leading coefficient is positive,
b. 0, −5 (multiplicity:2), 3
c. Sample answer: (−6, 108), (−1, 128), (1, −144), (4, 648)
d.
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a. The degree is 4, and the leading coefficient is 2. Because the degree is even and the leading coefficient is positive,
b.
0, −5 (multiplicity:2), 3
2-2 c.
Polynomial
Functions
Sample answer: (−6, 108), (−1, 128), (1, −144), (4, 648)
d.
42. f (x) = x5 + 3x4 – 10x3
SOLUTION: a. The degree is 5, and the leading coefficient is 1. Because the degree is odd and the leading coefficient is positive,
b.
The zeros are 0, −5, and 2. The zero 0 has multiplicity 3 since x is a factor of the polynomial three times.
c. Sample answer: Evaluate the function for a few x-values in its domain.
Choose x-values that fall in the intervals determined by the zeros of the function. Interval
(–∞, –5)
(–5, 0)
(0, 2)
(2, ∞)
x
−6
−3
1
3
f(x)
−1728
270
−6
216
d. Evaluate the function for several x-values in its domain.
x
−5
−4
−2
−1
0
2
f(x)
0
384
96
12
0
0
Use these points to construct a graph.
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−2
−1
0
2
96
12
0
0
2-2 Polynomial Functions
Use these points to construct a graph.
ANSWER: a. The degree is 5, and the leading coefficient is 1. Because the degree is odd and the leading coefficient is positive,
b. 0 (multiplicity: 3), −5, 2
c. Sample answer: (−6, −1728), (−3, 270), (1, −6), (3, 216)
d.
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