1. 4.2 Graphing Polynomial Functions
Graph polynomial functions.
Use the intermediate value theorem to determine whether a function has a real zero between two given real numbers.
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2. Graphing Polynomial Functions
If P(x) is a polynomial function of degree n, the graph of the function has:
 at most n real zeros, and thus at most n x- intercepts;
 at most n  1 turning points.
(Turning points on a graph, also called relative maxima and minima, occur when the function changes from decreasing to increasing or from increasing to decreasing.)
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3. Steps to Graph a Polynomial Function
1. Use the leading-term test to determine the end behavior.
2. Find the zeros of the function by solving f (x) = 0. Any real zeros are the first coordinates of the x-intercepts.
3. Use the x-intercepts (zeros) to divide the x-axis into intervals and choose a test point in each interval to determine the sign of all function values in that interval.
4. Find f (0). This gives the y-intercept of the function.
5. If necessary, find additional function values to determine the general shape of the graph and then draw the graph.
6. As a partial check, use the facts that the graph has at most n x-intercepts and at most n  1 turning points. Multiplicity of zeros can also be considered in order to check where the graph crosses or is tangent to the x-axis. We can also check the graph with a graphing calculator.
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Example
Graph the polynomial function
f (x) = 2x3 + x2  8x  4.
Solution:
1. The leading term is 2x3. The degree, 3, is odd, the coefficient, 2, is positive. Thus the end behavior of the graph will appear as:
2. To find the zero, we solve f (x) = 0. Here we can use factoring by grouping.
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Example continued
Factor:
The zeros are 1/2, 2, and 2. The x-intercepts are (2, 0), (1/2, 0), and (2, 0).
3. The zeros divide the x-axis into four intervals:
(, 2), (2, 1/2), (1/2, 2), and (2, ).
We choose a test value for x from each interval and find f(x).
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6. Example continued Above x-axis + 35 3 (2, ) Below x-axis  9 1
(1/2, 2) 1
(2, 1/2)
25 3
(, 2)
Location of points on graph
Sign of f(x)
Function value, f(x)
Test Value, x
Interval
4. To determine the y-intercept, we find f(0):
The y-intercept is (0, 4).
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Example continued
5. We find a few additional points and complete the graph.
6. The degree of f is 3. The graph of f can have at most 3 x-intercepts and at most 2 turning points. It has 3 x-intercepts and 2 turning points. Each zero has a multiplicity of 1; thus the graph crosses the x-axis at 2, 1/2, and 2. The graph has the end behavior described in step (1). The graph appears to be correct. 7
1.5
3.5
1.5 9
2.5
f(x) x
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Example
Graph the polynomial function
g (x) = x4 – 7x3 + 12x2 + 4x  16.
Solution:
1. The leading term is x4. The degree, 4, is even, the coefficient, 1, is positive. Thus the end behavior of the graph will appear as:
2. To find the zero, we solve g (x) = 0.
The zeros are 1, 2 and 4; 2 is of multiplicity 2; the other are multiplicity 1. The x-intercepts are (1, 0), (2, 0), and (4, 0).
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Example continued
3. The zeros divide the x-axis into four intervals:
(‒∞, ‒1), (‒1, 2), (2, 4), and (4, ∞)
Choose a test value for x from each interval.
Interval
(‒∞, ‒1)
(‒1, 2)
(2, 4)
(4, ∞)
Test Value ‒2 1 3 5
Function Value 96 ‒6 ‒4 54
Sign of g(x) + ‒
Location of Points on Graph
Above x-axis
Below x-axis
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Example continued
The test values and the corresponding function values also give us four points on the graph.
(‒2, 96), (1, ‒6), (3, ‒4), and (5,54)
4. To find the y-intercept, we find g(0).
The y-intercept is (0, ‒16)
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Example continued
5. We find a few additional points and complete the graph.
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12. Intermediate Value Theorem
For any polynomial function P(x) with real coefficients, suppose that for a  b, P(a) and P(b) are of opposite signs. Then the function has a real zero between a and b.
Example: Using the intermediate value theorem, determine, if possible, whether the function has a real zero between a and b.
a) f(x) = x3 + x2  8x; a = 4 b = 1
b) f(x) = x3 + x2  8x; a = b = 3
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Solution
We find f(a) and f(b) and determine where they differ in sign. The graph of f(x) provides a visual check.
f(4) = (4)3 + (4)2  8(4)
= 16
f(1) = (1)3 + (1)2  8(1)
= 8
By the intermediate value theorem, since f(4) and f(1) have opposite signs, then f(x) has a zero between 4 and 1.
y = x3 + x2  8x
zero
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Solution
f(1) = (1)3 + (1)2  8(1) = 6
f(3) = (3)3 + (3)2  8(3) = 12
By the intermediate value theorem, since f(1) and f(3) have opposite signs, then f(x) has a zero between 1 and 3.
y = x3 + x2  8x
zero
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