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G RAPHING P OLYNOMIAL F UNCTIONS. T HE P ROCESS Polynomials can be complicated functions, but there is a process you can use to make it easier to graph.

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Presentation on theme: "G RAPHING P OLYNOMIAL F UNCTIONS. T HE P ROCESS Polynomials can be complicated functions, but there is a process you can use to make it easier to graph."— Presentation transcript:

1 G RAPHING P OLYNOMIAL F UNCTIONS

2 T HE P ROCESS Polynomials can be complicated functions, but there is a process you can use to make it easier to graph them. 1. Use the leading coefficient test to determine what the graph does as its ends. 2. Find the zeroes of the equation. 3. Find the y-intercept of the equation. 4. Determine symmetry. 5. Determine how many turning points there can be. 6. Find any additional points that you need and graph the function.

3 E XAMPLE Let’s go through the entire process and graph the function x 3 – 2x 2 – 5x + 6. The first step is to find the end behavior of the function. The degree of the function is odd and its leading coefficient (1) is positive, so it increases without bound on the right side and decreases without bound on the left.

4 E XAMPLE The second step is to find the zeroes of the function. There are several ways to do so – synthetic division, estimation using the intermediate value theorem, and simple guess and check. However you choose to do so, the zeroes of x 3 – 2x 2 – 5x + 6 are -2, 1, and 3. Let’s plot these on our axes.

5 E XAMPLE Next, we find the y intercept of the function by setting x = 0. We obtain y = 6, so our y intercept is 6. Let’s add these to our plot.

6 E XAMPLE Next, we test for symmetry. Because of the x and x 3 term, f(-x) ≠ f(x), and because of the x 2 and constant terms, -f(x) ≠ f(x). There is no symmetry in this function.

7 E XAMPLE Next, we want to determine how many turning points there can be. We can have up to (degree – 1) turning points, so we can have up to 2.

8 E XAMPLE Finally, we want to find whatever other points we need for our graph. We’re not sure what the function does between -2 and 0 or between 1 and 3, so we’ll find f(-1) and f(2). f(-1) = 8 f(2) = -4 Now we can graph the function using all the information we’ve gathered about it.

9 F INAL G RAPH Our result is shown on the left. Note the two turning points and its behavior at the ends.


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