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### Section 5.3 : Graphing Polynomials

1. Sketch the graph of each of the following polynomial.

$f\left( x \right) = {x^3} - 2{x^2} - 24x$

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Start Solution

The first step is to determine the zeroes of the polynomial and the multiplicity of each zero. For this problem that means we’ll need to start off factoring the polynomial.

$f\left( x \right) = {x^3} - 2{x^2} - 24x = f\left( x \right) = x\left( {{x^2} - 2x - 24} \right) = x\left( {x - 6} \right)\left( {x + 4} \right)$

We have the following list of zeroes and multiplicities.

\begin{align*}x & = - 4 : {\mbox{ multiplicity 1}}\\ x & = 0 : {\mbox{ multiplicity 1}}\\ x & = 6 : {\mbox{ multiplicity 1}}\end{align*}

Because the multiplicity of each of the zeroes is odd we know that each will correspond to an $$x$$-intercept that will cross the $$x$$-axis (as opposed to just touching the $$x$$-axis without actually crossing).

Show Step 2

The $$y$$-intercept for the function is : $$\left( {0,f\left( 0 \right)} \right) = \left( {0,0} \right)$$. In this case the point is also an $$x$$‑intercept. This will happen on occasion so we’ll need to deal with it.

Show Step 3

The coefficient of the 3rd degree term is positive and so we know that the polynomial will increase without bound at the right end and decrease without bound at the left end.

Show Step 4

Next, we need to pick a couple of points to graph. We’ll pick one point on the left and right end and it doesn’t matter which points to pick here so just pick one close to the zero on either end. We’ll also pick a point between each pair of zeroes and again it doesn’t really matter which points to pick so we’ll just pick points that are near the middle of each pair of zeroes.

Here are the function evaluations for this polynomial. We’ll leave it to you to verify the evaluations.

$f\left( { - 5} \right) = - 55\hspace{0.25in}f\left( { - 2} \right) = 32\hspace{0.25in}f\left( 3 \right) = - 63\hspace{0.25in}f\left( 7 \right) = 77$ Show Step 5

Finally, here is a quick sketch of the polynomial.

Note that your graph may not look quite like this one.

This is a computer generated graph and as such is very accurate. Using only Algebra techniques it can be quite difficult to approach this kind of accuracy. As long as the basic behavior of your graph is correct and you’ve got the extra points correctly graphed then you should have something that is close enough to the actual graph to work for a quick sketch.

For those interested, there are several techniques from a Calculus class that allow for a much more accurate sketch of the polynomial!