# How to Analyze Polynomial Functions

Let’s look at an example of analysis of a polynomial function. The method is as follows:

Rule

### AnalyzingPolynomialFunctions

1.
Find the zeros.
2.
Find the stationary points.
3.
Find the inflection points.

Example 1

Analyze the function $f\left(x\right)={x}^{3}+6{x}^{2}+8x$

1.
Find the zeros of the function by setting $f\left(x\right)=0$. Since this is a polynomial function of degree 3, you cannot use the quadratic formula to solve it directly. However, as $f\left(x\right)$ does not have a constant term, we can factorize $x$ out of the rest of the function:
$\begin{array}{llll}\hfill f\left(x\right)& ={x}^{3}+6{x}^{2}+8x\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =x\phantom{\rule{-0.17em}{0ex}}\left({x}^{2}+6x+8\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{lll}\hfill f\left(x\right)={x}^{3}+6{x}^{2}+8x=x\phantom{\rule{-0.17em}{0ex}}\left({x}^{2}+6x+8\right)& \phantom{\rule{2em}{0ex}}& \hfill \end{array}$

This means that $x=0$ is a zero. You can then use the quadratic formula formula on ${x}^{2}+6x+8$ to find the other zeros: $\begin{array}{llll}\hfill x& =\frac{-6±\sqrt{\left({6}^{2}-4\cdot 1\cdot 8}}{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{-6±\sqrt{4}}{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{-6±2}{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Thus, $x=-4$ or $x=-2$. The zeros of $f\left(x\right)$ are thus $x=-4$, $x=-2$ and $x=0$.

2.
Find the maxima and minima by setting ${f}^{\prime }\left(x\right)=0$.

Find the derivative of $f\left(x\right)={x}^{3}+6{x}^{2}+8x$:

 ${f}^{\prime }\left(x\right)=3{x}^{2}+12x+8$

Then use the quadratic formula to find the maxima and minima of $f\left(x\right)$: $\begin{array}{llll}\hfill x& =\frac{-12±\sqrt{\left(1{2}^{2}-4\cdot 1\cdot 8}}{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{-12±\sqrt{112}}{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \approx \frac{-6±10.58}{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Thus, $x\approx -3.15$ or $x\approx -0.85$.

To find the points, you need to find their corresponding $y$-values. You find these by putting the $x$-values you found back into the main function $f\left(x\right)$: $\begin{array}{llll}\hfill y& =f\left(-3.15\right)\approx 3.08\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill y& =f\left(-0.85\right)\approx -3.08\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

You now need to determine which point is a maximum and which is a minimum. You do that by drawing a sign chart. Notice that the derivative can be factorized as:

 ${f}^{\prime }\left(x\right)=\left(x+3.15\right)\left(x+0.85\right)$

From this, you can see that the maximum is situated at $\left(-3.15,3.08\right)$ and the minimum is situated at $\left(-0.85,-3.08\right)$.
3.
Find the inflection points by setting ${f}^{″}\left(x\right)=0$.

First, you find the second derivative of $f\left(x\right)$ by differentiating ${f}^{\prime }\left(x\right)=3{x}^{2}+12x+8$:

 ${f}^{″}\left(x\right)=6x+12.$

Let ${f}^{″}\left(x\right)=0$ and solve the equation: $\begin{array}{llll}\hfill 6x+12& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 6x& =-12\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x& =-2\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Enter this $x$-value into the original function $f\left(x\right)$ to find the $y$-coordinate for the inflection point:

$\begin{array}{llll}\hfill f\left(-2\right)& ={\left(-2\right)}^{3}+6\cdot {\left(-2\right)}^{2}+8\cdot \left(-2\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-8+24-16\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{llll}\hfill f\left(-2\right)& ={\left(-2\right)}^{3}+6\cdot {\left(-2\right)}^{2}+8\cdot \left(-2\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-8+24-16\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

The inflection point is thus $\left(-2,0\right)$.

By making a sign chart for the second derivative, you can see where the graph of $f\left(x\right)$ is concave and where it is convex. Notice that ${f}^{″}\left(x\right)$ can be factorized as ${f}^{″}\left(x\right)=6\left(x+2\right)$: