Chapter 3.2, Problem 25E

To determine

To graph: the polynomial function

Explanation of Solution

Given information: the function is $P\left(x\right)=x^3\left(x+2\right){\left(x-3\right)}^2$ .

Graph:

The $x$ and $y$ intercepts of the graph of $P\left(x\right)=x^3\left(x+2\right){\left(x-3\right)}^2$ function is :

For $x$ intercept, solve equation $P\left(x\right)=0$ for $x$

  $\begin{array}{l}P\left(x\right)=0\\ x^3\left(x+2\right){\left(x-3\right)}^2=0\\ x^3=0\text{ or }\left(x+2\right)=0\text{ or }{\left(x-3\right)}^2=0\\ x=0\text{ or }(x+2)=0\text{ or (}x-3)=0\\ x=0\text{ or }x=-2\text{ or }x=3\\ \end{array}$

Therefore if $x=0$ and estimate $P\left(0\right)$ then the $y$ intercept is

  $\begin{array}{l}P(0)=0^3\left(0+2\right){\left(0-3\right)}^2\\ P\left(0\right)=0\left(2\right){\left(-3\right)}^2\\ P(0)=0\end{array}$

Now multiply the leading terms of each factor of polynomials to get leading term of polynomial $P\left(x\right)=x^3\left(x+2\right){\left(x-3\right)}^2$

Leading terms: $x^2\cdot x^3=x^5$

Now, $P$ is of odd degree and it leading coefficient is positive so it has the following end behaviours

  $y\to \infty \text{ as }x\to \infty \text{ and }y\to -\infty \text{ as }x\to -\infty$

Finally draw a graph of polynomial function plot the point $\left(-2,0\right),(0,0),\left(3,0\right)$ and connect them with smooth curve.

  

Interpretation:

From the above, it can be observed that the graph of $P\left(x\right)=x^3\left(x+2\right){\left(x-3\right)}^2$ with coordinates $\left(-2,0\right),(0,0),\left(3,0\right)$ .