Chapter 4.8, Problem 13E

To determine

To sketch: the graph of the function.

Explanation of Solution

Given information:

Given function,

  $h\left(x\right)=\left(x-3\right)\left(x^2+x+1\right)$

Calculation:

Consider the function $h\left(x\right)=\left(x-3\right)\left(x^2+x+1\right)$ :

Plot the $x$ intercepts.

Because 3is zeros of $h\left(x\right)$ , so plot points $\left(3,0\right)$ .

Plot points between and beyond the $x$ intercept.

$x$	$-1$	0	1	2
$h\left(x\right)$	$-4$	$-3$	$-6$	$-7$

Determine end behavior.

Write function as,

  $\begin{array}{c}h\left(x\right)=x\left(x^2+x+1\right)-3\left(x^2+x+1\right)\\ =x^3-2x^2-2x-3\end{array}$

It is a cubic function with positive leading coefficient.

So, $h\left(x\right)\to -\infty$ as $x\to -\infty$ and $h\left(x\right)\to \infty$ as $x\to \infty$ .

Draw the graph so that it passes through the plotted points and has the appropriate end behavior.