Chapter 2.2, Problem 38E

Solution

To draw the graph of the given polynomial.

  

Given:

  $f\left(x\right)=x^3$

Calculation:

Consider,

  $f\left(x\right)=x^3$

The degree is 3 and the leading coefficient is 1.

Here the degree is odd and the leading coefficient is positive.

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

When $x=-3$ ,

  $\begin{array}{c}f\left(3\right)={\left(-3\right)}^3\\ =-27\end{array}$

When $x=-2$ ,

  $\begin{array}{c}f\left(2\right)={\left(-2\right)}^3\\ =-8\end{array}$

When $x=-1$ ,

  $\begin{array}{c}f\left(1\right)={\left(-1\right)}^3\\ =-1\end{array}$

When $x=0$ ,

  $\begin{array}{c}f\left(0\right)=0^3\\ =0\end{array}$

When $x=1$ ,

  $\begin{array}{c}f\left(1\right)=1^3\\ =1\end{array}$

When $x=2$ ,

  $\begin{array}{c}f\left(2\right)=2^3\\ =8\end{array}$

When $x=3$ ,

  $\begin{array}{c}f\left(3\right)=3^3\\ =27\end{array}$

So the table of values is,

$x$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$
$y$	$-27$	$-8$	$-1$	$0$	$1$	$8$	$27$

Now the graph is,