Chapter 5.1, Problem 34PPSE

Solution

To Describe: The shape of the graph of each cubic function including end behaviour, turning points, and increasing/decreasing intervals.

The graph's end behaviour is rises to the left and falls to the right.

There are no turning points:

Increasing on: $\left(-\infty ,0\right),\left(0,\infty \right)$

Given:

  $y=10x^3+9$

Explanation:

Given the function $y=10x^3+9$

To create a table of values, substitute values to the equation.

Substitute $x=-1$ in $y=10x^3+9$ :

  $\begin{array}{c}y=10{\left(-1\right)}^3+9\\ =-1\end{array}$

Substitute $x=-0.5$ in $y=10x^3+9$ :

  $\begin{array}{c}y=10{\left(-0.5\right)}^3+9\\ =7.75\end{array}$

Substitute $x=0$ in $y=10x^3+9$ :

  $\begin{array}{c}y=10{\left(0\right)}^3+9\\ =9\end{array}$

Substitute $x=0.5$ in $y=10x^3+9$ :

  $\begin{array}{c}y=10{\left(0.5\right)}^3+9\\ =10.25\end{array}$

Substitute $x=0.5$ in $y=10x^3+9$ :

  $\begin{array}{c}y=10{\left(1\right)}^3+9\\ =19\end{array}$

$x$	$y$
$-1$	$-1$
$-0.5$	$7.75$
$0$	$9$
$0.5$	$10.25$
$1$	$19$

Graph of the function is as:

  

Find the end behaviour of $y=10x^3+9$ :

The leading coefficient in a polynomial is the coefficient of the leading term which is $10$ .

Since the degree is odd, the ends of the function will point in the opposite directions.

Since the leading coefficient is positive, the graph rises to the right.

The graph's end behaviour is rises to the left and falls to the right.

To find the turning points, set $y^{\prime }\left(x\right)=0$

  $\begin{array}{l}y=10x^3+9\\ y^{\prime }=30x^2\end{array}$

Set the equation to $0$ :

  $\begin{array}{c}30x^2=0\\ x^2=0\end{array}$

Thus, there are no turning points.

Determine the interval of increasing/decreasing:

By using the graph, the graph is

Increasing on: $\left(-\infty ,0\right),\left(0,\infty \right)$