Chapter 3.7, Problem 47E

To determine

To find the intercept, asymptotes, domain, range and sketch the graph of the rational function.

Answer to Problem 47E

The rational function has $x$ -intercept $2$ , $y$ -intercept is $2$ , vertical asymptote is $x=4\text{ and }x=-1$ , horizontal asymptote is $y=0$ , domain is $\left\{x|x\ne -1,4\right\}$ and range is $ℝ$ .

Explanation of Solution

Given information :

The rational function is $s\left(x\right)=\frac{4x-8}{\left(x-4\right)\left(x+1\right)}$

Calculation:

Intercept:

The $x$ -intercept are the zeros of the numerator,

  $\begin{array}{l}4x-8=0\\ 4x=8\\ x=2\end{array}$

So, $x$ -intercept is $2$ .

To find $y$ -intercept, substitute $x=0$ into original form of the function,

  $\begin{array}{l}s\left(0\right)=\frac{4\left(0\right)-8}{\left(0-4\right)\left(0+1\right)}\\ s\left(0\right)=\frac{4\left(0\right)-8}{\left(0-4\right)\left(0+1\right)}\\ s\left(0\right)=\frac{-8}{-4}\\ s\left(0\right)=2\end{array}$

The $y$ -intercept is $2$ .

Vertical asymptote:

The vertical asymptote occurs where the denominator is 0, that is, where the function is undefined.

  $\left(x-4\right)\left(x+1\right)=0$

Either,

  $\begin{array}{l}x-4=0\\ x=4\end{array}$

Or,

  $\begin{array}{l}x+1=0\\ x=-1\end{array}$

The vertical asymptote is a line $x=4$ and $x=-1$ .

Horizontal asymptote:

Here, the degree of the numerator is $n=1$ and denominator is $m=2$ .

Since, $n<m$ ,

So, the horizontal asymptote is the line $y=0$ .

Use the above information together with some additional values which is show in table below to sketch the graph:

$x$	$r\left(x\right)$
$-8$	$-0.48$
$-6$	$-0.64$
$-4$	$-1$
$-2$	$-2.67$
$6$	$1.14$
$8$	$0.67$

The graph is obtained as:

  

From the above graph, it can be observed that the domain of the rational function is $\left\{x|x\ne -1,4\right\}$ and range is $ℝ$ .

To check the result graph the rational function using graphing calculator,

  

Hence,

The rational function has $x$ -intercept $2$ , $y$ -intercept is $2$ , vertical asymptote is $x=4\text{ and }x=-1$ , horizontal asymptote is $y=0$ , domain is $\left\{x|x\ne -1,4\right\}$ and range is $ℝ$ .