Chapter 3.7, Problem 51E

To determine

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

Answer to Problem 51E

The rational function has $x$ -intercept is $-2$ , $y$ -intercept is $-\frac{3}{4}$ , vertical asymptote is $x=2\text{ and }x=-4$ , horizontal asymptote is $y=0$ , domain is $\left\{x|x\ne 2,-4\right\}$ and range is $ℝ$ .

Explanation of Solution

Given information :

The rational function is $t\left(x\right)=\frac{3x+6}{x^2+2x-8}$

Calculation:

Intercept:

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

  $\begin{array}{l}3x+6=0\\ 3x=-6\\ x=-2\end{array}$

So, $x$ -interceptis $-2$ .

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

  $\begin{array}{l}t\left(0\right)=\frac{3\left(0\right)+6}{{\left(0\right)}^2+2\left(0\right)-8}\\ t\left(0\right)=\frac{6}{-8}\\ t\left(0\right)=-\frac{3}{4}\end{array}$

The $y$ -intercept is $-\frac{3}{4}$ .

Vertical asymptote:

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

  $x^2+2x-8=0$

Use Quadratic formula to solve the above equation,

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Here, $a=1,b=2\text{ and }c=-8$

  $\begin{array}{l}x=\frac{-\left(2\right)\pm \sqrt{{\left(2\right)}^2-4\left(1\right)\left(-8\right)}}{2\left(1\right)}\\ x=\frac{-2\pm \sqrt{4+32}}{2}\\ x=\frac{-2\pm \sqrt{36}}{2}\\ x=\frac{-2\pm 6}{2}\end{array}$

Either,

  $\begin{array}{l}x=\frac{-2+6}{2}\\ x=\frac{4}{2}\\ x=2\end{array}$

Or,

  $\begin{array}{l}x=\frac{-2-6}{2}\\ x=\frac{-8}{2}\\ x=-4\end{array}$

The vertical asymptote is a line $x=2\text{ and }x=-4$ .

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 (by choosing some test point) to sketch the graph.

To check the result graph the rational function using graphing calculator and all the result calculated is correct.

The graph is obtained as:

  

From the above graph, it can be observed that the domain of the rational function is $\left\{x|x\ne 2,-4\right\}$ and range is $ℝ$ .

Hence,

The rational function has $x$ -interceptis $-2$ , $y$ -intercept is $-\frac{3}{4}$ , vertical asymptote is $x=2\text{ and }x=-4$ , horizontal asymptote is $y=0$ , domain is $\left\{x|x\ne 2,-4\right\}$ and range is $ℝ$ .