Chapter 3, Problem 79RE

To determine

To graph the rational function and show the x-andy-intercepts and asymptotes.

Explanation of Solution

Given information:

The rational function is $r\left(x\right)=\frac{x-2}{x^2-2x-8}$ .

Graph:

Factor: $r\left(x\right)=\frac{x-2}{(x+2)\left(x-4\right)}$

x-intercepts:the x-intercept are the zeros of the numerator,

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

Sox-intercept is $2$ .

y-intercepts: to find y-intercept, substitute $x=0$ in the original form of the function,

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

So, y-intercept is $\frac{1}{4}$ .

Horizontal asymptote: here the degree of the denominator is greater than numerator,

So horizontal asymptote is $y=0$ .

Vertical asymptote: the vertical asymptote is occurs where denominator is zero,

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

So the vertical asymptotes is $x=4,-2$

Slant asymptote: the slant asymptote is occurs when the numerator degree is one more than denominator,here the degree of the denominator is greater than numerator,

So slant asymptote does not exist.

Use the above information together with some additional values which is show in table below

To sketch the graph,

x	y
-1	-0.6
0	-0.25
1	0.11
2	0

The graph is obtained as:

Interpretation:

From the above graph it can be observed that the x-intercept is $2$ , y-intercept is $\frac{1}{4}$ , horizontal asymptote $y=0$ and vertical asymptote is $x=-2,4$ and slant asymptote does not exist.