Chapter 4, Problem 31RE

To determine

To calculate: The domain of rational function. Also, to find out the horizontal, vertical , or oblique asymptote.

Answer to Problem 31RE

The domain is $\left\{x|x\ne -3,3\right\}$ .

The vertical asymptote is $x=-3,3$ .

The horizontal asymptote is $y=0$.

Explanation of Solution

Given information:

The rational function $R(x)=\frac{x+2}{x^2-9}$ .

Formula used:

Vertical Asymptote: A rational function $R(x)=\frac{p(x)}{q(x)}$ , in lowest terms, will have a

asymptote $x=r$ if $r$ is a real zero of the denominator q .

Horizontal Asymptote: If a rational function is proper , the line $y=0$ is a asymptote of the graph.

Oblique Asymptote : it occur when the degree of the denominator of a rational

function is one less than the degree of the numerator. The asymptote will be along line

  $y=mx+b$ where $m\ne 0$

Calculation:

Consider ,

  $R(x)=\frac{x+2}{x^2-9}$

The domain of R is the set of all real numbers except $x=3,-3$ .

Now, R is in the lowest terms.

Denominator:-

  $\begin{array}{l}=x^2-9=0\\ =x^2=9\\ =x=\sqrt{9}\\ =x=\pm 3\end{array}$

The line $x=-3$ and $x=3$ are the vertical asymptote of the graph of R .

Since the degree of the numerator ,1,is less than the degree of denominator 2.The rational function is proper. The line $y=0$ is the horizontal asymptote of the graph of R .

Hence, the domain of R is $\left\{x|x\ne 3,-3\right\}$ .

The function has vertical and horizontal asymptote.

Vertical Asymptote at $x=-3,3$ and Horizontal Asymptote at $y=0$ .