Chapter 4, Problem 38RE

To determine

To analyze: The graph of Rational function $G(x)=\frac{x^2-4}{x^2-x-2}$ .

Answer to Problem 38RE

The domain is $\left\{x|x\ne -1\right\}$ .

The vertical asymptote is $x=-1$ .

The horizontal asymptote is $y=1$.

Explanation of Solution

Given information:

The rational function $G(x)=\frac{x^2-4}{x^2-x-2}$ .

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 ,

  $G(x)=\frac{x^2-4}{x^2-x-2}$

Numerator:-

  $\begin{array}{l}=x^2-4\\ ={(}^x-{(}^2\\ =(x-2)(x+2)\end{array}$

Denominator:-

  $\begin{array}{l}=x^2-x-2\\ =x^2-2x+x-2\\ =x(x-2)+1(x-2)\\ =(x-2)(x+1)\end{array}$

Therefore,

  $G(x)=\frac{(x-2)(x+2)}{(x-2)(x+1)}$

  $=\frac{(x+2)}{(x+1)}$

Step 1.

The domain of G is the set of all real numbers except $x=-1$ .

Step 2.

Now, G is in the lowest terms.

Step 3.

Now, the intercepts of the graph G are as follows:

x -intercept of the graph G

  $=(x+2)=0$

  $=x=-2$

The x -intercept is $x=-2$ .

y -intercept:- Since 0 is included in domain.

  $\begin{array}{l}G(x)=\frac{(x+2)}{(x+1)}\\ G(0)=\frac{(0+2)}{(0+1)}\end{array}$

  $=2$

Thus, y -intercept is $y=2$ .

Step 4.

Denominator:-

  $=(x+1)=0$

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

The real zeros of the denominator G is $x=-1$ . So, the graph of the G has one vertical asymptote $x=-1$ .

Step 5.

Since the degree of the numerator ,2,is equal to the degree of denominator 2.

The quotient of G is the leading coefficient of the numerator,1, and the leading coefficient of the denominator ,1.

The horizontal asymptote of the graph G is $y=1$ .

Now, solving the equation we get,

  $G(x)=\frac{x^2-4}{x^2-x-2}$

  $\begin{array}{l}=\frac{x^2-4}{x^2-x-2}=1\\ =x^2-4=x^2-x-2\\ =x^2-4-x^2+x+2=0\\ =x-2=0\\ =x=2\end{array}$

.

This implies the graph intersect the line $y=1$ at $x=2$ is a horizontal asymptote.

Hence, $\left(2,1\right)$ is a point on the graph of G .

Step 6.

  

Figure 1.

Step 7.

Using the information gathered from Step 1 − Step 5 the graph of G has been drawn. The green color shows the graph G(x). The graph has the vertical asymptote at $x=-1$ and horizontal asymptote at $y=1$ . The graph intersect at x -intercept and y -intercept point at $\left(-2,0\right)$ and $\left(0,2\right)$ respectively.