Chapter 4, Problem 35RE

To determine

To analyze: The graph of Rational function $R(x)=\frac{x^2+x-6}{x^2-x-6}$ .

Answer to Problem 35RE

The domain is $\left\{x|x\ne -2,3\right\}$ .

The vertical asymptote is $x=-2,3$ .

The horizontal asymptote is $y=1$.

Explanation of Solution

Given information:

The rational function $R(x)=\frac{x^2+x-6}{x^2-x-6}$ .

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 ,

  $R(x)=\frac{x^2+x-6}{x^2-x-6}$

Numerator:-

  $\begin{array}{l}=x^2+x-6\\ =x^2+3x-2x-6\\ =x(x+3)-2(x+3)\\ =(x-2)(x+3)\end{array}$

Denominator:-

  $\begin{array}{l}=x^2-x-6\\ =x^2-3x+2x-6\\ =x(x-3)+2(x-3)\\ =(x-3)(x+2)\end{array}$

Therefore,

  $R(x)=\frac{(x-2)(x+3)}{(x+2)(x-3)}$

Step 1.

The domain of R is the set of all real numbers except $x=-2,3$ .

Step 2.

Now, R is in the lowest terms.

Step 3.

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

x -intercept of the graph R

  $=(x-2)(x+3)=0$

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

The x -intercept is $x=2,-3$ .

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

  $\begin{array}{l}R(x)=\frac{(x-2)(x+3)}{(x+2)(x-3)}\\ R(0)=\frac{(0-2)(0+3)}{(0+2)(0-3)}\end{array}$

  $\begin{array}{l}=\frac{(-2)(3)}{(2)(-3)}\\ =1\end{array}$

The y -intercept is $y=1$ .

Step 4.

Denominator:-

  $=(x+2)(x-3)=0$

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

The real zeros of the denominator R is $x=-2,3$ . So, the graph of the R has two vertical asymptote $x=-2,3$ .

Step 5.

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

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

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

Now, solving the equation we get,

  $R(x)=\frac{x^2+x-6}{x^2-x-6}$

  $\begin{array}{l}=\frac{x^2+x-6}{x^2-x-6}=1\\ =x^2+x-6=x^2-x-6\\ =x^2+x-6-x^2+x+6=0\\ =2x=0\\ =x=0\end{array}$

.

This implies the graph intersect the line $y=1$ at $x=0$ is a horizontal asymptote.

Hence, $\left(0,1\right)$ is a point on the graph of R .

Step 6.

  

Figure 1.

Step 7.

Using the information gathered from Step 1 − Step 5 the graph of R has been drawn. The green color shows the graph R(x). The graph has the vertical asymptote at $x=-2,3$ and horizontal asymptote at $y=1$ .The line $y=1$ intersects at $x=0$ . Also, the graph intersects

x -intercept at point $\left(-3,0\right)$ , $\left(2,0\right)$ .