91Ó°ÊÓ

The given function is $R\left(x\right)=\frac{x^2+x-6}{x^2-x-6}$

Factor the numerator and denominator.

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

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

Since 0 is in the domain, find the y-intercept.

$$\begin{gathered} R\left(0\right)=\frac{(0+3)(0-2)}{(0-3)(0+2)} \\ =\frac{-6}{-6} \\ =1 \end{gathered}$$

Thus, the y-intercept is 1

The real zero of the numerator satisfies the equation$x+3=0,x-2=0$

Hence, the only x-intercept is $-3and2$.

Determine the behavior of the graph near the x-intercept by substituting $-3$in the denominator.

$$\begin{gathered} R\left(x\right)=\frac{(x+3)(x-2)}{(-3-3)\left(\right(-3+2)} \\ =\frac{(x+3)(x-2)}{6} \\ =\frac{1}{6}(x+3)(x-2) \end{gathered}$$

Next substitute $x=2$in the denominator.

$$\begin{gathered} R\left(x\right)=\frac{(x+3)(x-2)}{(2-3)(2+2)} \\ =\frac{(x+3)(x-2)}{-4} \\ =-\frac{1}{4}(x+3)(x-2) \end{gathered}$$

Plot the point $(-3,0)$and indicate a line with slope $\frac{1}{6}$on the graph and also the point $(2,0)$and indicate the line with slope role="math" localid="1646214852708" $-\frac{1}{4}$through that point.

The real zeroes of the denominator are the real solutions of the equation $(x-3)(x+2)=0$

The two lines $x=3,x=-2$are the two vertical asymptotes of the graph.

The degree of numerator is equal to the degree of denominator. So, the graph has a horizontal asymptote.

The leading coefficient of numerator and denominator is 1.

Thus, the horizontal asymptote is$y=1$

Substitute $R\left(x\right)=1$to determine whether the graph intersects the horizontal asymptote.

$$\begin{gathered} \frac{(x+3)(x-2)}{(x-3)(x+2)}=1 \\ (x+3)(x-2)=(x-3)(x+2) \\ {x}^2+x-6=x^2-x-6 \\ x-6=-x-6 \\ x=-x \end{gathered}$$

This condition is satisfied only when the x takes the value 0. Thus, only solution is $x=0$So, the graph intersects the horizontal asymptote at $(0,1)$

Construct the table and determine the values in the five intervals.

Plot the points and lines on the graph.