91Ó°ÊÓ

The given function is$G\left(x\right)=\frac{x^2-4}{x^2-x-2}$

Factor the numerator and denominator as follows,

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

The domain of the function is $\left\{x|x≠-1,2\right\}$

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

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

Thus, the y-intercept is 2

The real zero of the numerator is $x=-2$

Hence, the x-intercept is $-2$.

Determine the behavior of the graph near the x-intercept by substituting $-2$in the denominator.

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

Plot the point $(-2,0)$and indicate a line with slope $-1$on the graph.

The real zeroes of the denominator are the real solutions of the equation $x+1=0$

The line $x=-1$is vertical asymptote of the graph.

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

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

$$\begin{gathered} \frac{x+2}{x+1}=1 \\ x+2=x+1 \\ 2≠1 \end{gathered}$$

Thus, there is no solution for x.

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

Plot the points and lines on the graph.