91Ó°ÊÓ

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

The numerator and denominator is in factored form.

The domain is $\left\{x|x≠0,x≠2\right\}$

The value 0 does not lies in the domain, there is no y-intercept.

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

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

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

$$\begin{gathered} H\left(x\right)=\frac{x+2}{-2(-2-2)} \\ ≈\frac{x+2}{-2(-4)} \\ ≈\frac{1}{8}(x+2) \end{gathered}$$

Plot the point $(-2,0)$and indicate a line with slope $\frac{1}{8}$on the graph.

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

The two lines $x=0andx=2$are the two vertical asymptotes of the graph.

The degree of numerator is less than the degree of denominator. So, the function is proper and graph has horizontal asymptote $y=0$

Substitute $H\left(x\right)=0$to determine whether the graph intersects the horizontal asymptote.

$$\begin{gathered} \frac{x+2}{x(x-2)}=0 \\ x+2=0 \\ x=-2 \end{gathered}$$

The only solution is $x=-2$So, the graph intersects the horizontal asymptote at $(-2,0)$

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

Plot the points and lines on the graph.