Answer
The vertical asymptote is the line $x=\pm 2$.
The horizontal asymptote is the line $y=0$.
Work Step by Step
Given $g(x)=\frac{2}{(x-2)(x+2)}=\frac{2}{x^2-4}$
$x^2-4=0 $
$x=\pm 2$
The vertical asymptote is the line $x=\pm 2$.
The horizontal asymptote is the line $y=0$.
Make a list of values:
$x=−2→ y=0$
$x=−1→ y=-\frac{2}{3}$
$x=0→ y=-\frac{1}{2}$
$x=1→ y=-\frac{2}{3}$
$x=2→ y=0$
