Answer
Vertical asymptotes: $x=-2,x=2$
Horizontal asymptote: $y=1$
No oblique asymptote
Work Step by Step
We are given:
$y=\dfrac{x^2-9}{x^2-4}$
Determine the vertical asymptotes:
$x^2-4=0$
$x^2=4$
$x=\pm 2$
There are two vertical asymptotes: $x=-2,x=2$
Because the degree of the numerator equals the degree of the denominator, there is a horizontal asymptote, and there is no oblique asymptote:
$y=\dfrac{1}{1}=1$
