Question

How do you find the vertical, horizontal or slant asymptotes for `(2x^2+x-3)/(x^2-4)`?

Answer 1

see below

Explanation:

for vertical asymptotes, look for when the denominator is zero. here that requires `x^2 - 4 = 0 \implies x = pm 2`

next, exploring `x \to \pm \infty`, we should note that `(2x^2 + x - 3) / (x^2 - 4) = (2 + 1/x - 3\x^2) / (1 - 4/x^2)` , obtained by dividing top and bottom by `x^2`

and it's not hard to see that `[(2 + 1/x - 3\x^2) / (1 - 4/x^2) ]_{x \to pm \infty} = 2` which is a horizontal asymptote