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

1 Answer
Jun 21, 2016

Answer:

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