How do you find the vertical, horizontal or slant asymptotes for #(2x^2+x-7)/(x^2-1)#?

1 Answer
Nov 2, 2016

Answer:

The vertical asymptotes are #x=-1# and #x=1#
The horizontal asymptote is #y=2#

Explanation:

The denominator #(x^2-1)=(x+1)(x-1)#
As we cannot divide by zero, the vertical asymptotes are #x=-1# and #x=1#
As the degree of the numerator and denominator are the same, we don't have slant asymptotes
limit #(2x^2+x-7)/(x^2-1)=(2x^2)/x^2=2#
#x->+-oo#
So #y=2# is a horizontal asymptote
graph{(2x^2+x-7)/(x^2-1) [-10, 10, -5, 5]}