How do you find the vertical, horizontal or slant asymptotes for #y=(x^2-1)/(2x^2 + 3x-2)#?

1 Answer
Oct 23, 2016

Answer:

The vertical asymptotes are #x=1/2# and #x=-2#

The horizontal asymptote is #y=1/2#

Explanation:

We facorise #y=((x+1)(x-1))/((2x-1)(x+2))#
As we cannot divide by #0#
So for #x=1/2# and #x=-2# we have vertical asymptotes
To look for horizontal asymptotes, we rewrite
#y# as #y=(1-1/x^2)/(2+3/x-2/x^2)#

the limit #y=(1-0)/(2+0+0)=1/2#
#y->-oo#

the limit #y=(1-0)/(2+0+0)=1/2#
#y->+oo#
So #y=1/2# is a horizontal asymptote