How do you find the asymptotes for #f(x)=(3x^2+2) / (x^2 -1)#?

1 Answer
Jan 21, 2016

Answer:

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

The horizontal asymptote is #y=3#

Explanation:

The asymptotes occur where the denominator approaches zero, and where #x# becomes very large, either positively or negatively.

#f(x) = (3x^2+2)/(x^2-1)#

In this case the denominator is the difference of two squares so the function can be rewritten as
#f(x)= (3x^2+2)/((x-1)(x+1))#

The denominator is zero when either #x=1# or # x=-1# and these are therefore the vertical asymptotes.

AS #x# becomes very large, either positively or negatively,
#lim_(x->oo) =(3cancel(x^2) )/cancel(x^2)#

The horizontal asymptote is therefore #y=3#