How do you find any asymptotes of #f(x)=(x+3)/(x^2-4)#?

1 Answer
Oct 16, 2016

Answer:

There are two vertical asymptotes at #x=-2# and #x=2#
There is a horizontal asymptote at 0 as #x->oo#
There is a horizontal asymptote at 0 as #x->-oo#

Explanation:

# f(x)=(x+3)/(x^2-4) #
# :. f(x)=(x+3)/(x^2-2^2) #
# :. f(x)=(x+3)/((x+2)(x-2)) #

Vertical Asymptotes
These occur when the denominator is zero
# :. (x+2)(x-2) = 0 #
# x=+-2 ##

Horizontal Asymptotes
We also need to examine the behaviour as #x->+-oo#
Now for large #x#,
# x+3 ~~ x # and # x+2 ~~ x# and #x-3 ~~ x#
We have, # f(x)=(x+3)/((x+2)(x-2)) #
and so, for large #x#, # f(x) ~~ x/x^2 ~~ 1/x#
so, as # x-> -oo => f(x) -> 0^- #
and, as # x-> oo => f(x) -> 0^+ #

Summary
There are two vertical asymptotes at #x=-2# and #x=2#
There is a horizontal asymptote at 0 as #x->oo#
There is a horizontal asymptote at 0 as #x->-oo#

The graph validates this;
graph{(x+3)/((x+2)(x-2)) [-10, 10, -5, 5]}