How do you find the asymptotes for #(x+3)/(x^2-9)#?

1 Answer
Feb 15, 2016

Answer:

To find the asymptotes, you need to find out where the denominator approaches zero for the vertical asymptotes and then establish what happens as #x -> +-oo#

Explanation:

The denominator in this case equals the difference of two squares, #x^2# and #3^2#, so
#(x+3)/(x^2-9) = (x+3)/((x-3)(x+3))#
The #(x+3)# terms cancel out so there is a removable asymptote at #x=-3#
There is a vertical asymptote at #x=3#

As #x->+-oo#, # (x+3) ->x# and #(x^2 - 9) -> x^2#

#(x+3)/(x^2-9) -> x/x^2 = 1/x# which gives the slant asymptotes.