What are the vertical and horizontal asymptotes of #y = (x+3)/(x^2-9)#?

1 Answer
Jun 20, 2018

vertical asymptote at #x=3#

horizontal asymptote at #y=0#

hole at #x=-3#

Explanation:

#y = (x+3)/(x^2-9)#

First factor:

#y = ((x+3))/((x+3)(x-3))#

Since the factor #x+3# cancels that is a discontinuity or hole, the factor #x-3# does not cancel so it is a asymptote:

#x-3=0#

vertical asymptote at #x=3#

Now let's cancel out the factors and see what the functions does as x gets really big in the positive or negative:

#x -> +-oo, y ->?#

#y = cancel((x+3))/(cancel((x+3))(x-3))=1/(x-3)#

As you can see the reduced form is just #1# over some number #x#, we can ignore the #-3# because when #x# is huge it is insignificant.

We know that: #x ->+-oo, 1/x -> 0# hence, our original function has the same behavior:

#x ->+-oo, ((x+3))/((x+3)(x-3)) -> 0#

Therefore, the function has a horizontal asymptote at #y=0#

graph{y = (x+3)/(x^2-9) [-10, 10, -5, 5]}