How do you find the asymptotes for #f(x) =(x^2 - 4)/(9 - x^2)#?

1 Answer
May 6, 2016

Answer:

Vertical asymptotes at #x=3# and #x=-3# and horizontle asymptote at #y=-1#

Explanation:

To find all the asymptotes for function #y=(x^2-4)/(9-x^2)# let us first start with vertical asymptotes, which are given by putting denominator equal to zero i.e. here #9-x^2=0# i.e. #(3-x)(3+x)=0#. Hence vertical asymptotes are #x=3# and #x=-3#.

Further as in #y=(x^2-4)/(9-x^2)#, highest degree in numerator is for #x^2# and for denominator #-x^2#, hence we have a horizontal asymptote at #y=x^2/(-x^2)=-1#

graph{(x^2-4)/(9-x^2) [-10, 10, -5, 5]}