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

1 Answer
Jan 29, 2017

Answer:

Vertical asymptote at #x=-2#

Explanation:

As #(2x-4)/(x^2-4)=(2(x-2))/((x+2)(x-2))=2/(x+2)#

As #x->-2# from right, #2/(x+2)->oo#

and as #x->-2# from left, #2/(x+2)->-oo#

graph{2/(x+2) [-11.71, 8.29, -5.16, 4.84]}

Hence, we have a vertical asymptote at #x=-2#

Note that at #x=2#, we have a hole but no asymptote as #(x-2)# cancels out in numerator and denominator.