Question

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

Answer 1

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.