Question

How do you find the asymptotes for `h(x) = (x+6 )/(x^2 - 36)`?

Answer 1

Vertical asymptote at `x=6`.
Horizontal asymptote at `y=0`.

Explanation:

By factorizing the denominator as a difference of 2 squares, we get that `h(x)=(x+6)/((x+6)(x-6))=1/(x-6)`.

Hence a vertical asymptote occurs at points leading to division by zero in the denominator, ie when `x=6`.

Note that `x=-6` is a removable singularity and not an asymptote.

Horizontal asymptotes occur at `y=lim_(x->+-oo)h(x)=0`.

The plotted graph verifies the existence of these asymptotes.

graph{(x+6)/(x^2-36) [-3.71, 16.29, -5.6, 4.4]}