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

1 Answer
Feb 28, 2016

Answer:

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]}