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

1 Answer
Dec 14, 2015

Find the x-value where the equation becomes undefined.

Explanation:

When an equation is undefined, that means that the denominator equals zero.

Set the denominator equal to zero and solve:

#x^2-36=0#
#x^2=36#
#(x=-6)# #(x=6)#

The equation is undefined when #x=6# and #x=-6#, so there are vertical asymptotes at #x=6# and #x=-6#.

Also, since the degree of the denominator is greater than the degree of the numerator, there is a horizontal asymptote at #y=0#.