Question

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

Answer 1

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`.