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

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Answer 1 · 2015-12-14T05:57:42

Find the x-value where the equation becomes undefined.

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 $y=0$ .

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