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

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Answer 1 · 2016-02-28T19:50:13

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

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]}

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