How do you find the asymptotes for $(2x - 4) / (x^2 - 4)$ ?

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Answer 1 · 2017-01-29T11:14:08

Vertical asymptote at $x=-2$

As $(2x-4)/(x^2-4)=(2(x-2))/((x+2)(x-2))=2/(x+2)$

As $x->-2$ from right, $2/(x+2)->oo$

and as $x->-2$ from left, $2/(x+2)->-oo$

graph{2/(x+2) [-11.71, 8.29, -5.16, 4.84]}

Hence, we have a vertical asymptote at $x=-2$

Note that at $x=2$ , we have a hole but no asymptote as $(x-2)$ cancels out in numerator and denominator.

How do you find the asymptotes for (2x - 4) / (x^2 - 4) (2x - 4) / (x^2 - 4)?