How do you find the horizontal asymptote for $y = (x-4)^2/(x^2-4)$ ?

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Answer 1 · 2016-02-26T03:54:21

Horizontal asymptote is $y=1$

$y=(x−4)^2/(x^2−4)$ or $y=(x^2-8x+16)/(x^2−4)$

can also be simplified as $y=(x−4)^2/((x+2)(x-2))$

Hence vertical asymptote are $x=2$ and $x=-2$

Fuether, as highest degree of numerator divided by highest degree of denominator is same, and ratio is given by

$x^2/x^2=1$

Horizontal asymptote is $y=1$

How do you find the horizontal asymptote for y = (x-4)^2/(x^2-4)y = (x-4)^2/(x^2-4)?