How do you find the asymptotes for $\frac{x^{2} - 4 x - 32}{x^{2} - 16}$ $(x^2-4x-32)/(x^2-16)$ ?

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Answer 1 · 2015-11-20T20:55:38

Always simplify first before solving the problem ...

First, factor ...

$y = \frac{(x - 8) (x + 4)}{(x + 4) (x - 4)}$ $y=[(x-8)(x+4)]/[(x+4)(x-4)]$

Cancel out $(x + 4)$ $(x+4)$ ...

$y = \frac{x - 8}{x - 4}$ $y=(x-8)/(x-4)$

[Note: The function is undefined at $x = - 4$ $x =-4$ , but no asymptote. It is called a removable discontinuity .]

Now, the denominator is zero when $x = 4$ $x=4$ resulting in a vertical asymptote at $x = 4$ $x=4$ .

There is a horizontal asymptote at $y = 1$ $y=1$ . This is found by dividing the coefficients on x in the numerator and denominator.

hope that helped

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How do you find the asymptotes for (x^2-4x-32)/(x^2-16)x2−4x−32x2−16(x^2-4x-32)/(x^2-16)?