How do you find the asymptotes for #(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=[(x-8)(x+4)]/[(x+4)(x-4)]#

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

#y=(x-8)/(x-4)#

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

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

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

hope that helped

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