How do you find the vertical, horizontal and slant asymptotes of: #(8x^2-x+2)/(4x^2-16)#?

1 Answer
Nov 7, 2016

Answer:

Vertical asymptotes: #x=-2# and #x=2#
Horizontal asymptote: #y=2#
No slant asymptotes, as slant asymptotes only occur when the degree on the numerator is greater than the degree on the denominator.

Explanation:

Let #f(x)=(8x^2-x+2)/(4x^2-16)#

Vertical Asymptotes:
First, factor as much as possible. The numerator cannot be factored. The denominator can be factored into #4(x+2)(x-2)#
#f(x)=(8x^2-x+2)/((4)(x+2)(x-2))#

Vertical asymptotes occur when the denominator is equal to zero. In this case, that is when #x=-2# and when #x=2#.

Horizontal Asymptote:
Horizontal asymptotes are the value that the function approaches when the x value approaches #+-oo#.
#limx->+-oo[(8x^2-x+2)/(4x^2-16)] = 8/4=2#