Question

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

Answer 1

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`