Question

How do you find the asymptotes for `(8x^2-x+2)/(4x^2-16)`?

Answer 1

vertical asymptotes x = ± 2
horizontal asymptote y = 2

Explanation:

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation/s set the denominator equal to zero.

solve: `4x^2-16 = 0 → 4(x^2-4) = 0 → 4(x-2)(x+2) = 0`

`rArr x = -2 , x = 2 " are the asymptotes "`

Horizontal asymptotes occur as `lim_(xto+-oo) f(x) to 0`

divide all terms on numerator/denominator by `x^2`

`rArr ((8x^2)/x^2 - x/x^2 + 2/x^2)/((4x^2)/x^2 - 16/x^2)=(8-1/x+2/x^2)/(4-16/x^2)`

as`xto+-oo , 1/x , 2/x^2" and " 16/x^2 to 0`

and `y to (8-0+0)/(4-0) = 8/4 = 2`

`rArr y = 2 " is the asymptote "`
graph{(8x^2-x+2)/(4x^2-16) [-10, 10, -5, 5]}