Question

How do you find the Vertical, Horizontal, and Oblique Asymptote given `f(x)=(x-4)/(x^2-3x-4)`?

Answer 1

vertical asymptote x = -1
horizontal asymptote y = 0

Explanation:

The first step here is to factorise and simplify f(x)

`f(x)=(cancel(x-4))/(cancel((x-4)) (x+1))=1/(x+1)`

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

solve : x + 1 = 0 → x = -1 is the asymptote

Horizontal asymptotes occur as `lim_(xto+-oo),f(x)to0`

divide terms on numerator/denominator by x

`(1/x)/(x/x+1/x)=(1/x)/(1+1/x)`

as `xto+-oo,f(x)to0/(1+0)`

`rArry=0" is the asymptote"`

Oblique asymptotes occur when the degree of the numerator > degree of denominator. This is not the case here (numerator-degree 0, denominator-degree 1) Hence there are no oblique asymptotes.
graph{1/(x+1) [-10, 10, -5, 5]}