Question

How do you find the vertical, horizontal or slant asymptotes for `f(x)=( x-4)/ (x^2-1)`?

Answer 1

vertical asymptotes x = ± 1
horizontal asymptote y = 0

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 : `x^2-1=0rArrx^2=1rArrx=±1`

`rArrx=-1,x=1" are the asymptotes"`

Horizontal asymptotes occur as

`lim_(xto+-oo),f(x)toc" (a constant)"`

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

`(x/x^2-4/x^2)/(x^2/x^2-1/x^2)=(1/x-4/x^2)/(1-1/x^2)`

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

`rArry=0" is the asymptote"`

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (numerator-degree 1, denominator-degree 2 ) Hence there are no slant asymptotes.
graph{(x-4)/(x^2-1) [-10, 10, -5, 5]}