Question

How do you find the vertical, horizontal or slant asymptotes for `(x-4)/(x^2-3x-4)`?

Answer 1

The vertical asymptote is `x=-1`
The horizontal asymptote is `y=0`
There is no slant asymptote

Explanation:

We can factorise the denominator `x^2-3x-4=(x+1)(x-4)`
`:.(x-4)/(x^2-3x-4)=cancel(x-4)/((x+1)cancel(x-4))=1/(x+1)`

So the vertical asymptote is `x=-1` as we cannot divide by `0`
Limit `1/(x+1)=0^-`
`x->-oo`

Limit `1/(x+1)=0^+`
`x->+oo`

So the horizontal asymptote is `y=0`
And the intercept with the y axis is `(0,1)`
As the degree of the numerator is `<` the degree of the denominator, there is no slant asymptote
graph{1/(x+1) [-12.66, 12.65, -6.33, 6.33]}