Question

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

Answer 1

The vertical asymptotes are `x=-1` and `x=4`
No slant asymptote
The horizontal asymptote is `y=0`

Explanation:

Let's factorise the denominator

`x^2-3x-4=(x+1)(x-4)`

Let `f(x)=(2x+4)/((x+1)(x-4))`

The domain of `f(x)` is `D_f(x)=RR-{-1,4}`

As we cannot divide by `0`, `x!=-1` and `x!=4`

The vertical asymptotes are `x=-1` and `x=4`

The degree of the numerator is `<` than the degree of the denominator, so there is no slant asymptote

To calculate the limits as `x->+-oo`, we take the terms of highest degree in the numerator and the denominator.

`lim_(x->-oo)f(x)=lim_(x->-oo)(2x)/x^2=lim_(x->-oo)2/x=0^(-)`

`lim_(x->+oo)f(x)=lim_(x->+oo)(2x)/x^2=lim_(x->+oo)2/x=0^(+)`

The horizontal asymptote is `y=0`

graph{(2x+4)/(x^2-3x-4) [-11.25, 11.25, -5.625, 5.625]}