Question

How do you identify all asymptotes or holes for `f(x)=(2x-2)/(x^2-2x-3)`?

Answer 1

The vertical asymptotes are `x=-1` and `x=3`
No slant asymptote
The horizontal asymptote is `y=0`
No holes

Explanation:

Let's factorise the denominator

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

The domain of `f(x)` is `D_(f(x))=RR-{-1,3}`

As you cannot divide by `0`, `x!=-1` and `x!=3`

The vertical asymptotes are `x=-1` and `x=3`

The degree of the numerator `<` than the degree of the denominator, thereis no slant asymptote

We calculate the limits of `f(x)`, we take only the terms of highest degree.

`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-2)/(x^2-2x-3) [-8.89, 8.89, -4.444, 4.445]}