Question

How do you find any asymptotes of `g(x)=(2x-3)/(x^2-6x+9)`?

Answer 1

The vertical asymptote is `x=3`
There is no oblique asymptote

Explanation:

Let's factorise the denominator
`x^2-6x+9=(x-3)^2`
Therefore `g(x)=(2x-3)/(x-3)^2`
As you cannot divide by `0`, so `x!=3`
So `x=3` is a vertical asymptote
Limit `g(x)=2/x=0^-`
`x->-oo`
Limit `g(x)=2/x=0^+`
`x->+oo`

graph{(2x-3)/(x-3)^2 [-10, 10, -5, 5]}