Question

What are the values and types of the critical points, if any, of `f(x) = (x-5)/(x-3)^2`?

Answer 1

The vertical asymptote is `x=3`
The horizontal asymptote is `y=0`
The intercepts are `(0,-5/9)` and `(5,0)`

Explanation:

As we cannot divide by `0`, the vertical asymptote is `x=3`

The degree of the numerator is `<` the dergee of the denominator, there is no oblique asymptote.
`lim_(x->-oo)f(x)=lim_(x->-oo)x/x^2=lim_(x->-oo)1/x=0^(-)`
`lim_(x->+oo)f(x)=lim_(x->+oo)x/x^2=lim_(x->+oo)1/x=0^(+)`
The horizontal asymptote is `y=0`

The intercepts are
on the y-axis, `x=0` `=>` `f(0)=-5/9`
on the x-axis, `y=0` `=>` `0=x-5` `=>` `x=5`
graph{(x-5)/(x-3)^2 [-10, 10, -5.005, 4.995]}