Question

What are the asymptote(s) and hole(s), if any, of `f(x) =(x^2-3x+2)/(x-3)`?

Answer 1

Vertical asymptote `x=3` and oblique / slanting asymptote `y=x`

Explanation:

As `f(x)=(x^2-3x+2)/(x-3)=((x-1)(x-2))/(x-3)` and as `(x-3)` in denominator does not cancel out with numeraor, we do not ave a hole.

If `x=3+delta` as `delta->0`, `y=((2+delta)(1+delta))/delta` and as `delta->0`, `y->oo`. But if `x=3-delta` as `delta->0`, `y=((2-delta)(1-delta))/(-delta)` and as `delta->0`, `y->-oo`.

Hence `x=3` is a vertical asymptote.

Further `y=(x^2-3x+2)/(x-3)=(x^2-3x)/(x-3)+2/(x-3)`

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

Hence as `x->oo`, `y->x` and we have an oblique or slant asymptote `y=x`
graph{(y-(x^2-3x+2)/(x-3))=0 [-17.34, 22.66, -8.4, 11.6]}