Question

How to find the asymptotes of `f(x) = (x^2+x-2)/( x^3-3x^2+2x)`?

Answer 1

`f(x)` has vertical asymptotes `x=0` and `x=2`, a horizontal asymptote `y=0` and a hole at `(1, -3)`

Explanation:

Start by factoring both numerator and denominator and cancelling common factors:

`f(x) = (x^2+x-2)/(x^3-3x^2+2x) = ((x+2)color(red)(cancel(color(black)((x-1)))))/(x(x-2)color(red)(cancel(color(black)((x-1))))) = ((x+2))/(x(x-2))`

with exclusion `x != 1`

Notice that when `x=1`, both the numerator and denominator are zero in the given rational function. So `f(1)` is undefined, but the simplified expression is well defined when `x=1`:

`(color(blue)(1)+2)/(color(blue)(1)(color(blue)(1)-2)) = 3/(-1) = -3`

So `f(x)` has a hole (removable singularity) at `(1, -3)`

The remaining values of `x` at which the denominator is zero are `x=0` and `x=2`. In both of these cases the numerator is non-zero. So there are vertical asymptotes at each of these values of `x`.

Finally, since the degree of the numerator is greater than the denominator, `f(x) -> 0` as `x->+-oo`. That is: `f(x)` has horizontal asymptote `y=0`.

It has no slant (oblique) asymptotes. Such slant asymptotes can only occur if the degree of the numerator is `1` greater than the denominator.

Here's a graph of `f(x)` with the vertical asymptotes...

graph{(y-(x+2)/(x(x-2)))(0.9999x-2)(0.9999x+0.0001y) = 0 [-8.71, 11.29, -5.76, 4.24]}