Question

How do you find the vertical, horizontal and slant asymptotes of: `f(x)= (4x^2+ 4x-24)/(x^4- 2x^3 - 9x^2+ 18x)`?

Answer 1

`f(x)` has vertical asymptotes `x=0` and `x=3`, a horizontal asymptote `y=0` and holes (removable singularities) at `x=2` and `x=-3`.

Explanation:

The numerator factors like this:

`4x^2+4x-24 = 4(x^2+x-6) = 4(x+3)(x-2)`

The denominator factors like this:

`x^4-2x^3-9x^2+18x`

`= x((x^3-2x^2)-(9x-18))`

`= x(x^2(x-2)-9(x-2))`

`= x(x^2-9)(x-2)`

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

Note the common factors `(x-2)` and `(x+3)` in the numerator and denominator, so `f(x)` can be simplified, noting that `x=2` and `x = -3` are excluded from the domain:

`f(x) = (4x^2+4x-24)/(x^4-2x^3-9x^2+18x)`

`= (4color(red)(cancel(color(black)((x+3))))color(red)(cancel(color(black)((x-2)))))/(x(x-3)color(red)(cancel(color(black)((x+3))))color(red)(cancel(color(black)((x-2)))))`

`= (4)/(x(x-3))`

So when `x=0` or `x=3`, the denominator of `f(x)` is `0` and the numerator is non-zero, resulting in a vertical asymptote.

As `x->+-oo` the denominator `->+oo` resulting in a horizontal asymptote `y=0`.

`f(x)` has removable singularities (holes) at `x=2` and `x=-3`

graph{(4x^2+4x-24)/(x^4-2x^3-9x^2+18x) [-10, 10, -5, 5]}