Question

How do you find the asymptotes for `f(x) = (x^3 + x^2 - 6x) /( 4x^2 - 8x - 12)`?

Answer 1

Asymptotes are `x=3` , `x=-1` and `x/4`

Explanation:

Start by finding the factors for both numerator and denominator.
`f(x) = (x^3+x^2-6x)/(4x^2-8x-12)`

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

`f(x) = (x(x+3)(x-2))/(4(x-3)(x+1))`

This confirms that there are no removable asymptotes as there are no factors that cancel out.

The denominator approaches zero, and therefore the function value `->oo` as `x->3` or `x->-1` so there are vertical asymptotes at `x=3` and `x=-1`

As `x->oo` the function `f(x) -> (xcancel(x^2+x - 6))/(4cancel(x^2-2x-3)) -> x/4`

So there is a slant asymptote of `x/4`