Question

How do you find vertical, horizontal and oblique asymptotes for `f(x)= (x^3 - 5x^2 + 6x) / (x^2 + x -6)`?

Answer 1

First, notice that

`f(x) = (x^3-5x^2+6x)/(x^2+x-6) = (x(x-3)(x-2))/((x+3)(x-2)) = (x^2-3x)/(x+3)`

Using long division, we find that

`f(x) = x-6 + 18/(x+3)`,

from which we deduce that the vertical asymptote is `x=-3` and the oblique asymptote is `y=x-6`.

The oblique asymptote is found by letting `x` get arbitrarily larger i,e, as `x -> infty`, `18/(x+3)->0`, thus `y -> x-6`.