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

Jun 6, 2016

First, notice that

$f \left(x\right) = \frac{{x}^{3} - 5 {x}^{2} + 6 x}{{x}^{2} + x - 6} = \frac{x \left(x - 3\right) \left(x - 2\right)}{\left(x + 3\right) \left(x - 2\right)} = \frac{{x}^{2} - 3 x}{x + 3}$

Using long division, we find that

$f \left(x\right) = x - 6 + \frac{18}{x + 3}$,

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 $x \to \infty$, $\frac{18}{x + 3} \to 0$, thus $y \to x - 6$.