What are the asymptotes for (x + 4)/(x^2+x-6)?

Jul 27, 2015

In cases like this, you must often rewrite (factorise)

Explanation:

$= \frac{x + 4}{\left(x - 2\right) \left(x + 3\right)}$

Since the numerator may not be $= 0$ we conclude:
$x \ne 2 \mathmr{and} x \ne - 3$

So $x = 2 \mathmr{and} x = - 3$ are the vertical asymptotes.

As $x$ gets larger, the $4 , - 2 \mathmr{and} 3$ makes less and less of a difference, so it begins to look like:
$\frac{x}{x \cdot x} = \frac{1}{x}$ which will get smaller and smaller.

So $y = 0$ is the horizontal asymptote.
graph{(x+4)/(x^2+x-6) [-10, 10, -5, 5]}