# What are the asymptotes and removable discontinuities, if any, of f(x)= (x^3 + x^2 - 6x )/( 4x^2 - 8x - 12 )?

Jan 25, 2018

There are two asymptotes: $\text{ " x=3 " "and " } x = - 1$

There are no removable discontinuities (holes)

#### Explanation:

We need to start by factoring everything as much as possible:

$f \left(x\right) = \frac{{x}^{3} + {x}^{2} - 6 x}{4 {x}^{2} - 8 x - 12}$

$f \left(x\right) = \frac{x \left({x}^{2} + x - 6\right)}{4 \left({x}^{2} - 2 x - 3\right)}$

$f \left(x\right) = \frac{x \left(x - 2\right) \left(x + 3\right)}{4 \left(x - 3\right) \left(x + 1\right)}$

Now, let's remember the rules for finding asymptotes and holes:

If the factor is ONLY on the bottom, then there is an asymptote when it equals 0.

If the factor is on BOTH the top and bottom, then there is a hole when it equals 0.

It looks like we don't have any factors on both the top AND bottom. So, we just need to look at the bottom.

There will be an asymptote when $x - 3 = 0$

Therefore, the asymptote is $x = 3$

There will be an asymptote when $x + 1 = 0$

Therefore, the asymptote is $x = - 1$