# How do you find the asymptotes for g(x) = (2x^2 - 8x) / (x^2 - 6x + 8)?

Jan 17, 2016

$x = 2$ is a vertical asymptote and $y = 2$ is a horizontal asymptote.

#### Explanation:

Start by factorizing the numerator and denominator to obtain

$g \left(x\right) = \frac{2 x \left(x - 4\right)}{\left(x - 4\right) \left(x - 2\right)} = \frac{2 x}{x - 2}$.

Since we cannot divide by zero as it is undefined, the function may not have inputs that lead to division by zero.

Hence $x = 2$ is a vertical asymptote.

Since the other factor cancelled out, it implies that $x = 4$ is a removable discontinuity.

There are no even square roots to consider so no more vertical asymptotes.

Now to find horizontal asymptotes, we investigate the limit of the function at positive and negative infinity :

${\lim}_{x \to \pm \infty} g \left(x\right) = 2$, since the highest powers dominate the function at infinity.

Hence $y = 2$ is a horizontal asymptote.

The graph of the function verifies this :

graph{(2x^2-8x)/(x^2-6x+8) [-14.33, 17.71, -6.43, 9.59]}