# How do you find the asymptotes for h(x) = (2x - 1)/ (6 - x)?

Mar 27, 2018

vertical asymptote: $x = 6$
horizontal asymptote: $y = - 2$

#### Explanation:

To find the vertical asymptote, set the denominator of the function equal to zero and solve for x:

$6 - x = 0$
$- x = - 6$
$x = 6$

To find the horizontal asymptote(s), find the following limits:

$y = \setminus {\lim}_{x \setminus \to \infty} \left(\frac{\frac{2 \left(x\right)}{\left(x\right)} - \frac{1}{\left(x\right)}}{\frac{6}{\left(x\right)} - \frac{\left(x\right)}{\left(x\right)}}\right) = \left(\frac{2 - 0}{0 - 1}\right) = \frac{2}{- 1} = - 2$

$y = \setminus {\lim}_{x \setminus \to - \infty} \left(\frac{\frac{2 \left(- x\right)}{\left(- x\right)} - \frac{1}{\left(- x\right)}}{\frac{6}{\left(- x\right)} - \frac{\left(- x\right)}{\left(- x\right)}}\right) = \left(\frac{2 + 0}{0 - 1}\right) = \frac{2}{- 1} = - 2$