How do you find the asymptotes for $h (x) = \frac{2 x - 1}{6 - x}$ $h(x) = (2x - 1)/ (6 - x)$ ?

Question

1877 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-27T00:37:58

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

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

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

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

$y = \lim_{x\to infty}(frac{frac{2(x)}{(x)}-frac{1}{(x)}}{frac{6}{(x)}-frac{(x)}{(x)}}) = (frac{2-0}{0-1}) = frac{2}{-1} = -2$

$y = \lim_{x\to -infty}(frac{frac{2(-x)}{(-x)}-frac{1}{(-x)}}{frac{6}{(-x)}-frac{(-x)}{(-x)}}) = (frac{2+0}{0-1}) = frac{2}{-1} = -2$

How do you find the asymptotes for h(x) = (2x - 1)/ (6 - x)h(x)=2x−16−xh(x) = (2x - 1)/ (6 - x)?