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

1 Answer
Mar 27, 2018

Answer:

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 = \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#