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

Jan 25, 2016

vertical asymptote at $x = - \frac{1}{2}$
horizontal asymptote at $y = \frac{1}{2}$

#### Explanation:

Vertical asymptotes are found when the denominator of

the rational function is zero.

This will occur when 2x + 1 = 0 , hence 2x = - 1

vertical asymptote is : $x = - \frac{1}{2}$

[ Horizontal asymptotes can be found when the degree of

the numerator and the degree of the denominator are equal.]

In this question they are both of degree 1 and so a

horizontal asymptote exists.

The asymptote is found by taking the ratio of leading
coefficients.

horizontal asymptote is ; $y = \frac{1}{2}$

graph{(x+6)/(2x+1) [-20, 20, -10, 10]}