# How do you find the asymptotes for f(x)=(1-5x) /( 1+2x)?

Feb 9, 2016

vertical asymptote $x = - \frac{1}{2}$
horizontal asymptote $y = - \frac{5}{2}$

#### Explanation:

vertical asymptotes occur when the denominator of a rational function tends to zero.

To find the equation

solve 1 + 2x = 0 $\Rightarrow x = - \frac{1}{2}$

horizontal asymptotes occur as  lim_(x→±∞) f(x) → 0

If the degree of the numerator an denominator are equal then the equation can be found by taking the ratio of leading coefficients

in this case they are both of degree 1

I'll rewrite f(x) to assist in finding leading coefficients

f(x) $= \frac{- 5 x + 1}{2 x + 1}$

equation of asymptote: $y = - \frac{5}{2}$

here is the graph of f(x)
graph{(1-5x)/(1+2x) [-10, 10, -5, 5]}