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

Jul 10, 2016

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

#### Explanation:

The denominator of f(x) cannot equal zero. This would give division by zero which is undefined. Setting the denominator equal to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

solve: 1 + 2x =0 $\Rightarrow x = - \frac{1}{2} \text{ is the asymptote}$

Horizontal asymptotes occur as

${\lim}_{x \to \pm \infty} , f \left(x\right) \to c \text{ (a constant)}$

divide terms on numerator/denominator by x

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

as $x \to \pm \infty , f \left(x\right) \to \frac{0 - 5}{0 + 2}$

$\Rightarrow y = - \frac{5}{2} \text{ is the asymptote}$
graph{(1-5x)/(1+2x) [-10, 10, -5, 5]}