# How do you find the vertical, horizontal and slant asymptotes of: y = (1-5x)/(1+2x)?

Sep 14, 2016

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

#### Explanation:

The denominator of y cannot be zero as this would make y undefined. Equating the denominator 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: $1 + 2 x = 0 \Rightarrow x = - \frac{1}{2} \text{ is the asymptote}$

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by x

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

as $x \to \pm \infty , y \to \frac{0 - 5}{0 + 2}$

$\Rightarrow y = - \frac{5}{2} \text{ is the asymptote}$

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here ( both of degree 1 ) Hence there are no slant asymptotes.
graph{(1-5x)/(1+2x) [-10, 10, -5, 5]}