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

Jun 28, 2016

vertical asymptote $x = 5$
horizontal asymptote $y = 2$

#### Explanation:

For this rational function (fraction) the denominator cannot be zero. This would lead to division by zero which is undefined. By setting the denominator equal to zero and solving for $x$ we can find the value that $x$ cannot be. If the numerator is also non-zero for such a value of $x$ then this must be a vertical asymptote.

solve : $x - 5 = 0 \Rightarrow x = 5$ is the asymptote

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by $x$

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

as $x \to \pm \infty , y \to \frac{2}{1 - 0}$

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

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here ( both degree 1 ) Hence there are no slant asymptotes.
graph{(2x)/(x-5) [-20, 20, -10, 10]}