# How do you find the vertical, horizontal or slant asymptotes for y = (2e^x) / (e^x - 5)?

Oct 21, 2016

We have a vertical asymptote at $x = \ln 5 = 1.6094$

and horizontal asymptotes at $y = 0$ and $y = 2$

#### Explanation:

We have $y = \frac{2 {e}^{x}}{{e}^{x} - 5}$

As $x \to - \infty$, $y \to \frac{0}{0 - 5} = 0$

dividing numerator and denominator by ${e}^{x}$, we get

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

As such, when $x \to \infty$, $y \to \frac{2}{1 - 0} = 2$

Hence, horizontal asymptotes are $y = 0$ and $y = 2$

also $y {e}^{x} - 5 y = 2 {e}^{x}$ or $y {e}^{x} - 2 {e}^{x} = 5 y$

i.e. ${e}^{x} = \frac{5 y}{y - 2}$ or $x = \ln \left(\frac{5 y}{y - 2}\right) = \ln \left(\frac{5}{1 - \frac{2}{y}}\right)$

and as $y \to \pm \infty$, $x \to \ln 5 = 1.6094$,

we have a vertical asymptote at $x = \ln 5 = 1.6094$

graph{(2e^x)/(e^x-5) [-10, 10, -5, 5]}