# How do you find the asymptotes for y = (5e^x)/ ((e^x) - 8)?

Feb 24, 2017

Vertical asymptote at $x = \ln 8$
Horizontal asymptotes at: $y = 0$ and $y = 5$

#### Explanation:

The vertical asymptote is found when $D \left(x\right) = 0$:
${e}^{x} - 8 = 0$ so ${e}^{x} = 8$

Solve for $x$ by logging both sides of your equation:

$\ln {e}^{x} = \ln 8$

Vertical asymptote at $x = \ln 8$

Finding horizontal asymptotes :

1. N(x) = 5e^x = 0; e^x = 0; ln e^x = ln 0; x = ln 0 which is undefined. By definition logarithms are positive: $x > 0$ This means there is a horizontal asymptote at $y = 0$

2. $\lim x \to \infty \frac{5 {e}^{x} / {e}^{x}}{{e}^{x} / {e}^{x} - \frac{8}{e} ^ x} = \frac{5}{1 - 0} = 5$ so there is a horizontal asymptote at $y = 5$

Summary:
Vertical asymptote at $x = \ln 8$
Horizontal asymptotes at: $y = 0$ and $y = 5$