# How do you find the vertical, horizontal or slant asymptotes for (3e^x) /( e^x-2)?

Dec 25, 2016

Horizontal;) y=0 and y = 3. Vertical: $x = \ln 2 = 0.6931$, nearly.

#### Explanation:

graph{3e^x/(e^x-2) [-20, 20, -10, 10]}

$y = 3 {e}^{x} / \left({e}^{x} \left(1 - 2 {e}^{- x}\right)\right) = \frac{3}{1 - 2 {e}^{- x}}$

As $x \to \infty , y \to 3$. As $x \to - \infty , y \to 0$.

So, y - 0 and y = 3 are asymptotes.

By actual division,

$y = 3 + \frac{6}{{e}^{x} - 2}$

Here, as ${e}^{x} \to 2 , y \to \infty$.

So, x = ln 2 gives yet another asymptote.