# How do you find the vertical, horizontal or slant asymptotes for 9/(1+2e^-x)?

Dec 7, 2016

Horizontal asymptotes: $y = 0 \leftarrow \mathmr{and} y = 9 \rightarrow$. See the illustrative graph.

#### Explanation:

$x \in \left(- \infty , \infty\right) \mathmr{and} y \in \left(0 , 9\right)$

y-intercept ( x = 0 ) ; 3

y to 0$, a s$ x to -oo and y to 9$, a s$x to oo#.

graph{y(1+2e^(-x))-9=0 [-35.48, 35.47, -17.66, 17.81]}