How do you find the vertical, horizontal or slant asymptotes for (e^x)/(7+e^x)?

May 6, 2016

We do not have any other asymptote, just two horizontal asymptotes, $x = 0$ and $x = 1$

Explanation:

Dividing numerator and denominator by ${e}^{x}$ in ${e}^{x} / \left(7 + {e}^{x}\right)$

we get $\frac{1}{7 {e}^{- x} + 1}$

As $x \to \infty$, ${e}^{- x} \to \frac{1}{e} ^ \infty = 0$

Hence ${e}^{x} / \left(7 + {e}^{x}\right) = \frac{1}{7 {e}^{- x} + 1} \to \frac{1}{0 + 1} = 1$

Hence the asymptote is $y = 1$

When $x \to - \infty$, ${e}^{x} / \left(7 + {e}^{x}\right) = {e}^{- \infty} / \left(7 + {e}^{- \infty}\right) = \frac{0}{0 + 7} = 0$

Hence other asymptote is $y = 0$

We do not have any other asymptote, just two horizontal asymptotes

graph{e^x/(7+e^x) [-10, 10, -2, 2]}