How do you find the vertical, horizontal or slant asymptotes for $\frac{200}{1 + 10 (e^{- 0.5 x})}$ $200/(1+10(e^(-0.5x)))$ ?

Question

1500 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-06T13:57:34

Two horizontal asymptotes: $y = 0 \leftarrow and y = 100 \to$ $y = 0 larr and y=100rarr$

$0 \leq y \leq 200$ $0<=y<=200$

y-intercept ( x = 0 ): $\frac{200}{11} = 18.18$ $200/11=18.18$ , nearly.

$y \to 0$ $y to 0$ , as, $x \to \infty and y \to 200$ $x to oo and y to 200$ , as $x \to - \infty$ $x to -oo$ .

See zoomed graphs for clarity.

graph{y(1+10e^(-.5x))-200=0 [-660, 659.5, -330, 329.5]}

graph{y(1+10e^(-.5x))-200=0 [-41.26, 41.2, -20.7, 20.52]}

How do you find the vertical, horizontal or slant asymptotes for 200/(1+10(e^(-0.5x)))2001+10(e−0.5x)200/(1+10(e^(-0.5x)))?