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

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Answer 1 · 2016-12-07T03:15:06

Horizontal asymptotes: $y=0 larr and y=9 rarr$ . See the illustrative graph.

$x in (-oo, oo) and y in (0, 9)$

y-intercept ( x = 0 ) ; 3

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

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

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