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

1 Answer
Dec 6, 2016

Answer:

Two horizontal asymptotes: # y = 0 larr and y=100rarr #

Explanation:

#0<=y<=200#

y-intercept ( x = 0 ): #200/11=18.18#, nearly.

#y to 0#, as, # x to oo and y to 200#, as #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]}