posted by Lawn .
Sorry to bother you, I was wondering, why is it that for the graph of 1/(1+e^x), 1 is the horizontal asymptote?
I had thought the HA should be something that makes the reciprocal function undefined, so I kind of thought the HA should be -1.
Please explain simply, thank you very much! Really appreciate the help.
were you allowed to use a graphing calculator? because if you were, you would be able to find the HA by looking at the graph
The horizontal asymptote indicates what happens to y when x gets large.
As x gets huge negative, e^x --> 0, so we wind up with y = 1/1 = 1
As x gets huge positive, e^x --> oo, so we wind up with y=1/oo = 0
So, there are two horizontal asymptotes, at y=0 and y=1