Hi everyone,
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
No problem at all! I'd be happy to help explain why the horizontal asymptote of the graph 1/(1+e^x) is 1 instead of -1.
To understand this, let's first review what a horizontal asymptote (HA) is. A HA is a straight line that the graph of a function approaches as x approaches positive or negative infinity. In this case, we're interested in finding the HA as x approaches positive or negative infinity.
The function 1/(1+e^x) is a rational function, where the denominator is 1 + e^x. As x approaches positive or negative infinity, the value of e^x also approaches infinity. This means that the denominator 1 + e^x becomes very large.
Now, let's consider the behavior of 1/(1+e^x) as the denominator becomes very large. The numerator remains constant (1), but the denominator gets larger and larger. When the denominator becomes significantly larger than the numerator, the fraction becomes very close to zero. This indicates that the graph of the function approaches zero as x approaches positive or negative infinity.
However, in this case, there is a crucial detail to note. The denominator can never actually be equal to zero because e^x is always positive. Therefore, the graph of 1/(1+e^x) can never equal zero.
Now, as x approaches positive or negative infinity, the denominator becomes much larger than both 1 and the numerator. As a result, the fraction becomes very close to zero, but it never actually reaches zero. Instead, it gets closer and closer to zero without ever getting there.
So, we can conclude that the graph of 1/(1+e^x) approaches zero (but never reaches it) as x approaches positive or negative infinity. In other words, zero is the horizontal asymptote of the graph.
To answer your original question, the horizontal asymptote is 1 because the graph approaches 1 from below (since the fraction never actually equals zero). The asymptote is located above zero because the numerator is always 1.
I hope this explanation helps! If you have any further questions, feel free to ask.