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I had recently asked the question:
Logistic function passing through two points?
which received a very helpful answer. However, I'd like to ask your help again as the formulation of the problem has slightly changed and i'm lost...

Problem:
Given two points: $(x_l, y_l)$ and $(x_u, y_u)$ with: $x_l < x_u$ and $y_l < y_u$,
and given that the higher asymptote is one ($\lim_{x\to+\infty}f(x)=1$)
what's the logistic function that passes through the two points and the origin $(0, 0)$?

Thanks!

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Mino
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    Your requirements for the function to pass through the origin and for it to be asymptotic to $y=0$ are incompatible. Think hard on why. – J. M. ain't a mathematician Oct 31 '10 at 01:08
  • You are right. Even removing that meaningless constraint (and having lower asymptote somewhere < 0), I've no idea how to make it work... Any help? – Mino Oct 31 '10 at 09:11
  • You might want to edit your question. So, to clarify and shorten: it has to pass through the origin, and $y=1$ must be an asymptote? – J. M. ain't a mathematician Oct 31 '10 at 09:49
  • Correct. And obviously, pass through the two points. I'll now edit the question. – Mino Oct 31 '10 at 10:36
  • So, it should pass through three points? Note that the logistic function as shown in John's answer to your previous question has only two parameters... – J. M. ain't a mathematician Oct 31 '10 at 10:48
  • Yes, through the three points. That's why I'm stuck... – Mino Oct 31 '10 at 11:11
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    John's answer has two parameters because it implicitly assumes the asymptotes $y=0$ and $y=1$, so you can't just throw out the $y=0$ constraint without changing the form of the function. So you really need to say more about what kind of a function you want. –  Oct 31 '10 at 16:54
  • Yes, I understand that. But it is still a logistic function. Just it passes through the origin and two other points, and it has higher asymptote y=1. – Mino Oct 31 '10 at 16:56

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If you do not insist on $y=0$ as lower asymptote, but instead want the graph to pass through $(0,0)$, the general form of logistic function becomes $$y=1-\frac{b}{1+b\,e^{ax}}$$ where $a,b$ are determined from the two given points, and $b>0$. It's easy to eliminate $b$ from one of two equations: $$b= \frac{1-y_\ell}{1+(y_\ell-1)e^{ax_\ell}}$$ but plugging this into the second to find $a$, we get something that calls for numeric solution.

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