# How do I find a logistic function from its graph?

Oct 6, 2014

Hi there. A logistic graph is like an exponential with an upper limit, so it has two horizontal asymptotes, usually y=0 and y=B, as in the "spread of infection" graph here:

The curve is the solution to the diff eqn $\frac{\mathrm{dy}}{\mathrm{dt}} = r y \left(1 - \frac{y}{B}\right)$ with initial point $\left(t , y\right) = \left(0 , {y}_{0}\right) ,$which can be solved by separation of variables and partial fractions! (Think of the starting point at the lower left.) The solution curve is given by

$y = \frac{B {y}_{0}}{{y}_{0} + \left(B - {y}_{0}\right) {e}^{- r t}}$

If you have the graph, you can read off the $\left(0 , {y}_{0}\right)$, the upper limit $B$, and the inflection point $\left({t}_{\text{.inflect}} , \frac{B}{2}\right)$. (Infection point?)

The next part is to solve for $r$ using the inflection point: that's where the second derivative is 0, so take the derivative of $\frac{\mathrm{dy}}{\mathrm{dt}}$
or the second derivative of the equation for y, and solve!

That part I'll leave for you. You're welcome, from \dansmath/ ;-}

[[Added comment from dansmath: I think the prevailing notation is
y=B/(1 + (B/y_0 - 1)e^(-rt) which is the same equation, just divide top and bottom by ${y}_{0}$.]]