I'm trying to find a function that will go through $(0,1)$, $(13,0)$, with another point in between, say at $(6,p)$ where $0\le p\le 1$, that I can vary up and down to make the curve similar to a straight line, or as steep as possible when x is near 0.
Isn't it just a general exponential function?
I'm trying to find a function that enables me to 'flex' the curve as described above, by modifying only one variable. I'm sure I saw a function like that once, but maybe I'm wrong.
An exponential function won't work unless you add in a constant term, because exponential functions are never $0$.
With three points, the simplest choice for fitting would be a quadratic function, but there are lots of other possibilities. Which one you use should depend on some knowledge about where your function is supposed to represent -- but in the absence of such information I'd just go with the polynomial.