TITLE: What is the formula for this curve?
QUESTION [5 upvotes]: Three years of calculus in college have served me nothing, apparently, since I can't for the life of me remember even the basics. I'm working on a small software project where I have a table with say 20 cells, and I want the cells' opacity to go down as the index goes up.
Currently, I'm doing it linearly with $\textrm{opacity} = 1 - (\textrm{index}/20)$, or $y = 1-x$. The curve I'm going for is something where at the beginning I have a high value for the opacity, 1, but then it starts dropping like a roller-coaster, non-linearly. The best I can describe it is it looks like half of a 'C' draw on the positive xy axis.
The closest I got was $y =  e^x$, but that plot goes up. Can anyone tell me the name of what I'm looking for?
Edit: Ok it turns out what I'm looking for is a hyperbola, $y = 1/x$. However, since this is a opacity value, the range needs to be from $0$ to $1$, while the domain is also $0$ to $1$. But I'm getting some large values for small inputs.

REPLY [3 votes]: As an extension to Marvis' answer, any function of the form
$$x^p+y^p=1\implies y=(1-x^p)^{1/p}$$
where $0<p<1$ will work, with more curvature the lower $p$ is. Another alternative is to use a circular arc:
$$(x-1)^2+(y-1)^2=1\implies  y=1+\sqrt{1-(x-1)^2}$$
If you don't need it to be vertical at $x=0$, polynomials like $y=(x-1)^{2n}, n=1,2,3...$ might work. You get a steeper drop with larger $n$. If efficiency is an issue and you don't need to match a plot exactly, this is probably less expensive, thought it might not "curve as evenly" as you'd like it to.