What's the simplest curve that has an inflection point at zero and is asymptotic at 100.0?

Question

I know I should remember this from high school geometry, but what's the simplest curve that is crosses (0,0) and (as x increases) rises to be asymptotic to a horizontal line? (I'm not sure whether I want an inflection point or not -- I'm just trying to convert an open-ended number into a bounded "score".)

Thanks!

I've got to say that you guys are a lot easier to deal with than the Stack Overflow crowd. I've come in with two "dumb" questions and gotten considerate answers in both cases. Much appreciated!!

Answer 1

You mention in comments to Zev's answer that you want f(1/2)=50 and f(50)=99. Together with the other constraints, we can use this to design an appropriate function. The most elementary way to get an asymptote is by division, so suppose f had the form $$f(x) = 100 - \frac{100}{g(x)}$$ where $g$ is a some function to be designed. We can then fix three values for g, namely g(0)=1, g(1/2)=2 and g(50)=100. The simplest way to get a function to pass through three known point is to make it a quadratic polynomial, so we have $$g(x) = ax^2+bx+c$$ for constants a, b, and c to be determined. Inserting our known points gives us three equations with three unknowns, which can be solved to get $$a = \frac{-1}{2475}, b = 2-\frac{a}{2}, c=1$$ A negative a won't do (it would cause the function to begin dropping away from f=100 at very large x), but luckily the exact computed a is so small that we don't lose much by just setting it to 0. Thus the final design will be just $$f(x) = 100 - \frac{100}{2x + 1}$$ which makes f(0)=0, f(0.5)=50, f(50)=99.01. And luckily it also happens to be strictly increasing (and convex) for positive $x$.

Answer 2

Modifying J.M.'s suggestion a bit to fit what's in the title, one class of functions you might like is $$f_a(x)=100(1-e^{-x/a})$$ where $a>0$ is some parameter you can vary as you see fit. The larger $a$ is, the more spread out the function will be. Here is a plot of $f_a$ for $5\leq a\leq 20$:
$\hskip 1in$

(from Wolfram Alpha)

However, note that the nature of an asymptote is that it will never actually reach that value - that is, no $f_a$ will ever take on the value 100. If you want your function $f$ to satisfy $f(0)=0$ and $f(100)=100$, you can modify this suggestion as follows:

$$g_a(x)=100\left(\frac{1-e^{-x/a}}{1-e^{-100/a}}\right)$$

Then $g_a(0)=0$ and $g_a(100)=100$ for any $a>0$ you want to choose, but it retains the curved shape of the function $f_a$. The functions $g_a$ are also all bounded; just not by 100. If you plug in any $x>100$, then $g_a(x)>100$ too.