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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!!

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    "I've got to say that you guys are a lot easier to deal with than the Stack Overflow crowd." - interesting...2011-08-31

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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 $x$.

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    That's looking pretty good. I fudged the 2 factor to 2.5 (moved the 0.5 point to about 55), added a small linear component (to get closer to 100), and the capped the result at 99.9. It's smooth and "looks right", at least to my eyes (can never guess what the customer will think). I'll call it good.2011-08-31
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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 $5\leq a\leq 20$:
$\hskip 1in$ enter image description here

(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.

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    Gl$a$d th$a$t my answer helped! Thank you for the kind words.2012-02-09