# What is the equation I can use to create the inverse expoential graph?

I am seeking an equation I can apply to provide a score based on a measurable ability rating. It would be great to understand how the equation works so I can alter the curvature of this flexibly based on different scoring matrix.

Perhaps it's

$$f(x) = 100 \cdot \big(1 - e^{-\alpha x}\big)$$

or perhaps it's

$$f(x) = 100 \cdot \big(1 - \tfrac{1}{1+\beta x}\big)$$

or perhaps it's

$$f(x) = 100 \cdot \big(\tfrac{2}{\pi}\arctan(\gamma x)\big)$$

• No. $\log(x)$ doesn't have an asymptotic maximum. It could be any of these. Commented Jul 19, 2023 at 7:11

Maybe you can try the rational polynomial functions as:

$$f(x)=\dfrac{a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0}{b_mx^m+b_{m-1}x^{m-1}+...+b_{1}x+b_0}$$

You can control the limit in $$\infty$$ with $$n, m, a_n$$ and $$b_m$$. For this case $$n=m$$ and $$\dfrac{a_n}{b_m}=100$$. Also, you could control the point around $$0$$ with $$a_0$$ and $$b_0$$ and curvature with the rest of the coefficients.

You're searching for an asymptotic function. The simplest asymptotic function I can think of is $$y=\frac{1}{x}$$:

This already looks a bit like what we want. However, we can tweak it a bit. First of all, we want $$y$$ to increase with increasing $$x$$, so let's negate one of the two sides:

$$y=-\frac{1}{x}$$

Now, when $$x$$ approaches infinity, both sides will go towards 0. We want $$y$$ to approach $$100$$ in that situation, so let's fix that. We now have:

$$y-100=-\frac{1}{x}$$.

We also want the curve to pass through the origin, where $$x=y=0$$. We can achieve this by changing the right-hand side slightly:

$$y-100=-\frac{100}{x+1}$$

Now, when $$x$$ is small, the right hand side goes towards $$-100$$, whereas for large $$x$$ the denominator is dominant and it behaves quite like the $$-1/x$$ we had before.

One last tweak we can do is put a constant (call it $$a$$) in front of $$x$$ to control how fast the function approaches $$100$$ with increasing $$x$$. While we're at it, let's also add 100 to both sides. This results in our final function:

$$y=-\frac{100}{ax+1}+100$$

This is the curve for $$a=0.2$$:

### Sigmoid

$$f(x) = \dfrac{M}{1+e^{-kx}}$$

where:

$$M$$ - the asymptotic value

$$k$$ - controls how steep is the curve

### Power and exponential function

$$f(x)=M\left(1-e^{-kx^p}\right)$$

where:

$$M$$ - the asymptotic value

$$k$$ - controls how steep is the curve

$$p$$ - controls the bend of the knee

For example $$p=1$$:

$$p=2$$