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I want a graph that goes from $[0, \infty) \Rightarrow (0, 1]$ a bit like $e^{-x}$, but less severe. By that I mean I do not want all the action around $0$.

The $y$ axis must start at $1$ and decay to, but not reach $0$.

I could obviously scale the $x$-axis, but it is still severe with most movement around $0$.

UNAN
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    What do you mean by "action around 0?" – While I Am Aug 12 '21 at 15:57
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    With the same idea as @Mithrandir's answer, but simple and algebraic, why not try $f(x) = \dfrac{1}{1+ax^2}$, or, for that matter, $\dfrac{1}{1+ax}$. With the positive $a$ close to $0$ you can have as flat a negative slope as you want. Then again, you can even try $\dfrac{1}{1+a \log(x+1)}$, or use log log ..., but that brings us back in transcendental territory. – Torsten Schoeneberg Aug 12 '21 at 16:19

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$f(x)=-\frac{2}{\pi}\arctan(ax)+1$

The rate of convergence can be controlled with parameter $a$. In general, this function will converge to $0$ much slower than $e^{-x}$, since the derivative of $\arctan$ is $\frac{1}{1+x^2}$. The $-\frac{2}{\pi}$ and $+1$ ensure that the graph meets your stipulated conditions.

enter image description here

Edit: Start by thinking of functions with horizontal asymptotes which converge more slowly then the exponential function. Then, adjust so they meet the parameters you're looking for. Another example would be $y=\frac{1}{x}$. Adjusting to meet your requirements: $f(x)=\frac{1}{ax+1}$. Choose $a\in(0,1)$ as desired.

Mithrandir
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