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I think this is simple but my math skills are limited. I have a basic exponential growth formula: $$y=x \cdot (1-p)^n$$ and I have $y$ and $x$ and $n$ values and I need value of $p$. Then when I solve for $p$, I have to calculate $y$ with different values of $n$. It's easy for me to do that but than I get $y$ values that decreases fast and than slow like on this graph:

enter image description here

But I want to decrease slower and than faster like on this graph

enter image description here

How to change the beginning formula to get what I said?

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  • $\begingroup$ Maybe you want $y=x(1-p)^{1/n}$? $\endgroup$
    – MattW
    Commented Nov 25, 2023 at 2:57

3 Answers 3

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If one starts with $f(x) = 2^x$

Step 1: Reflect across the $x$ axis: $-f(x)$

Step 3: Translate by required amount $T_{-a, b}$: $-f(-x - a) + b$

enter image description here

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From your equation and graph I believe that your independent variable (the one on the horizontal axis) is $n$, and $0<p<1$.

One way to get the kind of graph you want is $$y=b-(1+p)^n$$ where $b$ is the $y$-intercept of the horizontal asymptote above the graph. You will probably need to shift the graph to the right, as in $$y=b-(1+p)^{n-a}$$ where $a$ is the amount of shift to the right.

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  • $\begingroup$ Thanks for the inputs, but I don't know how to solve this. Let me put some more info here. I have: y=x*(1-p)^n y= 0.0089 and x= 0.0588 n=11 (n goes from 1 to 11) If I use exponetional growth formula I get the following: 0.0089=0.0588*(1-p)^11 and from this p=0.15 When I have P it's easy for me to calculate y for each n (n1, y for n2 etc.)But in this case the values of y for each n in the begining decreases fast and than slover, and I want it to be slower and than faster.The graps I used are in question are just examples to discribe my problem. $\endgroup$
    – bacuto d.
    Commented Nov 9, 2014 at 16:16
  • $\begingroup$ I think what @Rory Daulton said is totally correct in your case. If you graph it online you get exactly what you wanted, with the $p=0.15$ (change the other values as you wish). $\endgroup$
    – Jzbach
    Commented Nov 5, 2021 at 9:40
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By changing a, n and q, using $$a-\frac{(x+n)^{2}}{q}$$ you can get your curve that "decreases slow and then fast". I got results of q=1600000, a=-0.012 and n=0.enter image description here

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