Graph Calculation

T

Timelens

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Aug 16, 2022
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Hello. I'm looking for a formula to somewhat fit the graph attached: 1660657596943.png
It needs to start off shallow, be steep in the middle and then become shallow again towards the end. I would like to put in a number from 0 to 1 and have the formula return roughly where it would be on the graph.
If anyone can solve this, thank you.
 
Timelens said:
Hello. I'm looking for a formula to somewhat fit the graph attached: View attachment 33720
It needs to start off shallow, be steep in the middle and then become shallow again towards the end. I would like to put in a number from 0 to 1 and have the formula return roughly where it would be on the graph.
If anyone can solve this, thank you.
Click to expand...
Can you provide the exact coordinates of your red dots?
 
Timelens said:
Hello. I'm looking for a formula to somewhat fit the graph attached: View attachment 33720
It needs to start off shallow, be steep in the middle and then become shallow again towards the end. I would like to put in a number from 0 to 1 and have the formula return roughly where it would be on the graph.
If anyone can solve this, thank you.
Click to expand...
Looks like you want some kind of Sigmoidal Curve. (Have a look here & here.)

The function:  \(\displaystyle f(x)=(1+0.5e^{-6x})^{-14}\)  will produce a graph like this:-

Sigmoidal-Curve.jpg
​

If you go into desmos and play around with the parameters (0.5, -6 & -14) you may get a line that more closely corresponds to yours.
 
Last edited:
The Highlander said:
Looks like you want some kind of Sigmoidal Curve. (Have a look here & here.)

The function:  \(\displaystyle f(x)=(1+0.5e^{-6x})^{-14}\)  will produce a graph like this:-

Sigmoidal-Curve.jpg
​

If you go into desmos and play around with the parameters (0.5, -6 & -14) you may get a line that more closely corresponds to yours.
Click to expand...
That is exceedingly helpful. My final result is:
(1 + (e^(-7 * x)))^-10

It's formatted like that since I'm using it for a piece of code I am writing. I expect I will be fine-tuning it, but I've got what I was looking for so I can't thank you enough!
 
A straight line going from the 1st point (0,0) to the last point (~1,1). So y=x somewhat fits the graph.
 
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