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I've got a graph image, but I need the formula used the create this graph. This image is being used to read the result corresponding values manually, but I want to automate this by using a formula.

The X axis is a known value 'som van de faktoren', the Y axis 'tijd in jaren' should be the resulting value.

The only value that is 100% sure is 0 (x) = 15 (y). However, as you can see 5 (x) = 12 (y) should be 99% correct too.

Is it possible to connect a formula based on the visual graph? I'm both curious for the formula as for the way of resolving it :-)

I'm not allowed to post images yet, so here is a url to the graph; https://www.dropbox.com/s/5u2vq6i5s0ld84x/Screenshot%25202015-07-13%252012.57.56.png?dl=0

Diego Voors
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1 Answers1

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The graph looks like a graph of an exponential function: $y=a\cdot e^{-bx}$

You have identified two points: $P_1(0/15)$ and $P_2(5/12)$. You can insert the values and you get two equations:

$15=a\cdot e^{-b\cdot 0}$

$12=a\cdot e^{-b\cdot 5}$

The first equation becomes $15=a\cdot 1$. Now you can insert the value of a in the second equation and calculate the value of b.

The graph with the values of a and b looks like

enter image description here

callculus42
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  • thanks for your answer, the resulting graph looks very similar to mine! Is it possible, with this information, to create a formula where I just enter $x$ and it gives me $y$? I will need this to make a computed calculation and display the number as a result, not a graph. So basically I will need a formula where I can put any value for $x$ resulting in the correct $y$ value for the answer. Thanks for your efforts so far, I appreciate it! – Diego Voors Jul 13 '15 at 20:48
  • Sure. After you have calculated the value of $\beta_v$ the formula is $y=a_v\cdot e^{-\beta_v \cdot x}$. I add the index v to highlight, that these are specific values. After that you can insert the values for x and you will get the corresponding y-values. I hope I have understood you right. Do you mean a table like this ?: $\begin{array}{|c|c|c|c|c|} \hline x &0 & \ldots & 5 &\ldots \ \hline y &15 & \ldots & 12 &\ldots \ \hline\end{array}$ – callculus42 Jul 13 '15 at 22:43