Count the number of points in each graph. The polynomial that fits exactly will be of order one less than the number of points. Then you can set up a series of equations to solve simultaneously.
Can anyone please show me how to determine the formula of the plot dots in the graph, the dots I think is divided into three parts, I put a line between the parts to avoid confusion....
A looks something like a x^3 graph
B looks like a sine curve
C looks weird.....
Can anyone please help me out?
Thanks in advance, the files are attached
There are 9 points so you can fit a unique 8th degree polynomial through those nine points. One way to do that is what Prove It suggested: . Putting the x and y values of the 9 points into that gives 9 equations to solve for the 9 numbers a, b, c, d, e, f, g, h, and i.
Try WolframAlpha.
Hmm. What software is available to you? Mathematica? MATLAB? Maple? Excel? I think all of these packages could solve a 9 x 9 system of linear equations for you. I think even my TI-85 calculator could do that, though I grant you that it would be exceptionally laborious to do by hand.
I have a very many (nine to be precise) simultaneous equations, I want to solve them to get a,b,c,d,e,f,g,h and i. I tried everything from websites to trying myself, but couldn't :-(
it is attached if you want to see it, please if anyone has a method of finding these values let me know....
Thanks in advance
The following commands in Mathematica solved the system:
f[x_]:= Aye x^8 + Bee x^7 + Cee x^6 + Dee x^5 + Eee x^4 + Eff x^3 + Gee x^2 + H x + Eye;
Solve[{f[2] == 426.8, f[3] == 470.2, f[4] == 503.4, f[5] == 557.3, f[6] == 564.7, f[7] == 575.4, f[8] == 579.8, f[9] == 624.7, f[10] == 669.9}, {Aye, Bee, Cee, Dee, Eee, Eff, Gee, H, Eye}]
It did claim that the coefficient matrix was badly conditioned. You might be better off plugging in different values for x, unless you don't have any control over where you sample your function.