# Thread: Find a formula given data points

1. ## Find a formula given data points

I need to find a formula for this data set:

f(3) = 1
f(2) ~ .7
f(1) ~.57
f(0) = 0

the ~ indicates an approximate value (it does not need to be exact, just close to that value). The formula f(x)=(x^2+10)/19 works well for f(1:3), but it does not go to zero like I need. Can anyone give me a better formula to fit this data?

2. ## Re: Find a formula given data points

Originally Posted by Cbas
I need to find a formula for this data set:

f(3) = 1
f(2) ~ .7
f(1) ~.57
f(0) = 0

the ~ indicates an approximate value (it does not need to be exact, just close to that value). The formula f(x)=(x^2+10)/19 works well for f(1:3), but it does not go to zero like I need. Can anyone give me a better formula to fit this data?
$f(x) = \frac{61}{600} x^3 - \frac{21}{40} x^2 + \frac{149}{150} x$

3. ## Re: Find a formula given data points

Skeeter's point is that given any "n" (x, y) points, there exist an n-1 degree polynomial through those points. For example, here you have four points (taking those "~"s as "="s) so there exist a cubic equation that will give those points. That is true because the general cubic, $y= ax^3+ bx^2+ cx+ d$ has four coefficients. Replacing the x and y by those values gives four equations, $0= a(0^3)+ b(0^2)+ c(0)+ d= d= 0$, $a(1^3)+ b(1^2)+ c(1)+ d= a+ b+ c+ d= .57$, $a(2^3)+ b(2^2)+ c(2)+ d= 8a+ 4b+ 2c+ d= .7$, and $a(3^3)+ b(3^2)+ c(3)+ d= 27a+ 9b+ 3c+ d= 1$, four equations to solve for a, b, c, and d.

4. ## Re: Find a formula given data points

Perfect. I had no idea you could do that

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### equation finder given dat

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