Question

What's the formula given three points to come up with the quadratic equation were your give points

(x1, y1 ) (x2, y2) (x3, y3)

I remember seeing some really long equation were you just plug the numbers into it and get your answer but I have forgotten the equaiton

Answer 1

I think it required zeroing some things out

it had three parts i think with fraction in both of them

Answer 2

Well, here is a way to find a polynomial in x^n for any (n+1) points:

If you have points (X1, Y1), (X2,Y2), (X3,Y3) .... (Xn+1 , Yn+1)

then a function that goes through one of those points (Xp,Yp) and is zero at all of the others is:
Yp [ (X-X1)... (X-Xp-1)(Xp-Xp+1) ... (X-Xn+1) ] / [(Xp-X1)(Xp-X2)(Xp-Xp-1)...(Xp-Xp+1) ...(Xp-Xn+1)
note - skips (X-Xp) on top and (Xp-Xp) on bottom) for each point p so function is zero at all given points except p but is Yp at p.
Then if you add those functions you get a polynomial that goes through each point.

Answer 3

Now if n = 3

y= Y1[ (x-X2)(x-X3)] / [(X1-X2)(X1-X3)]
+ Y2[ (x-X1)(x-X3)] / [(X2-X1)(X2-X3)]
+ Y3[ (x-X1)(x-X2)] / [(X3-X1)(X3-X2)]

Answer 4

I suspect that is what you mean.