You can solve this with a computer easily by using Cramer's rule as the algorithm.
Note: As line as (x1,y1),(x2,y2),(x3,y3) are not on the same line the determinant is non-zero.
I am trying to solve a system of quadratic equations.
Given:
y1 = a*x1^2 + b*x1 + c
y2 = a*x2^2 + b*x2 + c
y3 = a*x3^2 + b*x3 + c
(x1, y1), (x2, y2), (x3, y3)
I need to solve for a,b,c.
Trying to use Gaussian Elimination does not seem to give any success.
Does anyone have any idea where to find a solution to this? I have checked out the web on solving for the three-point quadratic, but nothing shows the algorithm.
Hello, Hydra!
I am trying to solve a system of quadratic equations.
Given: .
. .
I need to solve for .
First of all, this system of equations is not quadratic.
We have a linear system: .
As ThePerfectHacker suggested, Cramer's Rule would be the simplest approach.
Curious . . . Do you REALLY want a general formula for a 3-by-3 system?
. . It is truly ugly . . .
He was talking about the solution for to the above system. Any way you do it is will lead to a long equation answer.
The only simple think to compute is the determinant which is the Virtumonde Determinant.
After reading the various replies to this query, I do not think I stated my intentions clearly. I am trying to solve for a parabola that passes through three points.
Given 3 points (x1, y1), (x2, y2), (x3, y3), where x1=x, x2=x+1, x3=x+2
I want to solve the three equations simultaneously to get one generic equation for those three points. So obviously the three equations to solve simultaneously for would be
y1 = a*x1^2 + b*x1 + c = a*x^2 + bx + c
y2 = a*x2^2 + b*x2 + c = a*(x+1)^2 + bx + b + c = ax^2 + 2ax + a + bx + b + c
y3 = a*x3^2 + b*x3 + c = a*(x+2)^2 + bx + 2b + c = ax^2 + 4ax + 4a + bx + 2b + c
Sorry for any confusion caused.
y1 = a*x1^2 + b*x1 + c = a*x^2 + bx + c
y2 = a*x2^2 + b*x2 + c = a*(x+1)^2 + bx + b + c = ax^2 + 2ax + a + bx + b + c
y3 = a*x3^2 + b*x3 + c = a*(x+2)^2 + bx + 2b + c = ax^2 + 4ax + 4a + bx + 2b + c
These are 3 very friendly equations, because they all have + c on the end. By doing 2nd - 1st and 3rd - 2nd you get rid of the c at a stroke and have two eqns with two unknowns.
These 2 equations both have + b on the end so you can take one away from the other and find "a".