How do i find a quadratic equation using only 3 coordinates?

Question

Im pretty sure 3 points are all you need to define a quadratic, so how do i do it?
i know the basic equation will be ax^2+bx+c, how do i find the a,b,c factors using only three sets of coordinates?
(x1,y1),(x2,y2),(x3,y3)

I cant plug in my coordinates because they will be variables. I want to find the equations for a,b,c expressed in terms of x1,x2,x3 and y1,y2,y3. Im making a program for bending moments in beams and i need to be able to plug in three coordinates and then get the quadratic equation given to me.

Answer 1

You set up a system of linear equations:

x1²a+x1b+c=y1
x2²a+x2b+c=y2
x3²a+x3b+c=y3

Note that in this case, a, b, and c are your variables and x1, x2, x3, y1, y2, and y3 are coefficients. This is the opposite of the way you normally think about quadratic equations, so be careful.

Once you have the system set up, solve it by any method (since you have the equations in standard form already, I would normally use elimination in this instance). The resulting quadratic will be y=ax²+bx+c .

Answer 2

You're supposed to plug x -4 instead of y on the left hand side, so (x -4 )^2 = ..... then solve the resulting quadratic equation in x. Come on, have a go!

Answer 3

You plug in the values you have for x and y. You will be left with three equations (one for each point) and three unknowns. Just solve the resulting three equations. Example: If one of your points is (1,1), plug into ax^2 + bx + c = y. You get a + b + c = 1. That is your first equation. If (2,4) is your second point, you get 4a + 2b + c = 4. If (3,9) is your third point, you get 9a + 3b + c = 9. Just solve this set of equations and you are done.

Answer 4

Plug the three points in for x and y. This will give you three equations. Now solve for a, b and c.
You could use substitution, elimination or matrices.