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Given 3 points, $(x_1, y_1), (x_2, y_2), (x_3, y_3)$, how might I find an equation intersecting all of these points? Given just 2 points, to find a linear equation, this is the formula: $$y\ =\frac{\left(y_2-y_1\right)}{\left(x_2-x_1\right)}x+\frac{\left(y_1+y_2-\left(x_1\left(\frac{\left(y_2-y_1\right)}{\left(x_2-x_1\right)}\right)+x_2\left(\frac{\left(y_2-y_1\right)}{\left(x_2-x_1\right)}\right)\right)\right)}{2}$$ In the same form of $y=...$, what is the formula for a quadratic equation with 3 points?

MazeOfEncryption
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  • Do you know any linear algebra? – DKS Feb 19 '18 at 15:33
  • Look up Lagrange polynomials. These are designed to handle interpolation. – Joel Feb 19 '18 at 15:35
  • there's a similar question already on the site: https://math.stackexchange.com/questions/88881/getting-a-standard-form-quadratic-from-a-set-of-points-3-points?rq=1 is this that you are looking for? – agneau Feb 19 '18 at 15:36
  • As satisfying as the closed form might be, it is conceptually clearer to notice that in the case of 2 points $(x_1,y_1)$, $(x_2,y_2)$, to find the coefficients of a linear equation $y=ax+b$ that passes through those two points you plug the coordinates of those 2 points into that equation to get a system of two equations $$y_1 = a x_1 + b$$ $$y_2 = a x_2 + b$$ and then you solve that system of equations for $a,b$. This will give you insight to the answer of @DrSonnhardGraubner. – Lee Mosher Feb 19 '18 at 15:41

2 Answers2

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Note that by Lagrange's polynomial

$$L(x) = {y_1}\cdot{x - x_2 \over x_1 - x_2}\cdot{x - x_3 \over x_1 - x_3}+ {y_2}\cdot{x - x_1 \over x_2 - x_1}\cdot{x - x_3 \over x_2 - x_3}+ {y_3}\cdot{x - x_1 \over x_3 - x_1}\cdot{x - x_2 \over x_3 - x_2}$$

See some example of application here.

user
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you can make the ansatz $$y=a^2+bx+c$$ and plug in all your points in this equation and then you must solve the system $$y_1=ax_1^2+bx_1+c$$ $$y_2=ax_2^2+bx_2+c$$ $$y_3=ax_3^2+bx_3+c$$ for $a,b,c$

Dr. Sonnhard Graubner
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