Quadratic equation formula for a,b,c from 3 points

Question

I can solve for a, b, c given three points for a parabola for example (1,1)(2,4)(3,9) but i need to create a program which returns a,b,c in the form:

$$y = ax^2 + bx + c$$

What is a formula that will find out a, b, c? Someone on another forum mentioned Lagrange but being a high school student i don't know much about Uni level physics. Also i can work with matrices if anyone has a matrix formula - but please explain your formulas as to why it works!

Any help appreciated!

Thanks, Itechmatrix

Answer 1

When you plug in the three points, you have three equations in three unknowns, where the unknowns are A,B,and C. Use matrix arithmetic to solve via simple program.

By Matrix, I mean using Cramer's rule.

Answer 2

Assuming ($x_1, y_1$), ($x_2, y_2$) and ($x_3, y_3$) are three points that satisfy the equation:

$y= ax^2 + bx +c$

we get the three equations:

$y_1= ax_1^2 + bx_1 +c$

$y_2= ax_2^2 + bx_2 +c$

$y_3= ax_3^2 + bx_3 +c$

now $a, b, c$ are the unknowns and we have three equations in three variables. Now you could use Cramer's rule as suggested in the other answer.

In Matrix from:

$ Y = \left[ {\begin{array}{c} y_1 \\ y_2 \\ y_3 \end{array} } \right] $

$ X= \left[ {\begin{array}{ccc} x_1^2 & x_1 & 1 \\ x_2^2 & x_2 & 1 \\ x_3^2 & x_3 & 1 \\ \end{array} } \right] $

and

$ A = \left[ {\begin{array}{c} a \\ b \\ c \end{array} } \right] $

which gives

$Y = X A$

I might be missing a transpose. Here you know $Y$ and you know $X$. Inverting we get

$X^{-1} Y = X^{-1} X A = I A$.

Thus $A = X^{-1} Y $

Answer 3

\begin{equation} y = ax^2 + bx + c \implies a = -\frac{b x + c - y}{x^2} \implies b = -\frac{a x^2 + c - y}{x} \end{equation} Given coordinates $ \qquad(1,1)(2,4)(3,9)$

\begin{equation} (1,1)\implies a = -\frac{1b + c - 1}{1}\implies b = -\frac{1a + c - 1}{1}\\ (2,4)\implies a = -\frac{2b + c - 4}{4}\implies b = -\frac{4a + c - 4}{2}\\ (3,9)\implies a = -\frac{3b + c - 9}{9}\implies b = -\frac{9a + c - 9}{3} \end{equation}

We do not have a complete solution but $a$ and $b$ appear to be complimentary and $c$ has always been a guess in many math operations. If we eliminate $c$, we can always add it back into the original equation form.

\begin{equation} (1,1)\implies a = \frac{1 - b}{1}\implies b = 1(1 - a)\\ (2,4)\implies a = \frac{2 - b}{2}\implies b = 2(1 - a)\\ (3,9)\implies a = \frac{3 - b}{3}\implies b = 3(1 - a) \end{equation}

\begin{equation} \implies a=\frac{x-b}{x}\qquad b=x(1-a)\\ \implies(a,b)\in\{(x,0).(x,-x)\}\quad\text{for}\quad y=x(ax+b) \end{equation}