# How to find orignal equations of type $y=ax^2+bx+c$. given 3 coordinate points?

Ok, simple question, having trouble understanding this in school. So given a set of 3 points (xy-plane), such as

(40,30) (60,28) (20,25)

i have to find the equation of the parabola. I understand that you need to use the inverse of the matrix of the values of a,b and c and multiplied by your answers to the 3 equations. However my problem is that i cannot figure out or understand how to make the original equations used for the matrix to solve it. So can anyone explain how i come up with a system of equations to use to solve the matrix, given these 3 points?

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Your first problem is in the title. The function you're looking for is $y=ax^2+bx+c$. And you have 3 points on that parabola. So just plug them in.

$1600a+40b+c=30$

$3600a+60b+c=28$

$400a+20b+c=25$

I assume you can take it from here?

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honestly no, my problem is that i can't understand how you get c –  Link Dec 21 '11 at 1:45
wolframalpha.com/input/… –  FiniteA Dec 21 '11 at 1:57
thanks that helped alot! –  Link Dec 21 '11 at 2:55

in general, if a polynomial $\sum_{i=1}^n a_ix^i$ goes through the $n$ points $(x_j,y_j)$ (different $x$-coordinates) then $$\left( \begin{array}{cccc} x_1^n&x_1^{n-1}&...&1\\ .&&&\\ &.&&\\ &&.&\\ x_n^n&x_n^{n-1}&...&1\\ \end{array} \right) \left( \begin{array}{c} a_n\\ a_{n-1}\\ .\\ .\\ .\\ a_0\\ \end{array} \right) = \left( \begin{array}{c} y_1\\ .\\ .\\ .\\ y_n\\ \end{array} \right)$$

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Perhaps an easier way will be to use the Lagrange Interpolation Formula. Given $n$ points such that no two of them lies on a same vertical line, the formula guarantee the existence of a polynomial $P(x)$ such that $P(x)$ passes through these $n$ points and the degree of $P(x)$ is at most $n-1$.

Since 3 points are given, by using the formula, the polynomial formed will be at most degree of 2. Here is how the formula works:

Suppose the 3 points are $p_1 = (x_1 = 40, y_1 = 30), p_2 = (x_2 = 60, y_2 = 28), p_3 = (x_3 = 20, y_3 = 25)$. Now, form 3 polynomial $R_1(x), R_2(x), R_3(x)$ such that $R_1(x_1)=1, R_1(x_2)=R_1(x_3)=0$, $R_2(x_2)=1, R_2(x_1)=R_2(x_3)=0$ and $R_3(x_3)=1, R_3(x_1)=R_3(x_2)=0$. It is easy to see that
$$R_1(x)=\frac{(x-60)(x-20)}{(40-60)(40-20)}=-\frac{x^2-80x+1200}{400}$$ $$R_2(x)=\frac{(x-40)(x-20)}{(60-40)(60-20)}=\frac{x^2-60x+800}{800}$$ $$R_3(x)=\frac{(x-40)(x-60)}{(20-40)(20-60)}=\frac{x^2-100x+2400}{800}$$ This $P(x)$ that we want to find is
$$P(x)=30R_1(x)+28R_2(x)+25R_3(x)$$ And the rest are just algebra.

Generally, given $n$ points, $p_1, p_2, \cdots, p_n$, such that $p_i=(x_i, y_i)$, a polynomial $P(x)$ that passes through all these points is
$$P(x)=\sum^n_{i=0}y_i\prod_{j\ne i}\frac{x-x_j}{x_i-x_j}$$

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