Find at least two ways to find $a, b$ and $c$ in the parabola equation I've been fighting with this problem for some hours now, and i decided to ask the clever people on this website. 
The parabola with the equation $y=ax^2+bx+c$ goes through the points $P, Q$ and $R$. How can I find $a, b$ and $c$ in at least two different ways, when
$P(0,1), Q(1,0)$ and $R(-1,3)$?
 A: Here is one way. Plug the coordinates in the equation for each point. Then solve the system. You get, for $P,Q$ and $R$ respectively
$$
\matrix{a\cdot0^2+b\cdot 0+c=1\\ a\cdot1^2+b\cdot 1+c=0\\a\cdot(-1)^2+b\cdot (-1)+c=3}\quad\Leftrightarrow\quad \matrix{c=1\\ a+b+c=0\\a-b+c=3}
$$
Note that this has a unique solution $(a,b,c)$. You already have $c$ from the first equation. Now find $a$ and $b$ from the last two. You may add them to get $a$, and subtract them to get $b$.
A: You can find normal equation using least square method
First you have to obtain sum of the square of deviation say S and then put the values zero of partial differentiation of S with respect to a,b and c to get three normal equation then solve three normal equation to get a,b and c
A: Let $y=A(x-\alpha)(x-\beta)+B(x-\beta)(x-\gamma)+C(x-\gamma)(x-\alpha)$ where $\alpha,\beta,\gamma$ are the abscissa of $P,Q,R$ 
So,  $y=Ax(x-1)+B(x-1)(x+1)+C(x+1)x$ putting $\alpha=0,\beta=1,\gamma=-1$
As the parabola passes through $P(0,1)$ so it will satisfy the equation of the parabola 
$\implies 1=B(0-1)(0+1)\implies B=-1$
Similarly, for $Q(1,0), 0=C(1+1)\implies C=0$ and  for $R(-1,3), 3=A(-1)(-1-1)\implies A=\frac32$
So, $y=\frac32x(x-1)+(-1)(x-1)(x+1)+0(x+1)x=\frac{x^2}2-\frac32x+1$
But $y=ax^2+bx+c$
Now compare with the coefficients of the different powers of $x$ 
