Parabola of Best Fit but starting from Origin? I know that formula for Parabola of Best Fit is this:
$$
\begin{bmatrix}
∑x_i^4 & ∑x_i^3 & ∑x_i^2 \\
∑x_i^3 & ∑x_i^2 & ∑x_i \\
∑x_i^2 & ∑x_i & n \\
\end{bmatrix}*\begin{bmatrix}
a \\
b \\
c \\
\end{bmatrix} = \begin{bmatrix}
∑x_i^2y_i \\
∑x_iy_i \\
∑y_i \\
\end{bmatrix}
$$
(where a, b, c are coefficients of $ax^2+bx+c=0$ that I'm looking for).
However, the formula above finds a parabola that does not necessarily go through the origin.
For example, it may find that for a particular set of (x,y) points the best fit parabola has a=200, b=-4432, c=49.
Is there a formula to calculate the "Parabola of Best Fit" that starts in the Origin (0,0)?
It may be less "best fit" but I care about controlling the point where is it's peak.
WHY:
I have 4 sets of numbers.
I want to calculate the parabola of best fit through each set.
And then I want to compare their slopes.
The goal is to find out if those sets of points grow at more or less the same pace.
Or if one set is growing significantly faster/slower than the others.
I can't compare four parabolas if they start God knows where.
I will be grateful if you could give me any other mathematical tools on how I can approach this task.
Thanks.
 A: You have the given dataset with each point $x_i$ corresponding to the value $y_i$ for $i=1\cdots N$. Denote $\mathbf x$ as the (vector) containing $x_1\cdots x_n$ and similarly with $\mathbf y$.
Now, linear regression corresponds to forming the feature matrix $$
X=[\mathbf x;\mathbf 1]
$$
So that the approximation $
\hat {\mathbf y}=X\beta
$
where $\beta =[b,c]^T$ is optimal for the Mean squared error.

The previous statements corresponds to finding a linear estimate $\hat y=bx+c$  which you can extend to parabolas/quadratic by concatenating a column of $\mathbf x^{*2}$ to $X$ (Note. $*2$ denotes elementwise squaring)
The feature matrix is now $$X=[\mathbf x^{*2};\mathbf x; \mathbf 1]$$
However, you want to the solution to be always containing the point at the origin.
You can simply remove the bias in the feature matrix: $$
X=[\mathbf x^{*2};\mathbf x]
$$
Then compute the coefficients to the ordinary least squares estimate by $$
\beta =[a,b]^T=(X^TX)^{-1}X^Ty
$$
These coefficients corresponds to the estimate $\hat y=ax^2+bx$
