By using the method of least squares, find the best parabola through the points:
$(1, 0)$,   $(-1, -2)$,   $(2, 2)$,   $(0, -1)$

Step 1. The general equation of a parabola is $c_0 + c_1x + c_2x^2 = y$. Plugging the data points into this formula gives a matrix equation $\mathrm{A}\mathbf{c} = \mathbf{y}$.
$\left[\Rule{0pt}{4.8em}{0pt}\right.$$\left]\Rule{0pt}{4.8em}{0pt}\right.$ $\left[\begin{matrix}c_0\\c_1\\c_2\end{matrix}\right]$ $=$ $\left[\Rule{0pt}{4.8em}{0pt}\right.$$\left]\Rule{0pt}{4.8em}{0pt}\right.$

Step 2. The matrix equation $\mathrm{A}\mathbf{c} = \mathbf{y}$ has no solution, so instead we use the normal equation $\mathrm{A^T} \mathrm{A}\, \hat{\mathbf{c}} = \mathrm{A^T} \mathbf{y}$
$\mathrm{A^T} \mathrm{A} =$ $\left[\Rule{0pt}{3.6em}{0pt}\right.$$\left]\Rule{0pt}{3.6em}{0pt}\right.$

$\mathrm{A^T} \mathbf{y} =$ $\left[\Rule{0pt}{3.6em}{0pt}\right.$$\left]\Rule{0pt}{3.6em}{0pt}\right.$

Step 3. Solving the normal equation gives the answer
$\hat{\mathbf{c}} =$ $\left[\Rule{0pt}{3.6em}{0pt}\right.$$\left]\Rule{0pt}{3.6em}{0pt}\right.$
which corresponds to the formula
$y =$

You can earn partial credit on this problem.