given $y = a + bx + cx^2$ fits three given points, find and solve the matrix equation for the unknowns $a,b$, and $c$ 
Given $y = a + bx + cx^2$ fits three given points, find and solve the matrix equation for the unknowns $a$, $b$, and $c$.

the equation fits the points $(1,0), (-1, -4),$ and $(2, 11)$
I really don't know what I'm supposed to do here.  I tried setting it up like a normal matrix like
$$\begin{pmatrix}
1&-1&2\\
0&4&11
\end{pmatrix}
(x\quad y)=??$$
but this doesn't make any sense because the original equation has multiple x's in it and I feel like im not taking care of that. 
Can you help me get started with this? I just don't get it and I'm so frustrated.
Thanks!
 A: First note that $y = a + bx + cx^2$ can also be written as
$$(1 \quad x \quad x^2) \left(\begin{array}{c} a \\ b \\ c
\end{array}\right) = y$$
You are given three points, $(x_1,y_1) = (1,0)$, $(x_2,y_2) = (-1,-4)$ and $(x_3,y_3) = (2,11)$. So you can set up three of these equations:
$$\begin{align}
(1 \quad x_1 \quad x_1^2) \left(\begin{array}{c} a \\ b \\ c
\end{array}\right) &= y_1 \\
(1 \quad x_2 \quad x_2^2) \left(\begin{array}{c} a \\ b \\ c
\end{array}\right) &= y_2 \\
(1 \quad x_3 \quad x_3^2) \left(\begin{array}{c} a \\ b \\ c
\end{array}\right) &= y_3
\end{align}$$
Or, in one matrix form:
$$\left(\begin{array}{ccc} 1 & x_1 & x_1^2 \\ 1 & x_2 & x_2^2 \\ 1 & x_3 & x_3^2 \end{array}\right) \left(\begin{array}{c} a \\ b \\ c
\end{array}\right) = \left(\begin{array}{c} y_1 \\ y_2 \\ y_3
\end{array}\right)$$
You know the values of $x_1, x_2, x_3, y_1, y_2, y_3$, so filling them in you will be left with an equation of the form $Ax = b$, with $A$ and $b$ known and $x$ unknown.
Edit: For completeness, filling in those values we get
$$\left(\begin{array}{ccc} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 2 & 4 \end{array}\right) \left(\begin{array}{c} a \\ b \\ c
\end{array}\right) = \left(\begin{array}{c} 0 \\ -4 \\ 11
\end{array}\right)$$
You could do a Gaussian elimination, but this example has a simple solution which we can just calculate by hand. Subtracting the second row from the first we get 
$$(0 \quad 2 \quad 0) \left(\begin{array}{c} a \\ b \\ c
\end{array}\right) = 4 \quad \Longrightarrow \quad \fbox{b = 2}$$
Subtracting the first from the last we get
$$(0 \quad 1 \quad 3) \left(\begin{array}{c} a \\ b \\ c
\end{array}\right) = 11 \quad \Longrightarrow \quad b + 3c = 11 \quad \Longrightarrow \quad \fbox{c = 3}$$
Then finally you could use the first row to get
$$(1 \quad 1 \quad 1) \left(\begin{array}{c} a \\ b \\ c
\end{array}\right) = 0 \quad \Longrightarrow \quad a + b + c = 0 \quad \Longrightarrow \quad \fbox{a = -5}$$
(For the enthusiast, note that the matrix $A$ is a Vandermonde matrix.)
A: If the equation fits three points, then if we assign the x-axis value and $y$-axis value to the equation, we can have a row in the matrix.
Example, assign $(1,0)$ to the equation, we have $a + b+ c = 0$.
The same for the other two points. Then you have a matrix that can be solved using a calculator or by Gaussian calculation.
