Least square approximation where X vector is given. I want to find a curve of the form $y = a + b \sqrt{x}$ that best fits the points: $(3, 1.5)$, $(7, 2.5)$ and $(10, 3)$ by substituting the $x$ vector $= \sqrt{x}$
My understanding of the process to find the least squares approximation was that the $x$ vector contained the unknown variables so you could use guassian elimination to find the variables $a$,$b$ from: $A^t A x = A^t B$ but I am unsure how this would work when the $x$ vector is already given and you want to find $a$,$b$ still.
 A: The points are given as $(x_i, y_i)$ and one has the unknowns $a$, $b$ which need to be optimized:
$$
y_i = a + b \sqrt{x_i} \quad (i \in \{1,2,3\}) \iff \\
A x = y
$$
with
$$
x = (a, b)^t \in \mathbb{R}^2, \quad y = (y_i) \in \mathbb{R}^3, \quad A = (a_{ij}) \in \mathbb{R}^{3\times 2}, \quad a_{i1} = 1, \quad a_{i2} = \sqrt{x_i} \quad (i \in \{1,2,3\})
$$
gives
$$
A x = y \iff 
\left(
\begin{matrix}
1 & \sqrt{x_1} \\
1 & \sqrt{x_2} \\
1 & \sqrt{x_3}
\end{matrix}
\right)
\left(
\begin{matrix}
a \\
b
\end{matrix}
\right)
=
\left(
\begin{matrix}
y_1 \\
y_2 \\
y_3
\end{matrix}
\right)
$$
$A$ is overspecified (more equations than unknowns), so the matrix is not square and can not be inverted. 
However
$$
A^t A x = A^t y \Rightarrow x = (A^tA)^{-1} A^t y
$$
can be shown to give a solution with minimal error in the $2$-norm, which is the standard, Euclidean norm.
I get $x = (-0.31587, 1.05405)^t$ which gives $Ax = (1.5098, 2.4729, 3.0173)^t$ and $\delta = \lVert Ax-y \rVert = 0.033639$, which seems reasonable.

Graph of $f(x) = a + b \sqrt{x}=−0.31587+ 1.05405\sqrt{x}$.
