Write the equation of the image of the parabola $y=x^2-2x+1$ under a translation $\vec{p}=\hat{i}+3\hat{j}$ Write the equation of the image of the parabola $y=x^2-2x+1$ under a translation $\vec{p}=\hat{i}+3\hat{j}$

$y=(x-1)^2$ is a parabola whose vertex is $(1,0)$ and focus is $(1,\frac{1}{4})$.I dont know how to solve further.
 A: 
Write the equation of the image of the parabola $y=x^2-2x+1$ under a translation $\vec{p}=\hat{i}+3\hat{j}$

Translating $y=f(x)$ over $\vec{p}=\color{blue}{a}\hat{i}+\color{red}{b}\hat{j}$ corresponds to the equation $y-\color{red}{b}=f(x-\color{blue}{a})$ in the same coordinate system; so with $y=x^2-2x+1$ and $\vec{p}=\color{blue}{1}\hat{i}+\color{red}{3}\hat{j}$, you get:
$$y-\color{red}{3}=(x-\color{blue}{1})^2-2(x-\color{blue}{1})+1 \iff y=x^2 - 4 x + 7 \tag{$*$}$$

$y=(x-1)^2$ is a parabola whose vertex is $(1,0)$

After rewriting the parabola in this standard form, you could also simply shift the vertex towards $(1+\color{blue}{1},0+\color{red}{3})=\color{purple}{(2,3)}$, so the equation becomes:
$$y=(x-\color{purple}{2})^2+\color{purple}{3}\iff y=x^2 - 4 x + 7 \tag{$\star$}$$
Note that $(*)$ and $(\star)$ match.
A: $p(x,y)=(x,y)+(1,3)=(x+1,y+3), \quad p(x,y)^t=(x+1,y+3)^t$ so $p$ can be seen as a matrix 
$\pmatrix{1 & 0 & 1\\ 0 & 1 & 3 \\ 0 & 0 & 1}$ which is $\pmatrix{1 & 0 & 1\\ 0 & 1 & 3 \\ 0 & 0 & 1}\pmatrix{x \\ y \\ 1}=\pmatrix{x+1 \\ y+3 \\ 1}$ 
so for $y=(x-1)^2$ it is 
$\pmatrix{1 & 0 & 1\\ 0 & 1 & 3 \\ 0 & 0 & 1}\pmatrix{x \\ (x-1)^2 \\ 1}=\pmatrix{x+1 \\ (x-1)^2+3 \\ 1}$
so $y'=(x-1)^2+3=x^2-2x+4=(x'-1)^2-2(x'-1)+4=x'^2-4x'+7$
