Using translation and change of scale to sketch graphs of these quadratics I borrowed the following problems from MIT open courseware problem sets. (This is the very first problem in the problem set.) I understand what completing the square is, but what does it mean when they say “use translation” and “change of scale”. I am unable to solve this as I didn't understand the question. It would be nice if I could get the solution.

By completing the square, use translation and change of scale to sketch: 
a) $y = x^2 − 2x − 1$;
b) $y =3x^2 +6x +2$.

 A: For a) 
Here is a picture with the parabola $y=x^2$ (black) and the parabola $y=x^2−2x−1=(x−1)^2−2$ (blue) translated as André Nicolas indicated in a comment above. 

(Blue axes: $X=x-1$ is the dashed horizontal axis and $Y=y+2$ is the dashed vertical axis. See below) 
I present my own explanation.
If you had the parabola $y=x^{2}$ you would be able to sketch its graph. The
idea is to get a parabola in the same form, but translated with respect to it. How? 
By completing the square you write $y=x^2-2x-1$ as
$$\color{blue}{y}=x^2-2x+1-2=\color{blue}{\left( x-1\right) ^{2}-2}.$$
Now, if you translate the $x,y$-axes, by making the change of
variables $x=X+1$ and $y=Y-2$, you get
$$\color{blue}{Y=X^{2}},$$
which is a parabola centered at $(X,Y)=(0,0)$. Note that the point $(X,Y)=(0,0)$ is located at $\color{blue}{(x,y)=(1,-2)}$. In the figure the dash lines are the  $X$- and $Y$-axes, while the solid ones are the $x$- and $y$-axes.
In this case you have no need to change the scale, because the coefficient of $x^2$ is $1$.

Added: For b)
$$
\begin{eqnarray*}
\color{blue}y &=&3x^{2}+6x+2=3\left( x^{2}+2x+\frac{2}{3}\right)  \\
&=&3\left( x^{2}+2x+1-\frac{1}{3}\right) =3\left( \left( x+1\right) ^{2}-
\frac{1}{3}\right)  \\
&=&\color{blue}{3\left( x+1\right) ^{2}-1}.
\end{eqnarray*}
$$
If you put $x=X-1,y=Y-1$, you get $$\color{blue}{Y=3X^{2}}.$$ The point $(X,Y)=(0,0)$ is now 
$\color{blue}{(x,y)=(-1,-1)}$ and the factor $3$ is the coefficient of $x^2$.

Plots of $y=3x^2$ (black) and $y=3x^2+6x+2=3(x+1)^2-1$ (blue)
(Blue axes: $X=x+1$ is the dashed horizontal axis and $Y=y+1$ is the dashed vertical axis.) 
