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.)