Answer
$ a.\qquad$
New vertex:$\quad V'(-1,3)$
New focus: $ \quad F'(-1,2)$
New directrix: $ \quad y=4$
$ b.\qquad$
See graph.
Work Step by Step
$ a.\qquad$
$x^{2}=-4y=-4(1)y$
Which has the form:
$y=-\displaystyle \frac{x^{2}}{4p}$ ,$\qquad x^{2}=-4py\ \ $...opens down
Vertex = $(0,0)$
Focus = $(0,-p)=(0,-1)$
Directrix: $y=p=1$
New vertex: $\left\{\begin{array}{ll}
x'=x-1=0-1=-1 & \text{... shift left}\\
y'=y+3=0+3=3 & \text{... shift up}
\end{array}\right., \quad V'(-1,3)$
New focus: $\left\{\begin{array}{ll}
x'=x-1=0-1=-1 & \text{... shift left}\\
y'=y+3=-1+3=2 & \text{... shift up}
\end{array}\right., \quad F'(-1,2)$
New directrix (horizontal line, shifted up): $y=4$
$ b.\qquad$
See graph.
The vertex and focus are now on the line $x=-1$
(the y-axis is shifted left).
