Answer
vertex: $\quad (3,-1)$
focus: $\displaystyle \quad (3,-\frac{5}{4})$
directrix: $\displaystyle \quad y=-\frac{3}{4}$
Work Step by Step
Comparing the form of the equation with table 2,
$\quad \quad \quad (x-3)^{2}=-(y+1)\quad $has the form
$\begin{array}{|c|c|c|c|} \hline
{focus}&{directrix}&{equation}& ... opens\\ \hline
{(h,k-a)}&{y=k+a}&{(x-h)^{2}=-4a(y-k)}& down \\ \hline \end{array}$
$(x-3)^{2}=-4\displaystyle \cdot(\frac{1}{4})(y+1)\quad \Rightarrow\quad h=3, k=-1, a=\displaystyle \frac{1}{4}$
vertex: $\quad (h,k)=(3,-1)$
focus: $\displaystyle \quad (h,k-a)=(3,-1-\frac{1}{4})=(3,-\frac{5}{4})$
directrix: $\displaystyle \quad y=k+a\quad \Rightarrow\quad y=-\frac{3}{4}$
To graph, find the endpoints of the latus rectum
(line segment parallel to the directrix, containing the focus)
by setting $y=-\displaystyle \frac{5}{4}$
$(x-3)^{2}=-(-\displaystyle \frac{5}{4}+1)$
$(x-3)^{2}=\displaystyle \frac{1}{4}$
$x-3=\displaystyle \pm\frac{1}{2}$
$x=3\displaystyle \pm\frac{1}{2}$
Endpoints of the latus rectum: $(2.5,-1.25), (3.5,-1.25)$
