Answer
Vertex: $(3,-1)$
Focus: $\left(3,-\dfrac{5}{4}\right)$
Directrix: $y=-\dfrac{3}{4}$
See graph
Work Step by Step
We are given the parabola:
$(x-3)^2=-(y+1)$
The standard equation is:
$(x-h)^2=4p(y-k)$
Determine $h,k,p$:
$h=3$
$k=-1$
$4p=-1\Rightarrow p=-\dfrac{1}{4}$
Determine the vertex:
$(h,k)=(3,-1)$
Determine the focus:
$(h,k+p)=\left(3,-1+\left(-\dfrac{1}{4}\right)\right)=\left(3,-\dfrac{5}{4}\right)$
Determine the directrix:
$y=k-p$
$y=-1-\left(-\dfrac{1}{4}\right)$
$y=-\dfrac{3}{4}$
Determine the two points defining the latus rectum:
$y=-\dfrac{5}{4}$
$(x-3)^2=4\left(-\dfrac{1}{4}\right)\left(-\dfrac{5}{4}+1\right)$
$(x-3)^2=\dfrac{1}{4}$
$x-3=\pm \dfrac{1}{2}$
$x-3=-\dfrac{1}{2}\Rightarrow x_1=\dfrac{5}{2}$
$x-3=\dfrac{1}{2}\Rightarrow x_2=\dfrac{7}{2}$
$\Rightarrow \left(\dfrac{5}{2},-\dfrac{5}{4}\right),\left(\dfrac{7}{2},-\dfrac{5}{4}\right)$
Plot the points, draw the directrix, and graph the parabola:
