Question 544669
Given that a parabola has a directrix of y= -3.25 and focus (0,-2.75), find the vertex. 
Write the equation of the parabola then graph the parabola.
**
Standard form of equation of given parabola: (x-h)^2=4p(y-k), (h,k)=(x,y) coordinates of vertex.
..
Since the given x-coordinate of the focus=0, the axis of symmetry is x=0.
And since the focus is above the directrix, the parabola opens upwards.
x-coordinate of vertex=x-coordinate of the focus=0
y-coordinate of vertex is midway between focus and directrix on the axis of symmetry:
(-3.25-2.75)/2=-6/2=-3
vertex:(0,3)
p=distance from vertex to focus or to directrix on the axis of symmetry=-3 to-2.75=.25
4p=.25=1/4
p=1/16
Equation of given parabola:
(x-0)^2=(1/16)(y+3)
x^2=(y+3)/16
..
see graph below as a visual check on answers above:
y=16x^2-3
{{{ graph( 300, 300, -5, 5, -10, 10,16x^2-3) }}}