Question 617006
given that a parabola has a directrix of y=-3.25 and focus (0,2.75),...find the vertex 
then write the equetion of the parabola and graph the parabola 
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This is a parabola that opens upwards:
Its standard form of equation: (x-h)^2=4p(y-k), (h,k)=(x,y) coordinates of the vertex, A=multiplier which affects width or steepness of the curve.
axis of symmetry: y-axis or x=0
x-coordinate of vertex=0
y-coordinate of vertex=3 (halfway between focus and directrix on the axis of symmetry)(2.75+3.25)/2=3
vertex: (0,3)
p=.25 (distance from focus or directrix to vertex on the axis of symmetry
4p=1
Equation: x^2=y-3
see graph below:
 {{{ graph( 300, 300, -10,10, -10, 10,x^2+3) }}}