Question 618495
How do you find the vertex, focus, directrix, and axis of symmetry of the parabola? 
x^2-2x+8y+9=0
complete he square
(x^2-2x+1)+8y+9-1=0
(x-1)^2=-8y-8
(x-1)^2=-8(y+1)
This is an equation of a parabola that opens downwards
Form of equation: (x-h)^2=-4p(y-k), (h,k)=(x,y) coordinates of vertex
For given equation:
vertex:(1,-1)
axis of symmetry: x=1
4p=8
p=2
focus: (1,-1-p)=(1,-1-2)=(1,-3) (p units below vertex on axis of symmetry)
directrix: y=-1+p=-1+2=1 (p units above vertex on axis of symmetry)