SOLUTION: x^2+2x-4y+17=0 Find Vertex and Directrix of the parabola

Question 817124: x^2+2x-4y+17=0
Find Vertex and Directrix of the parabola
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
Find Vertex and Directrix of the parabola
x^2+2x-4y+17=0
complete the square:
(x^2+2x+1)-4y+17-1=0
(x+1)^2=4y-16
(x+1)^2=4(y-4)
This is an equation of a parabola that opens up.
Its basic form of equation: (x-h)^2=4p(y-k)
For given parabola:
vertex:(-1,4)
axis of symmetry:x=-1
4p=4
p=1
directrix:y=3 (p-distance below vertex on the axis of symmetry)
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