SOLUTION: graph the parabola and identify the vertex, focus, directrix: x^2+16y=0

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Question 932351: graph the parabola and identify the vertex, focus, directrix: x^2+16y=0
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
graph the parabola and identify the vertex, focus, directrix: x^2+16y=0
This is a parabola that opens downward.
Its basic equation: (x-h)^2=4p(y-k), (h,k)=coordinates of vertex
For given parabola:
x^2=-16y
vertex: (0,0)
axis of symmetry: x=0 or y-axis
4p=16
p=4
focus:(0,-4) (p-units below vertex on the axis of symmetry)
directrix:y=4 (p-units above vertex on the axis of symmetry)
see graph below:
+graph%28+300%2C+300%2C+-10%2C+10%2C+-10%2C+10%2C+-x%5E2%2F16%29+