SOLUTION: how to graph and identify the focus, directrix, and axis of symmetry of x=-y^2

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Question 773821: how to graph and identify the focus, directrix, and axis of symmetry of x=-y^2
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
how to graph and identify the focus, directrix, and axis of symmetry of x=-y^2
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Given equation is that of a parabola which opens leftward.
Its basic form: (y-k)^2=-4p(x-h), (h,k)=(x,y) coordinates of the vertex.
..
y^2=-x
vertex: (0,0)
axis of symmetry: y=0
4p=1
p=1/4
focus: (-1/4,0) (p distance to left of vertex on the axis of symmetry)
directrix: x=1/4 (p distance to right of vertex on the axis of symmetry)
see graph below:
+graph%28+300%2C+300%2C+-10%2C+10%2C+-10%2C+10%2C+sqrt%28-x%29%2C+-sqrt%28-x%29%29+