document.write( "Question 773821: how to graph and identify the focus, directrix, and axis of symmetry of x=-y^2
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Algebra.Com's Answer #472421 by lwsshak3(11628)\"\" \"About 
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how to graph and identify the focus, directrix, and axis of symmetry of x=-y^2
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\n" ); document.write( "Given equation is that of a parabola which opens leftward.
\n" ); document.write( "Its basic form: (y-k)^2=-4p(x-h), (h,k)=(x,y) coordinates of the vertex.
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\n" ); document.write( "y^2=-x
\n" ); document.write( "vertex: (0,0)
\n" ); document.write( "axis of symmetry: y=0
\n" ); document.write( "4p=1
\n" ); document.write( "p=1/4
\n" ); document.write( "focus: (-1/4,0) (p distance to left of vertex on the axis of symmetry)
\n" ); document.write( "directrix: x=1/4 (p distance to right of vertex on the axis of symmetry)
\n" ); document.write( "see graph below:
\n" ); document.write( "\"+graph%28+300%2C+300%2C+-10%2C+10%2C+-10%2C+10%2C+sqrt%28-x%29%2C+-sqrt%28-x%29%29+\"
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