Question 984896
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{{{4p(y-k)=(x-h)^2}}}
where (h,k) is the vertex and p is distance from vertex to focus
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Given equation:
{{{y=x^2}}}
{{{4(1/4)(y-0)=(x-0)^2}}}
So vertex=(h,k)=(0,0)
Focus is at p from vertex, +1/4 from vertex=(0,1/4)
Directrix is -p from vertex, y=-(1/4) 
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{{{ graph( 800, 800, -10, 10, -10, 10,  x^2/12) }}}