Question 481380
Identify the vertex and directrix of the parabola (x+4)^2=-1/8(y+3) 
and Write the equation of a parabola a directrix at x = 1 and a focus at (-3, 0).
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(x+4)^2=-1/8(y+3)
This is an equation of a parabola of the standard form: (x-h)^2=-4p(y-k), with (h,k) being the coordinates of the vertex, and the parabola opens downwards.
For given equation: 
vertex: (-4,-3)
4p=1/8
p=1/32
directrix:y=-3+1/32)
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directrix at x = 1 and a focus at (-3, 0)
Given data shows this is an equation of a parabola of the standard form: (y-k)^2=-4p(x-h)
vertex: (-1,0)
Equation: y^2=-8(x+1)
see graph as a visual check on the answer
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y=±(-8(x+1))^.5
{{{ graph( 300, 300, -10, 10, -10, 10, (-8(x+1))^.5,-(-8(x+1))^.5) }}}