document.write( "Question 481380: Identify the vertex and directrix of the parabola (x+4)^2=-1/8(y+3)\r
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\n" ); document.write( "\n" ); document.write( "Write the equation of a parabola a directrix at x = 1 and a focus at (-3, 0).\r
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Algebra.Com's Answer #329638 by lwsshak3(11628)\"\" \"About 
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Identify the vertex and directrix of the parabola (x+4)^2=-1/8(y+3)
\n" ); document.write( "and Write the equation of a parabola a directrix at x = 1 and a focus at (-3, 0).
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\n" ); document.write( "(x+4)^2=-1/8(y+3)
\n" ); document.write( "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.
\n" ); document.write( "For given equation:
\n" ); document.write( "vertex: (-4,-3)
\n" ); document.write( "4p=1/8
\n" ); document.write( "p=1/32
\n" ); document.write( "directrix:y=-3+1/32)
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\n" ); document.write( "directrix at x = 1 and a focus at (-3, 0)
\n" ); document.write( "Given data shows this is an equation of a parabola of the standard form: (y-k)^2=-4p(x-h)
\n" ); document.write( "vertex: (-1,0)
\n" ); document.write( "Equation: y^2=-8(x+1)
\n" ); document.write( "see graph as a visual check on the answer
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\n" ); document.write( "y=±(-8(x+1))^.5
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