document.write( "Question 596993: Find the vertex,focus,and directrix of the equation; x^2+4x+6y-2=0\r
\n" ); document.write( "\n" ); document.write( "(What I've tried)
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\n" ); document.write( " X^2+4x+6y-2=0\r
\n" ); document.write( "\n" ); document.write( " (x+2)^2 = 6y+2 (here's where i'm confusing myself)
\n" ); document.write( " (x+2)^2 = 6(y+2)? If that's right then it should be...
\n" ); document.write( " (x+2)^2 = 4(6/4)(y+2)
\n" ); document.write( " P=(6/4)
\n" ); document.write( " Vertex = (-2,-2)
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Algebra.Com's Answer #378012 by lwsshak3(11628)\"\" \"About 
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Find the vertex,focus,and directrix of the equation;
\n" ); document.write( "x^2+4x+6y-2=0
\n" ); document.write( "complete the square
\n" ); document.write( "(x^2+4x+4)+6y-2-4=0
\n" ); document.write( "(x+2)^2=-6y+6
\n" ); document.write( "(x+2)^2=-6(y-1)
\n" ); document.write( "This is an equation of a parabola that opens downwards of the standard form:
\n" ); document.write( "(x-h)^2=-4p(y-k), (h,k)=(x,y) coordinates of vertex
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\n" ); document.write( "For given equation:
\n" ); document.write( "vertex:(-2,1)
\n" ); document.write( "axis of symmetry: x=-2
\n" ); document.write( "4p=6
\n" ); document.write( "p=3/2
\n" ); document.write( "focus:(-2,1-3/2)=(-2,1/2) (p units below vertex on axis of symmetry)
\n" ); document.write( "directrix: y=1+3/2
\n" ); document.write( "directrix: y=5/2 (p units above vertex on axis of symmetry)\r
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