document.write( "Question 638489: y^2-4x+6y+29=0
\n" ); document.write( "Identify the conic section. It is a parabola, give the vertex.
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Algebra.Com's Answer #402318 by lwsshak3(11628)\"\" \"About 
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y^2-4x+6y+29=0
\n" ); document.write( "Identify the conic section. It is a parabola, give the vertex.
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\n" ); document.write( "y^2-4x+6y+29=0
\n" ); document.write( "y^2+6y-4x+29=0
\n" ); document.write( "complete the square
\n" ); document.write( "(y^2+6y+9)-4x=-29+9
\n" ); document.write( "(y+3)^2-4x+20=0
\n" ); document.write( "(y+3)^2-4(x-5)=0
\n" ); document.write( "conic section is that of a parabola that opens rightwards
\n" ); document.write( "Its standard form of equation: (y-k)^2=4p(x-h), (h,k)=(x,y) coordinates of the vertex
\n" ); document.write( "vertex: (5,-3)
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