document.write( "Question 638489: y^2-4x+6y+29=0
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document.write( "Identify the conic section. It is a parabola, give the vertex. \n" );
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Algebra.Com's Answer #402318 by lwsshak3(11628)![]() ![]() ![]() You can put this solution on YOUR website! y^2-4x+6y+29=0 \n" ); document.write( "Identify the conic section. It is a parabola, give the vertex. \n" ); document.write( "** \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) \n" ); document.write( " \n" ); document.write( " |