document.write( "Question 614352: Graph the parabola by finding the vertex, directrix, and opening. Equation given 4x-3=-ysq.+4y+1 \n" ); document.write( "
Algebra.Com's Answer #386522 by lwsshak3(11628)\"\" \"About 
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Graph the parabola by finding the vertex, directrix, and opening. Equation given
\n" ); document.write( "4x-3=-ysq.+4y+1
\n" ); document.write( "4x-3=-y^2+4y+1
\n" ); document.write( "complete the square
\n" ); document.write( "4x-3=-(y^2-4y+4)+1+4
\n" ); document.write( "4x-3=-(y-2)^2+5
\n" ); document.write( "4x-8=-(y-2)^2
\n" ); document.write( "-(y-2)^2=4x-8
\n" ); document.write( "(y-2)^2=-4(x-2)
\n" ); document.write( "This is an equation of a parabola that opens leftwards.
\n" ); document.write( "Its standard form: (y-k)^2=-4p(x-h), (h,k)=(x,y) coordinates of the vertex
\n" ); document.write( "For given equation: (y-2)^2=-4(x-2)
\n" ); document.write( "Vertex: (2,2)
\n" ); document.write( "axis of symmetry: y=2
\n" ); document.write( "4p=4
\n" ); document.write( "p=1
\n" ); document.write( "directrix: x=3 (one unit to the right of the vertex on the axis of symmetry)\r
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