document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1121358>Question 1121358</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find the standard form of the parabola:  x2 - 16x - 48y - 512 = 0<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #737215 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=737215\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Complete the square, adding 64 (half the 16 squared) to both sides<BR>\n" );
document.write( "x^2-16x+64-48y-512=64<BR>\n" );
document.write( "(x-8)^2-576=48y, rearranging.<BR>\n" );
document.write( "(1/12)(x-8)^2-12=y, dividing by 48 both sides\r<BR>\n" );
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document.write( "The vertex is at (8,-12)\r<BR>\n" );
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document.write( "Some consider the form ax^2+bx+c=0 to be standard form.<BR>\n" );
document.write( "This would be y=(1/12)x^2-(3/4)x-5/9\r<BR>\n" );
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