document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1159778>Question 1159778</A>: <I><SPAN STYLE=\"word-break: break-all;\"> How do i find the quadratic equation for this:<BR>\n" );
document.write( "\"You're designing a candy bowl that has a circular bottom surface. They want the diameter of the bowl in the opening above to be 12 cm and the height of the bowl to be 8 cm. Formulate a quadratic equation that, when rotating around the y-axis, describes a circular bottom surface. You can decide the diameter for the bottom yourself.\"\r<BR>\n" );
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document.write( "I have set the diameter to 2cm and decided that it will rotate from x=1 to x=-1. Shouldn't the equation be y = (x+1)(x-1)???\r<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 #782885 by </B><A HREF=/tutors/aboutme.mpl?userid=MowMow><B>MowMow(42)</B></A><img src=\"/images/stars/stars-07.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=MowMow><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=MowMow><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=782885\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> No.  The parabola has to come below the x-axis to have a circular base.<BR>\n" );
document.write( "So lets bring it down below the x-axis to (0,-2) for the vertex.<BR>\n" );
document.write( "Then the vertex equation is <BR>\n" );
document.write( "y = a(x-h)^2 + k<BR>\n" );
document.write( "(h,k) = (0,-2)<BR>\n" );
document.write( "and a point the parabola goes through above the x-axis is (6,8)<BR>\n" );
document.write( "so 8 = a(6-0)^2 -2<BR>\n" );
document.write( "   10=36a<BR>\n" );
document.write( "    a = 10/36 = 5/18<BR>\n" );
document.write( "The equation would be:  y = (5/18)(x)^2 -2\r<BR>\n" );
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