document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=320535>Question 320535</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The problem asks to find the vertex and graph the following quadratic function.<BR>\n" );
document.write( "y=-x+6x<BR>\n" );
document.write( " My professor says I need only turn in the coordinates. This is what I have tried... x=-b/2a = -6/2(-1) = -6/-2 =3  x=3\r<BR>\n" );
document.write( "\n" );
document.write( "y=-2(3)^2+6(3)= -2(9) + 18 = -18 + 18=0  y=0  My book says the vertex is (3,9) so I am totally confused. I come up with (0,0) for the y intercept but I have no idea how that relates to me finding the coordinates for this graph. I have a very hard time grasping this concept.  Thanks for your time.\r<BR>\n" );
document.write( "\n" );
document.write( "Tracy Jones </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #229549 by </B><A HREF=/tutors/aboutme.mpl?userid=scott8148><B>scott8148(6628)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=scott8148><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=scott8148><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=229549\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> y = -x^2 + 6x\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "you found the axis of symmetry OK (x = -b / 2a)\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "substituting back into the function ___ y = -(3^2) + 6(3) = -9 + 18 = 9\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "not sure where the x^2 coefficient of 2 comes from in your substitution? </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );