document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=240648>Question 240648</A>: <I><SPAN STYLE=\"word-break: break-all;\"> How do you find the turning point (vertex) of a Quadratic function, not using complete the square? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #176293 by </B><A HREF=/tutors/aboutme.mpl?userid=rapaljer><B>rapaljer(4671)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=rapaljer><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=rapaljer><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=176293\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> If the equation is y=ax^2 +bx +c, then the vertex is always at the value where x=-b/(2a).  After you find x, substitute this back into the original equation (or let the calcluator do it for you!!) to find y.\r<BR>\n" );
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document.write( "Dr. Robert J. Rapalje, Retired<BR>\n" );
document.write( "Seminole State College of Florida<BR>\n" );
document.write( "Altamonte Springs Campus  \r<BR>\n" );
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