document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=347826>Question 347826</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find the vertex, line of symmetry, the max/min value of the quadratic function and graph the function.\r<BR>\n" );
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document.write( "f(x)= -2x^2+2x+8\r<BR>\n" );
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document.write( "I have tried to solve this several times and I keep getting stuck. This is what I have tried:<BR>\n" );
document.write( "putting it into the form ax^2+bx-c to complete the square<BR>\n" );
document.write( "-2x^2+2x=-8<BR>\n" );
document.write( "Divide both sides by -2 to make x^2 a value of 1<BR>\n" );
document.write( "-2x^2+2x/-2 = -8/-2<BR>\n" );
document.write( "x^2+x = -4<BR>\n" );
document.write( "(1/2)^2 = 2/4<BR>\n" );
document.write( "1/4 = 1/2<BR>\n" );
document.write( "This is about as far as I can seem to go. I don't understand this stuf at all and I have reviewed many different sites as well as the information from my class and I am just lost. Any help will be very appreciated.  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #248689 by </B><A HREF=/tutors/aboutme.mpl?userid=solver91311><B>solver91311(24713)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=solver91311><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=solver91311><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=248689\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font face=”Garamond” size=”+2”>\r<BR>\n" );
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document.write( "The vertex of a parabola described by the function \r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE\ \ \ \ \ \ \ \ \ \ f(x)\ =\ ax^2\ +\ bx\ +\ c\">\r<BR>\n" );
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document.write( "has an <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large x\">-coordinate of <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large \frac{-b}{2a}\"> and a <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large y\">-coordinate of <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large f\left(\frac{-b}{2a}\right)\">.  The axis of symmetry is the vertical line <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large x\ =\ \frac{-b}{2a}\">.  If the lead coefficient is positive, the parabola opens upward.  Therefore the value of the function at the vertex (i.e. the <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large y\">-coordinate as discussed above) is the minimum value of the function.  If the lead coordinate is less than zero, then the parabola opens downward and the vertex is a maximum.\r<BR>\n" );
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document.write( "For your function, the <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large x\">-coordinate of the vertex is <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large \frac{-2}{2(-2)}\ =\ \frac{1}{2}\">.  You can calculate <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large f\left(\frac{1}{2}\right)\"> to get the <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large y\">-coordinate of the vertex.  The rest of your questions fall right in line once you have the coordinates of the vertex.\r<BR>\n" );
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document.write( "John<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE e^{i\pi} + 1 = 0\"><BR>\n" );
document.write( "My calculator said it, I believe it, that settles it<BR>\n" );
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