document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=8488>Question 8488</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Please help me.  I have no idea what they want.\r<BR>\n" );
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document.write( "5.  Put the following in turning point form (y = a(x-h)^2+k) and give the vertex and y-intercept.\r<BR>\n" );
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document.write( "a)  <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+y+=+x%5E2+%2B+2x+%2B+3+ ALIGN=MIDDLE  ALT=\"+y+=+x%5E2+%2B+2x+%2B+3+\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "c)  <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+y+=+3x%5E2+-+6x+%2B+4+ ALIGN=MIDDLE  ALT=\"+y+=+3x%5E2+-+6x+%2B+4+\"   DISABLED_style= \"max-width:90%; height:auto;\" > </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #4686 by </B><A HREF=/tutors/aboutme.mpl?userid=Earlsdon><B>Earlsdon(6294)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Earlsdon><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=4686\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> While I'm familiar with the form: <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y+=+a%28x-h%29%5E2+%2B+k ALIGN=MIDDLE  ALT=\"y+=+a%28x-h%29%5E2+%2B+k\"   DISABLED_style= \"max-width:90%; height:auto;\" >, I was not aware that this form was called the turning point form...you learn something new every day!\r<BR>\n" );
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document.write( "Anyway, putting your quadratic equations in this form you can quickly identify the location of the vertex of the parabola and other features.\r<BR>\n" );
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document.write( "So, step 1, get the quadratics into the \"turning point\" form, <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y+=+a%28x-h%29%5E2+%2B+k ALIGN=MIDDLE  ALT=\"y+=+a%28x-h%29%5E2+%2B+k\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "a)  <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y+=+x%5E2+%2B+2x+%2B+3 ALIGN=MIDDLE  ALT=\"y+=+x%5E2+%2B+2x+%2B+3\"   DISABLED_style= \"max-width:90%; height:auto;\" >  You will need to \"complete the square\". This is a process by which you can write any general quadratic function <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y+=+ax%5E2+%2B+bx+%2B+c ALIGN=MIDDLE  ALT=\"y+=+ax%5E2+%2B+bx+%2B+c\"   DISABLED_style= \"max-width:90%; height:auto;\" > in the form <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y+=+a%28x-h%29%5E2+%2B+k ALIGN=MIDDLE  ALT=\"y+=+a%28x-h%29%5E2+%2B+k\"   DISABLED_style= \"max-width:90%; height:auto;\" > where the vertex of its graph can be identified. Heres how:\r<BR>\n" );
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document.write( "Isolate the x-terms by subtracting 3 from both sides:\r<BR>\n" );
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document.write( "  <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y-3+=+x%5E2+%2B+2x ALIGN=MIDDLE  ALT=\"y-3+=+x%5E2+%2B+2x\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Add the square of one half the x-coefficient.  That's <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28%281%2F2%29%282%29%29%5E2 ALIGN=MIDDLE  ALT=\"%28%281%2F2%29%282%29%29%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" > = 1\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y-3%2B1+=+x%5E2+%2B+2x+%2B+1 ALIGN=MIDDLE  ALT=\"y-3%2B1+=+x%5E2+%2B+2x+%2B+1\"   DISABLED_style= \"max-width:90%; height:auto;\" >  Factor the trinomial as the square of a binomial.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y-2+=+%28x+%2B+1%29%5E2 ALIGN=MIDDLE  ALT=\"y-2+=+%28x+%2B+1%29%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" >  now add 2 to both sides.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y+=+%28x+%2B+1%29%5E2+%2B+2 ALIGN=MIDDLE  ALT=\"y+=+%28x+%2B+1%29%5E2+%2B+2\"   DISABLED_style= \"max-width:90%; height:auto;\" > Here's your quadratic in \"turning point\" form, <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y+=+a%28x-h%29%5E2+%2B+k ALIGN=MIDDLE  ALT=\"y+=+a%28x-h%29%5E2+%2B+k\"   DISABLED_style= \"max-width:90%; height:auto;\" >.\r<BR>\n" );
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document.write( "The coordinates of the vertex are at: (h, k)<BR>\n" );
document.write( "In this case, h = -1 and k = 2, so the vertex is located at (-1, 2)\r<BR>\n" );
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document.write( "To find the y-intercept, just set x = 0 in the original equation and solve for y.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y+=+x%5E2+%2B+2x+%2B+3 ALIGN=MIDDLE  ALT=\"y+=+x%5E2+%2B+2x+%2B+3\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y+=+0%5E2+%2B+2%280%29+%2B+3 ALIGN=MIDDLE  ALT=\"y+=+0%5E2+%2B+2%280%29+%2B+3\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y+=+3 ALIGN=MIDDLE  ALT=\"y+=+3\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "The y-intercept is located at: (0, 3)\r<BR>\n" );
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document.write( "Try this on problem c)\r<BR>\n" );
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