document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=916845>Question 916845</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Figuring out graphing of this polynomial: <BR>\n" );
document.write( "f(x) = 12x^3 - 12x^2 - 24x \r<BR>\n" );
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document.write( "what I have so far:<BR>\n" );
document.write( "I know I can factor it down, therefore 12x(x+1)(x-2)<BR>\n" );
document.write( "and the x-int would be (-1,0),(2,0)<BR>\n" );
document.write( "But why is (0,0) also a factor? I don't understand that. \r<BR>\n" );
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document.write( "also how would I find the degree and leading coefficient? very lost on that concept <BR>\n" );
document.write( "And when I plug this poly into a graphing calculator it has a steepness and a lowness at two points. I am in precalculus so I don't know the finer details of graphing/derivatives and I must be able to do this freehand how would I know how far up to make the graph and how far low to make the graph?  Am I just suppose to estimate them for precalculus? What if both sides bounces how do I tell which one is higher or lower or if they are both the same? \r<BR>\n" );
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document.write( "Thank you </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #556314 by </B><A HREF=/tutors/aboutme.mpl?userid=josgarithmetic><B>josgarithmetic(39617)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=josgarithmetic><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=556314\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> You also have another x-intercept:  (0,0), because of the factor, x.\r<BR>\n" );
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document.write( "The highest exponent of the variable is 3, from 12x^3.<BR>\n" );
document.write( "The degree of the function is 3.\r<BR>\n" );
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document.write( "You can test the intervals which the x-intercepts define and get the signs in each interval.  The intervals to check are these:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-infinity%3Cx%3C-1 ALIGN=MIDDLE  ALT=\"-infinity%3Cx%3C-1\"   DISABLED_style= \"max-width:90%; height:auto;\" >,<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-1%3Cx%3C0 ALIGN=MIDDLE  ALT=\"-1%3Cx%3C0\"   DISABLED_style= \"max-width:90%; height:auto;\" >;<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=0%3Cx%3C2 ALIGN=MIDDLE  ALT=\"0%3Cx%3C2\"   DISABLED_style= \"max-width:90%; height:auto;\" >;<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=2%3Cx%3Cinfinity ALIGN=MIDDLE  ALT=\"2%3Cx%3Cinfinity\"   DISABLED_style= \"max-width:90%; height:auto;\" >.\r<BR>\n" );
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document.write( "Finding the maximum and minimum values is something that you just must estimate, if you want these.  Beginning Calculus will teach how to determine them more accurately.\r<BR>\n" );
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document.write( "Based on the leading term of f, the values of f fall toward the left and rise to the right.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=graph%28300%2C300%2C-6%2C6%2C-22%2C8%2C12x%5E3-12x%5E2-24x%29 ALIGN=MIDDLE  ALT=\"graph%28300%2C300%2C-6%2C6%2C-22%2C8%2C12x%5E3-12x%5E2-24x%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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