document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1026704>Question 1026704</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find the equation of the normal line to the curve y = 3x^3 + 4x^2 - 5x + 1 at x = 1. Thank youuuu </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #641973 by </B><A HREF=/tutors/aboutme.mpl?userid=stanbon><B>stanbon(75887)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=stanbon><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=641973\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Find the equation of the normal line to the curve y = 3x^3 + 4x^2 - 5x + 1 at x = 1. Thank youuuu <BR>\n" );
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document.write( "slope = 9x^2 + 8x - 5<BR>\n" );
document.write( "f'(1) = 9 + 8 - 5 = 12<BR>\n" );
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document.write( "slope of normal:: -1/12<BR>\n" );
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document.write( "f(1) = 3 + 4 - 5 + 1 = 3<BR>\n" );
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document.write( "Find linear equation with m = -1/12 passing thru (1,3)<BR>\n" );
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document.write( "Form::<BR>\n" );
document.write( "3 = (-1/12)1 + b<BR>\n" );
document.write( "b = (37/12)<BR>\n" );
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document.write( "Equation::<BR>\n" );
document.write( "y = (-1/12)x + (37/12)<BR>\n" );
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document.write( "Cheers,<BR>\n" );
document.write( "Stan H.<BR>\n" );
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