document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1136446>Question 1136446</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find the equation of the normal line to the curve.\r<BR>\n" );
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document.write( "1. y=x^3-4x and that is parallel to the line x+8y-8=0 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #754214 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=754214\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> x+8y=8<BR>\n" );
document.write( "y=(-1/8)x+1.  Has a slope of -1/8<BR>\n" );
document.write( "The normal line to the curve has a slope of -1/8 as well<BR>\n" );
document.write( "It is perpendicular to the line tangent to the curve, and that has a negative reciprocal of -1/8 slope, or 8.<BR>\n" );
document.write( "so we want the tangent line to the curve y=x^3-4x, where the slope is 8<BR>\n" );
document.write( "Take the derivative, which is 3x^2-4, and set it equal to 8<BR>\n" );
document.write( "3x^2=12<BR>\n" );
document.write( "x^2=4<BR>\n" );
document.write( "x=-2 and 2.<BR>\n" );
document.write( "At x=both -2 and 2, y=0, so the tangent of the line to the curve at (-2, 0) and (2, 0) has a slope of 8.  From the point slope formula y-y1=m(x-x1), m slope and (x1, y1) point, y=8(x-2) or y=8x-16 and y=8(x+2) or y=8x +16. These are equations of the tangent line to the curve at those two points.\r<BR>\n" );
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document.write( "The normal line to the curve has slope -1/8 and goes through (-2, 0), and its equation is y=(-1/8)(x+2) or <BR>\n" );
document.write( "y=-(1/8)x-1/4. <BR>\n" );
document.write( "There is a second one that has equation y=(-1/8)(x-2) or y=(-1/8)(x)+1/4<BR>\n" );
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