document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=261766>Question 261766</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The equation of the tangent line to the curve y=x^3-6x^2 at its point of inflection is\r<BR>\n" );
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document.write( "(A) y= -12x+8<BR>\n" );
document.write( "(B) y= -12x+40<BR>\n" );
document.write( "(C) y= 12x-8<BR>\n" );
document.write( "(D) y= -12x+12<BR>\n" );
document.write( "(E) y= 12x-40\r<BR>\n" );
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document.write( "I know that the point of inflection relates to the second derivative, so do I just take the double derivative of the given equation? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #192874 by </B><A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><B>jim_thompson5910(35256)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><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=192874\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> First, derive <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y=x%5E3-6x%5E2 ALIGN=MIDDLE  ALT=\"y=x%5E3-6x%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" > to get <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=dy%2Fdx=3x%5E2-12x ALIGN=MIDDLE  ALT=\"dy%2Fdx=3x%5E2-12x\"   DISABLED_style= \"max-width:90%; height:auto;\" >. Derive <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=dy%2Fdx ALIGN=MIDDLE  ALT=\"dy%2Fdx\"   DISABLED_style= \"max-width:90%; height:auto;\" > to get the second derivative <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=d%5E2y%2Fdx%5E2=6x-12 ALIGN=MIDDLE  ALT=\"d%5E2y%2Fdx%5E2=6x-12\"   DISABLED_style= \"max-width:90%; height:auto;\" >. Recall that the point of inflection occurs when the second derivative is zero. So set the second derivative equal to zero to get the equation <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=6x-12=0 ALIGN=MIDDLE  ALT=\"6x-12=0\"   DISABLED_style= \"max-width:90%; height:auto;\" >. Solve for 'x' and use this 'x' value to find the corresponding 'y' value of the point of inflection. Then use the 'x' value and plug it into <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=dy%2Fdx=3x%5E2-12x ALIGN=MIDDLE  ALT=\"dy%2Fdx=3x%5E2-12x\"   DISABLED_style= \"max-width:90%; height:auto;\" > to find the slope of the tangent line at that x value.\r<BR>\n" );
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document.write( "Since you have the slope of the line (ie the slope of the tangent) and the point that the line goes through (the inflection point), you can the formula <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y-y%5B1%5D=m%28x-x%5B1%5D%29 ALIGN=MIDDLE  ALT=\"y-y%5B1%5D=m%28x-x%5B1%5D%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >, where 'm' is the slope and <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \left(x_1,y_1\right)\"> is the given point the line goes through, to find the equation of the tangent line.  </SPAN>\n" );
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