document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=986687>Question 986687</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Determine the gradient of the tangent to the following curve where x=2: y=(x^2 -3)(x+3)<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #607459 by </B><A HREF=/tutors/aboutme.mpl?userid=Cromlix><B>Cromlix(4381)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Cromlix><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=Cromlix><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=607459\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Hi there,<BR>\n" );
document.write( "y = (x^2 - 3)(x + 3)<BR>\n" );
document.write( "y = x^3 + 3x^2 - 3x - 9<BR>\n" );
document.write( "Differentiate:<BR>\n" );
document.write( "dy/dx = 3x^2 + 6x - 3<BR>\n" );
document.write( "Substitute x = 2<BR>\n" );
document.write( "= 3(2)^2 + 6(2) - 3<BR>\n" );
document.write( "= 12 + 12 - 3<BR>\n" );
document.write( "= 21<BR>\n" );
document.write( "Gradient of tangent = 21<BR>\n" );
document.write( "Hope this helps :-) </SPAN>\n" );
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