document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1090061>Question 1090061</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The question asks:<BR>\n" );
document.write( "The tangent line intersects the graph of f(x)=2x^3+10x^2&#8722;28x at another point. Determine the other point of intersection. I have already found my tangent line eq'n y=-34x+18 @ ×= -3. I substituted and factored :<BR>\n" );
document.write( "-34x+18=2x^3+10x^2-28x=0<BR>\n" );
document.write( "2x^3+10x^2+6x-18=0<BR>\n" );
document.write( "2(x^3+5x^2+3x-9)=0<BR>\n" );
document.write( "2(x-1)(x+3)^2=0<BR>\n" );
document.write( "I know that my x-axis is 1, how do I find y-coordinate? Or do I need to ? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #704476 by </B><A HREF=/tutors/aboutme.mpl?userid=natolino_2017><B>natolino_2017(77)</B></A><img src=\"/images/stars/stars-08.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=natolino_2017><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=704476\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> You can evaluate on f or in the tangent line.\r<BR>\n" );
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document.write( "y(1) = f(1) = -34 + 18 = 2 + 10 -28 = 16.\r<BR>\n" );
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document.write( "So the second point is (1,16) although it has a different slope on that point (f'(1) = 6+20-28= -2 which is not -34)\r<BR>\n" );
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document.write( "@natolino_  </SPAN>\n" );
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