document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=5314>Question 5314</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I need to know the steps to solve this problem.  Hope you can help me.\r<BR>\n" );
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document.write( "If f(x) = 3-x^2\r<BR>\n" );
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document.write( "how do I find the slope of the tangent line to the graph of f at (1,2)<BR>\n" );
document.write( "How do I find the slope-intercept equation of the tangent line to the graph of f at (1,2)\r<BR>\n" );
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document.write( "How do I graph f and the tangent line.\r<BR>\n" );
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document.write( "Thank you<BR>\n" );
document.write( "Sue </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #2699 by </B><A HREF=/tutors/aboutme.mpl?userid=longjonsilver><B>longjonsilver(2297)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=longjonsilver><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=longjonsilver><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=2699\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> gradient is found by differentiation.\r<BR>\n" );
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document.write( "f(x) = 3-x^2 so f'(x) = -2x --> this is the gradient at any point on the curve. You need to find the gradient at x=1, so then... f'(x) = -2(1) which is -2.\r<BR>\n" );
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document.write( "Now the tangent at a point (by definition) has the same gradient as the curve at that point, so the general form of the straight line is y=mx+c. We know m=-2, so we have y=-2x+c. We need to find c. To do that we need to know the x and y, which we do! (1,2)...so 2 = -2(1) + c --> c=4. So the tangent has equation y=-2x+4\r<BR>\n" );
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document.write( "When you say graph, you mean sketch it? or do it on graph paper?\r<BR>\n" );
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document.write( "Sketch --> need to know what shape the quadratic is..its x^2 term is -ve so that means it is a n-shape hump. Where does it cross the x-axis and the y-axis? You find out. \r<BR>\n" );
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document.write( "For the straight line... at the point on the curve where x=1, draw the tangent. the line should cross the y-axis at y=4. Remember this is a sketch, so not perfect artistic skills required, but at least try to get the general jist of the shapes and layout :-)\r<BR>\n" );
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document.write( "jon.<BR>\n" );
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