document.write( "Question 1035925: Find the equation of the line that passes through point (0,5) and is tangent to the curve \"+y=x%5E3+-+2x%5E2+%2B+1\".\r
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Algebra.Com's Answer #650762 by Edwin McCravy(20055)\"\" \"About 
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document.write( "Let the point of tangency be (p,q)\r\n" );
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document.write( "Then since it must satisfy y=x³-2x²+1 \r\n" );
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document.write( "*  The point of tangency is (p,p³-2p²+1)\r\n" );
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document.write( "Let the equation of the tangent line be y = mx+b\r\n" );
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document.write( "Since the tangent line goes through (0,5),\r\n" );
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document.write( "    5 = m(0)+b\r\n" );
document.write( "    5 = 0+b\r\n" );
document.write( "    5 = b\r\n" );
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document.write( "*  Equation of tangent line is y = mx+5\r\n" );
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document.write( "The slope of the tangent line is the derivative of\r\n" );
document.write( "y=x³-2x²+1 at the point of tangency:\r\n" );
document.write( "y'=3x²-4x \r\n" );
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document.write( "we substitute y'=m and x=p\r\n" );
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document.write( "m = 3p²-4p\r\n" );
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document.write( "So the equation of the tangent line y = mx+5 is now\r\n" );
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document.write( "*    y = (3p²-4p)x+5\r\n" );
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document.write( "Since the tangent line must go through the point\r\n" );
document.write( "of tangency (p,p³-2p²+1)\r\n" );
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document.write( "p³-2p²+1 = (3p²-4p)p+5\r\n" );
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document.write( "p³-2p²+1 = 3p³-4p²+5\r\n" );
document.write( "       0 = 2p³-2p²+4\r\n" );
document.write( "       0 = p³-p²+2\r\n" );
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document.write( "p = -1 is the only real solution.\r\n" );
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document.write( "We find the equation of the tangent line with p = -1\r\n" );
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document.write( "y = (3p²-4p)x+5\r\n" );
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document.write( "y = [3(-1)²-4(-1)]x+5\r\n" );
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document.write( "y = [3(1)+4]x+5\r\n" );
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document.write( "y = [3+4]x+5\r\n" );
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document.write( "y = 7x+5\r\n" );
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document.write( "   \r\n" );
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document.write( "Edwin
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