document.write( "Question 593835: Find the equation of line with a slope -1 that is the tangent to the curve y=1/x-1\r
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\n" ); document.write( "\n" ); document.write( "I am only in an advanced functions course and have not learned how to take the derivative of a function yet.\r
\n" ); document.write( "\n" ); document.write( "I think the solution should have two equations since the original function has two curves and both have vertical asymptotes with the y-axis and horizontal asymptotes as x=-1.\r
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Algebra.Com's Answer #376493 by AnlytcPhil(1806)\"\" \"About 
You can put this solution on YOUR website!
To see what we are doing let's draw the graph\r\n" );
document.write( "of y= \"1%2F%28x-1%29\"\r\n" );
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document.write( "Now I will draw a bunch of green lines which have slope -1.\r\n" );
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document.write( "You will notice that some of those green lines\r\n" );
document.write( "intersect the graph twice, some don't intersect\r\n" );
document.write( "it at all, and the two lines that are tangent\r\n" );
document.write( "to the graph intersect it at just one point.\r\n" );
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document.write( "We want the equations of those two green lines\r\n" );
document.write( "which intersect the graph at exactly one point only.\r\n" );
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document.write( "Let the lines that we want have equations of the form  y = mx + b,\r\n" );
document.write( "with the slope = m = -1 so substituting -1 for m, the lines will \r\n" );
document.write( "have equations of the form:\r\n" );
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document.write( "y = -1x + b\r\n" );
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document.write( "or\r\n" );
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document.write( "y = -x + b\r\n" );
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document.write( "To find points of intersection of the line and the curve, \r\n" );
document.write( "we solve the system:\r\n" );
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document.write( "\"system%28y=1%2F%28x-1%29%2Cy=-x%2Bb%29\"\r\n" );
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document.write( "by substituting (-x+b) for y in the first equations:\r\n" );
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document.write( "-x+b = \"1%2F%28x-1%29\"\r\n" );
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document.write( "Multply both sides by (x-1)\r\n" );
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document.write( "(-x+b)(x-1) = \"1%2F%28x-1%29\"(x-1)\r\n" );
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document.write( "-x² + x + bx - b = 1\r\n" );
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document.write( "Multiply by -1 to get the x² term positive:\r\n" );
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document.write( "x² - x - bx + b = -1\r\n" );
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document.write( "Get 0 on the right\r\n" );
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document.write( "x² - bx - x + b + 1 = 0\r\n" );
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document.write( "x² - (b + 1)x + (b + 1) = 0\r\n" );
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document.write( "We want there to be exactly one solution.\r\n" );
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document.write( "To guarantee that that quadratic equation\r\n" );
document.write( "has exactly one real solution, we find the discriminant \r\n" );
document.write( "and set it equal to 0:\r\n" );
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document.write( "discriminant = B²-4AC\r\n" );
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document.write( "(Note: I used capital letters to avoid conflict of\r\n" );
document.write( "notation with b and B):\r\n" );
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document.write( "where A=1, B=-(b+1),  C=b+1 \r\n" );
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document.write( "B²-4AC = [-(b+1)]² - 4(1)(b+1)] = 0\r\n" );
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document.write( "(b+1)² - 4(b+1) = 0\r\n" );
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document.write( "(b+1)[(b+1) - 4] = 0\r\n" );
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document.write( "(b+1)(b+1-4) = 0\r\n" );
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document.write( "(b+1)(b-3) = 0\r\n" );
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document.write( "b+1 = 0;  b-3 = 0\r\n" );
document.write( "  b = -1;   b = 3\r\n" );
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document.write( "So the tangent lines are the lines\r\n" );
document.write( "y = -x + b with these two values for b.\r\n" );
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document.write( "So they are \r\n" );
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document.write( "y = -x - 1 \r\n" );
document.write( "and\r\n" );
document.write( "y = -x + 3\r\n" );
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document.write( "And here are their graphs:\r\n" );
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document.write( "Edwin
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