Question 760583
Find the point(s) on the curve of y=1/(x-2) for which the slope of the tangent line is -1.
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y = (x-2)^-1
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y' = {{{(-1)*(x-2)^(-2)}}}
{{{-1/(x-2)^2 = -1}}}
{{{(x-2)^2 = 1}}}
x = 1, x = 3
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--> (1,-1) & (3,1)