# Find equation for each tangent to curve y = 1 / (x-1) that has slope -1?

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lim f(x+h)-f(x) / h = -1 / (x-2)^2

h-> 0

-1 / (x-2)^2 = -1

I don't know what im doing help??

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Use the quotient rule. Set the derivative equal to -1 to get the x value where the tanget is negative one. Then use y = mx+b to get the tangent line.

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thats what i did

x= 0,2

i just dont get the y = mx+b part

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Those are the x-coordinates of the points on the graph with slope -1. Then you find the y-coordinate using your equation [imath]y=\frac{1}{x-1}[/imath] so you get the complete coordinate. You can stick the x and y into y=mx+b, using -1 as the slope, and find out the equations (solving for b).

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Ok now get the derivative at x = 2 and x = 0 then find the value of the function at x = 2 and x =0 then find b such that y = mx + b remembering that the derivative gives the slope, M the original function gives Y and you already found X. There will be two sepret tangent lines and two seperate b's

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If the slope of the tangent and the slope of the derivative are the same thing then you are looking for when the derivative equals -1.

If I where you I would rewrite the equation as follows:

$\frac{1}{(x-1)}=(x-1)^{-1}$

and then just solve it using the chain rule once you have the derivative remember a few things like:

$\frac{dy}{dx}=m$

$y-y_{1}=m(x-x_{1})$

So once you know when the derivative is equal to -1 you already have m and x. So all you need to know is what y is at the x points you find.

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of course i forgot to include the -1

damn i make sure i actually dont get something before i ask it then i realize how dumb my mistakes are.. lol