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Math Help - Tangents to a curve

  1. #1
    Junior Member
    Joined
    Mar 2009
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    Tangents to a curve

    This is a University level homework question that I am having some difficulty with....

    Find the equation of the line with the slope -1 that is the tangent to the curve
     <br />
y=\frac{1}{x-1}<br />

    This is what I believe I should do:
    use the slope to get the equation
     <br />
y=-1x+k<br />

    Then set the y-values to equal eachother
     <br />
\frac{1}{x-1}=-1x+k<br />

    set it to equal 0:
     <br />
-1x^{2}+kx-k=0<br />

    Am I correct so far and where do I go from here?

    Thank You
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  2. #2
    Member
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    Apr 2009
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    You're working too hard. Find y'. That's the slope of any line tangent to y at that point.

    y' = \frac{-1}{(x-1)^2}

    Line has a slope of -1, set y' to -1 and solve for x

    -1 = \frac{-1}{(x-1)^2}

    Occurs @ x = 2

    Plug into original equation to get the point and you get (2, 1)

    Use y_1 = -1x + b, plug in your point

    1 = 1(2) + b

    Equation is: y_1 = -x - 1
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  3. #3
    Senior Member Pinkk's Avatar
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    From
    Uptown Manhattan, NY, USA
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    If you want a tangent line that has slope of 1, then you must have some x value where \frac{dy}{dx}=-1

    \frac{dy}{dx}=-\frac{1}{(x-1)^{2}}=-1

    Simply algebra will give you two values where that is true; x=0,x=2

    The equation of the line when x=0:

    y-\frac{1}{0-1}=-1(x-0)

    y=-x-1


    And a similar approach for x=2
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