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Math Help - Siplification Needed For Tangent Slope

  1. #1
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    Siplification Needed For Tangent Slope

    Find an equation of the tangent line to the curve at the given point:

    y=(x-1)/(x-2), (3,2)

    And I am using the the formula:

    lim as x approaches a = [f(x) - f(a)]/(x-a).

    So plugging in what we already know, we get:

    [x-1/x-2 - 2]/x-3 = slope of the tangent line

    Sooooo I'm wasn't sure how to procede. I tried a few different things and I ended up like this:

    So the first thing I did was make the 2 that was being subtracted from the first rational number: [2(x-2)]/(x-2). This would simplify the whole thing to:

    [(x-1)-2(x-2)/(x-2)]/x-3. Then I brought the x-3 up:

    [(x-1)-2(x-2)]/[(x-2)(x-3)]

    I'm stuck.

    Any help would be appreciated.
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  2. #2
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    Re: Siplification Needed For Tangent Slope

    Quote Originally Posted by ubom2 View Post
    Find an equation of the tangent line to the curve at the given point:

    y=(x-1)/(x-2), (3,2)

    And I am using the the formula:

    lim as x approaches a = [f(x) - f(a)]/(x-a).

    So plugging in what we already know, we get:

    [x-1/x-2 - 2]/x-3 = slope of the tangent line

    Sooooo I'm wasn't sure how to procede. I tried a few different things and I ended up like this:

    So the first thing I did was make the 2 that was being subtracted from the first rational number: [2(x-2)]/(x-2). This would simplify the whole thing to:

    [(x-1)-2(x-2)/(x-2)]/x-3. Then I brought the x-3 up:

    [(x-1)-2(x-2)]/[(x-2)(x-3)] <--- I'll take this line

    I'm stuck.

    Any help would be appreciated.
    All your calculations are OK. Unfortunately you have lost somehow that you are asked to calculate a limit.

    You only have to do one step more:

    \lim_{x\to3} \left(\frac{(x-1)-2(x-2)}{(x-2)(x-3)}\right) = \lim_{x\to3} \left( \frac{x-1-2x+4}{(x-2)(x-3)}\right) = \lim_{x\to3} \left( \frac{-x+3}{(x-2)(x-3)} \right)= \lim_{x\to3} \left( \frac{-(x-3)}{(x-2)(x-3)}\right)

    Now cancel!

    Determine the limit now.
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  3. #3
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    Re: Siplification Needed For Tangent Slope

    SOOOOO

    the equation for the tangent line would be y = -x + 5!!

    Thanks
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