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Math Help - equation of a tangent line

  1. #1
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    equation of a tangent line

    I am asked to find the equation of a line with a slope of -1 that is tangent to the curve
     y=\frac{1}{x-1}
    I know I am doing something wrong here, but I just can't figure out what.
    first, I have the equation for the slope, y=-1x+k
    next, I set the y-values to equal eachother
     y=\frac{1}{x-1}=-1x+k

    now, here's where I think I begin to mess this up. I am trying to rearrange this into a quadratic equation.
     1=-x^2+x+kx-k \\ 0=-x^2+x+kx-k-1
    a=-1 b=(1+k) c=(-k-1)
    Then I use  b^2-4ac

     1+k^2-4(-1)(-k-1) =0 \\ 1+k^2+4(-k-1)=0 \\  1+k^2-4k-4=0
    now I'm stuck and don't know where to go from here.
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  2. #2
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    Re: equation of a tangent line

    Quote Originally Posted by whit221 View Post
    a=-1 b=(1+k) c=(-k-1)
    Then I use  b^2-4ac

     1+k^2-4(-1)(-k-1) =0
    $b^2=(1+k)^2=1+2k+k^2\ne1+k^2$.

    Quote Originally Posted by whit221 View Post
    1+k^2+4(-k-1)=0 \\  1+k^2-4k-4=0
    now I'm stuck and don't know where to go from here.
    With the adjustment above, you are correct. Just solve this equation in $k$.
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  3. #3
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    Re: equation of a tangent line

    Thank you very much! Now it all makes sense!
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