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Math Help - Slope and Equation of the Line

  1. #1
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    Slope and Equation of the Line

    This is for my advanced functions course so NO CALCULUS is used. Please help find the equation.

    Find the equation of the line with slope -1 that is the tangent to the curve y=1/(x-1).

    This is my work until I get stuck.

    y=-1x+k
    -1x+k=1/(x-1)
    y=??

    I get stuck from here. Can someone please solve this completely and be exact with the steps. Thanks!
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Slope and Equation of the Line

    You are on the right track...you have:

    k-x=\frac{1}{x-1}

    Multiply through by x-1, write the resulting quadratic in standard form, and then use the condition on the discriminant for one repeated root (do you see why?). You will find two values for k.
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    Re: Slope and Equation of the Line

    Well that's what I was getting confused about. I ended up cross multiplying like you're suggesting. But then I don't know if the quadratic is correct, because it certainly looks wrong. And then the discriminant looks even worse.

    (k-x)(x-1)-1=0
    x^2+x(k+1)+(k-1)=0
    Is that the correct quadratic?
    I'm not certain how to get a value for k from the discriminant of this? I'm assuming the k doesn't equal 1.
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: Slope and Equation of the Line

    Your first equation is correct. The quadratic in standard form that you should wind up with is:

    x^2-(k+1)x+(k+1)=0

    Now, when a quadratic has 1 root, what do we then know about the discriminant?
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  5. #5
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    Re: Slope and Equation of the Line

    So does K=-1? The discriminant would only cross the x axis once.
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  6. #6
    MHF Contributor MarkFL's Avatar
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    Re: Slope and Equation of the Line

    For the quadratic ax^2+bx+c the discriminant is defined to be b^2-4ac. When the discriminant is zero, the there is only 1 repeated root for the quadratic. So, equating the discriminant in this case to zero, we find:

    (-(k+1))^2-4(1)(k+1)=0

    (k+1)((k+1)-4)=0

    (k+1)(k-3)=0

    Thus, we find two possible tangent lines:

    y=-x-1

    y=-x+3
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  7. #7
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    Re: Slope and Equation of the Line

    Thanks so much! I see what I was doing wrong.
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