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

  1. Feb 20, 2012 #1
    1. The problem statement, all variables and given/known data

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

    2. Relevant equations

    y=1/x-1

    3. The attempt at a solution

    Slope of -1 means y=-1x+k

    So...
    -1x+k = 1/x-1

    I don't know how to rearrange this into a quadratic equation so that I can solve this.
    I know I use b^2-4ac=0 and substituting for a, b and c.... I'm just stuck on rearranging the equation into the form: ax^2+bx+c=0
     
  2. jcsd
  3. Feb 20, 2012 #2

    Mark44

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    You're missing a very important piece: how to find the slope of the tangent line to y = 1/x - 1.

    I'm going to guess that you are in a class that has discussed how to find the tangent to a curve.
     
  4. Feb 20, 2012 #3

    Mark44

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    OTOH, maybe this actually is a precalculus-type problem. What does it mean to say that a line is tangent to a curve?

    For your equation, -x + k = 1/x - 1, what about multiplying both sides by x?
     
  5. Feb 20, 2012 #4
    (-x+k)(x) = -x^2+k(x)
    (1/x-1)(x) = ?

    You're right, but I'm confused as to how to multiply the second part? x/x^2-x ?
     
  6. Feb 20, 2012 #5
    Here's my second stab at it...

    -1x+k = 1/x-1

    Rearranged: -x^2+kx-1

    Using the discriminant of the quadratic formula:

    (b^2-4ac)

    a=-1 b=k^2 c=-1

    k^2-4(-1)(-1)=0
    Solving:
    k = -2

    Therefore the equation is y=-1x-2
    For the line with a slope of -1 that is tangent to the curve of 1/x-1
    Can anybody verify if I got this right?
     
  7. Feb 20, 2012 #6

    Mark44

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    Where did the = go? Also, you have an error.
     
  8. Feb 20, 2012 #7

    Mark44

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    What is x * (1/x)?
    What is x * (-1)?
     
  9. Feb 20, 2012 #8

    Mark44

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    From post #3:
     
  10. Feb 20, 2012 #9

    Ray Vickson

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    What you wrote means [itex] (1/x) - 1.[/itex] Is that what you meant, or did you really mean [itex] 1/(x-1)[/itex]? If you meant the latter, you need to use brackets, but if you meant the former then what you wrote is OK.

    RGV
     
  11. Feb 20, 2012 #10

    Mark44

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    I think he meant (1/x) - 1, but I'm not certain of it.
     
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