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Finding equation of derivative that is parallel to another line

  1. Oct 10, 2008 #1
    1. The problem statement, all variables and given/known data
    find equations of the tangent lines to the curve
    y= x-1/x+1


    that are parallel to the line x-2y = 2


    2. Relevant equations



    3. The attempt at a solution
    ok so i found the derivative, then set it equal to the slope (-1). then plugged x back into the original equation to get

    y = -x + 41/14 and y= -x + 17/10


    is this right? thanks in advance!
     
  2. jcsd
  3. Oct 10, 2008 #2

    gabbagabbahey

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    Are you sure that the slope of x-2y=2 is -1? ;0)
     
  4. Oct 10, 2008 #3
    oops
     
  5. Oct 10, 2008 #4
    so the slope is 1/2.....i plugg that in so my answer is now y = x-1. is this correct?
     
  6. Oct 10, 2008 #5

    gabbagabbahey

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    Your slope is right now, but you should be getting two lines; and the one you have is incorrect. Let's star with your solutions for x from setting dy/dx=1/2...what do you get for those?
     
  7. Oct 10, 2008 #6
    from there, my solutions for x = x^2 + 2X + 1 = 4.......then x^2 + 2x -3 = 0.

    is my dy/dx wrong? i get 2/(x+1)^2
     
  8. Oct 10, 2008 #7

    gabbagabbahey

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    your dy/dx is right; but what are the solutions to x^2+2x-3=0?
     
  9. Oct 10, 2008 #8
    my solutions are x = -2+or-4/2. from there one equation i get is y= x/2+7/2 and the other is y=x/2-1/2
     
  10. Oct 10, 2008 #9

    gabbagabbahey

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    how do you get x=-2 or -4/2?! (-2)^2+2(-2)-3=-3 not zero...did you try factoring the quadratic?
     
  11. Oct 10, 2008 #10
    i meant -2+or-4 all over 2.
     
  12. Oct 10, 2008 #11

    gabbagabbahey

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    Either way its still incorrect; x^2+2x-3=(x+3)(x-1) so x=1 or x=-3.
     
  13. Oct 10, 2008 #12
    ya tahts what i got.....i was tryin to say that x = -2+-4/2 which equaled 1,-3. my mistake.

    when i plug them in and put in the form i get y=x/2+7/2 and y= x/2-1/2. thanks for your help btw
     
  14. Oct 10, 2008 #13

    gabbagabbahey

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    Okay, looks good. :0)
     
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