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Math Help - Tangent

  1. #1
    Member
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    Tangent

    Determine the equation of the tangent to y=18/(x+2)^2 at the point (1,2).

    My answer:

    I found the derivative of the function and then I substituted 1 into the Mt.
    Then I substituted (1,2) into y=Mtx + b to find the y intercept.

    I didn't get the right answer with the textbook. Is mine right or theirs?

    f'(x)=-36 / (x+2) ^3
    Mt= -36/27

    y=-36/27x - 1.5

    final answer: 8x+6y +9=0

    Back of the book's answer:

    4x+3y-10=0
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  2. #2
    Junior Member StarlitxSunshine's Avatar
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    I got the book's answer. I was doing the same thing as you up until you found "b". Something went different after that. Here's what I did:

    y = 18/(x+2)^2

    y=mx+b

    m = dy/dx = (x+2^2)(0) - (18(2(x+2)(1))/(x+2)^4

    = 0-36(x+2)/(x+2)^4

    = -36/(x+2)^3

    = -36/27

    m = -4/3

    Substitute m into y=mx+b

    y = (-4/3)x+b

    Substitute (1,2) to find b

    2 = (-4/3)(1) + b

    b= 10/3

    Into the equation:

    y=mx+b

    y= -4/3x+10/3


    Multiply by 3 to get standard form:

    3y=-4x+ 10

    Set equal to zero:

    <br />
4x+3y-10 = 0<br />
    Last edited by StarlitxSunshine; November 7th 2009 at 09:18 AM. Reason: Sorry. Had to fix something. Something went wrong with the coding >_<
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