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Math Help - Tangent lines to the curve

  1. #1
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    Tangent lines to the curve

    Hello,

    I need to find out the tangent lines to the curve f (X) = 2X - 8X + 3,

    Which are parallels to line X + 2Y + 1 = 0.

    I have slope of line : m = -1/2,

    and derivative : f'X = 4X - 8,

    I made : m = f'X,

    and I obteined : X = 1 7/8,

    I replace in : f(X) and reach to Y = -4 31/32,

    With those values I get Y intersection = -129/32.

    Then I built my equation,
    but I think something is wrong, cause those are a very big fractions.
    But I cant find the mistake. Maybe someone could show me the error in this operation?
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    That's the correct answer.
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  3. #3
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    The point on the curve where the tangent is drawn is [tex](\frac{15}{8}, -\frac{1589}{32})/MATH]

    Equation of the line parallel to x + 2y + 1 = 0 is x + 2y + c = 0.

    Substitute the above point in the equation and find c.
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  4. #4
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    The point on the curve where the tangent is drawn is (15/8, -159/32})

    Equation of the line parallel to x + 2y + 1 = 0 is x + 2y + c = 0.

    Substitute the above point in the equation and find c.
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