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Math Help - Equation of tangent to the curve...

  1. #1
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    Equation of tangent to the curve...

    Please advise...

    Find the equation of tangent to curve g(x)=4x^3-2x^2+4 at the point (1;6)

    Find the derivative of g(x)
    g'(x) = 12x^2 -4x

    Substitute 1,6 into g'(x)

    12(1)^2 - 4 = rate of change at 1, 6 = 12 - 4 = 8

    y = 8x + c
    Substitute 1,6 into above

    6 = 8(1) + c
    c = -2

    Tangent line
    y = 8x - 2 = 2(4x-1)

    IS THIS CORRECT???
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  2. #2
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    Re: Equation of tangent to the curve...

    Quote Originally Posted by Scoplex View Post
    Please advise...

    Find the equation of tangent to curve g(x)=4x^3-2x^2+4 at the point (1;6)

    Find the derivative of g(x)
    g'(x) = 12x^2 -4x

    Substitute 1,6 into g'(x)

    12(1)^2 - 4 = rate of change at 1, 6 = 12 - 4 = 8

    y = 8x + c
    Substitute 1,6 into above

    6 = 8(1) + c
    c = -2

    Tangent line
    y = 8x - 2 = 2(4x-1)

    IS THIS CORRECT???
    Yes, it looks fine
    Thanks from Scoplex
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