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Math Help - Horizontal tangent?

  1. #1
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    Horizontal tangent?

    given function: f(x)=3x^2+6x-10

    how can i find horizontal tangent of function f(x)=3x^2+6x-10 ?
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  2. #2
    MHF Contributor Calculus26's Avatar
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    the horizontal tangent is where f ' = 0

    f ' = 6x +6 = 0

    at x= -1

    y = f '(-1)(x+1) +f(1) = f(1)

    y = -1
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  3. #3
    Senior Member DeMath's Avatar
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    Quote Originally Posted by Calculus26 View Post
    the horizontal tangent is where f ' = 0

    f ' = 6x +6 = 0

    at x= -1

    y = f '(-1)(x+1) +f(1) = f(1)

    y = -1
    Sorry, maybe y=-13 ?!!
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  4. #4
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    why y = -13 ?
    hm...
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  5. #5
    MHF Contributor Calculus26's Avatar
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    y = f '(-1)(x+1) +f(1) = f(1)

    My mistake
    should be f(-1) not f(1)
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  6. #6
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    so it should be

    f ' = 6x +6 = 0

    at x= -1

    y = f '(-1)(x+1) + f(-1)

    y = f '(-1)(-1+1) + f(-1)

    y = f(-1)

    f(x)=3x^2+6x-10

    y = 3(-1)^2 + 6(-1) - 10
    = 3 - 6 - 10
    = -13 ?

    thank you !!
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