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Math Help - equation of the tangent line to the curve at the given point

  1. #1
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    equation of the tangent line to the curve at the given point

    Find an equation of the tangent line to the curve at the given point.
    2x4 + 4x2 − 7x, (1, -1)

    I know i would need to use m = lim f(x)-f(a)
    x->a x-a

    but then i have lim 2x^4+4x^2-7x-1
    x->1 x-1

    i am not quite sure how to factor that limit so i can cancel the x-1 and get a equation that i can plug my x=1 value into and get my slope so i can use the formula:
    y-h = m(x-a)
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  2. #2
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    Quote Originally Posted by rhcp1231 View Post
    Find an equation of the tangent line to the curve at the given point.
    2x4 + 4x2 − 7x, (1, -1)

    I know i would need to use m = lim f(x)-f(a)
    x->a x-a

    but then i have lim 2x^4+4x^2-7x-1
    x->1 x-1

    i am not quite sure how to factor that limit so i can cancel the x-1 and get a equation that i can plug my x=1 value into and get my slope so i can use the formula:
    y-h = m(x-a)
    the gradient is m = \lim_{x \to a} \frac{f(x) - f(a)}{x - a} = f'(a)

    in this problem, m = f'(1) = 8 + 8 - 7 =9
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  3. #3
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    how did you get
    ?
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  4. #4
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    from f'(x) = 8x^3 + 8x - 7
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  5. #5
    Newbie blackcompe's Avatar
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    The derivative of the function at some point is the slope of the tangent line of the curve at that point. If y = 2x^4 + 4x^2 − 7x, and you want to find the tangent line equation at (1, -1), then the tangent line equation is: y = -1 + y'(1)(x - 1). Using the general derivative rules, we find that y' = (2x^4)' + (4x^2)' − (7x)' = 8x^3 + 8x - 7. So y'(1) = 8(1)^3 + 8(1) - 7 = 8 + 8 - 7 = 9.

    Therefore the tangent line at (1, -1) is y = -1 + 9(x - 1)
    Last edited by blackcompe; February 9th 2010 at 06:15 PM.
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  6. #6
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    oooo so the dervitive of the original equation is the slope of the tangent?
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  7. #7
    dMh
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    Quote Originally Posted by rhcp1231 View Post
    oooo so the dervitive of the original equation is the slope of the tangent?
    Yes.
    That's what a derivative of a function is.
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