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Math Help - [SOLVED] Slopes of Tangents and Secants

  1. #1
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    [SOLVED] Slopes of Tangents and Secants

    Find the slope of the tangent line of the parabola 2x - x^2 at point (2,0):

    So I have to use the forumla

    m = lim f(a + h) - f(a) / h
    h->0

    So we have to find f(a) which = 0.

    Then put in that to f(a + h) which = 2(0 + h) - (0+h)^2

    So then you get for the equation:

    (2h - 0) / h

    Which would then equal 2, but the answer is -2...can anyone help?
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  2. #2
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    Sorry it'd be (2h - h^2) on top wouldn't it? But that still doesn't work
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  3. #3
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    Quote Originally Posted by Dickson View Post
    Find the slope of the tangent line of the parabola 2x - x^2 at point (2,0):

    So I have to use the forumla

    m = lim f(a + h) - f(a) / h
    h->0

    So we have to find f(a) which = 0.

    Then put in that to f(a + h) which = 2(0 + h) - (0+h)^2

    So then you get for the equation:

    (2h - 0) / h

    Which would then equal 2, but the answer is -2...can anyone help?
    f(x) = 2x - x^2

    f(2+h) = 2(2+h) - (2+h)^2

    f(2) = 0

    \lim_{x \to 2} \frac{f(2+h) - f(2)}{h}

    \lim_{x \to 2} \frac{2(2+h) - (2+h)^2 - 0}{h}

    \lim_{x \to 2} \frac{4+2h -(4 + 4h + h^2)}{h}

    \lim_{x \to 2} \frac{4+2h -4 - 4h - h^2}{h}

    \lim_{x \to 2} \frac{-2h - h^2}{h}

    \lim_{x \to 2} \frac{h(-2 - h)}{h}

    \lim_{x \to 2} (-2 - h) = -2
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  4. #4
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    I thought you were supposed to use the value you got from f(a) and put that in for the A value in f(a + h)
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  5. #5
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    Quote Originally Posted by Dickson View Post
    I thought you were supposed to use the value you got from f(a) and put that in for the A value in f(a + h)
    no, it's the same "a" ... a = 2.
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  6. #6
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    Quote Originally Posted by skeeter View Post
    no, it's the same "a" ... a = 2.
    Okay thanks a lot
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