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Math Help - Tangents

  1. #1
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    Tangents

    Let f(x) = 12 - x^2 for x > 0 and f(x) is greater than or equal to 0. The line tangent to the graph of f at the point (k,f(k)) intercepts the x-axis at x = 4. What is the value of k?

    ------
    My attempt:

    f'(x) = -2x

    y = -2(4) + b = 0, b= 8

    y = -2x +8 --> tangent line


    I know it's wrong because the tangent line is y = -4x + 16 and is tangent at x = 2.

    How would one go about doing it correctly?
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  2. #2
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    Sorry for the bump, but I really need help.
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  3. #3
    Jen
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    Consider what you know about the derivative.

    You know that you are looking for a tangent line at the point (k, f(k)). You also know that the derivative of your function is linear so it will have a constant slope. Consider what is happening between the points (k,f(k)) and (4,0) where it intersects the x-axis.



    Tangents-capture.jpg

    Use what you know about the slope between (k,f(k)) and (4,0) and the fact that the derivative represents a slope to help you solve for k.

    Hope these hints help.
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