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Math Help - Tangent line derivatives

  1. #1
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    Tangent line derivatives

    Hi,

    I know this question is easy but I havent done this topic in awhile and I forget how to do it.

    f(x)=-0.1x^3 -0.26x^2 + 0.861x + 1.539
    I need to sketch the tangent lines to y=f(x) at x=-3 and x=2?
    How do I work out the tangent line?
    Please
    Last edited by wolfhound; October 10th 2010 at 10:06 AM.
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  2. #2
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    e^(i*pi)'s Avatar
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    To find the tangent you need to take the derivative of f(x)
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  3. #3
    MHF Contributor Unknown008's Avatar
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    And when you found f'(x), you just replace x be the values given to you.
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  4. #4
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    Hi, thats what I done, I got -0.279

    So on my graph when x=-3, the corresponding y value is -0.279 ?

    it looks like the tangent will cut through the curve?
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  5. #5
    MHF Contributor Unknown008's Avatar
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    No, the gradient at the x coordinate x = -3 is -0.279.

    To find the y value, you need to replace in f(x).

    But since you replaced in f'(x), you get the value of the gradient.
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  6. #6
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    ok so when i plug in f(-0.279) I get a value of 1.28 for y.

    so I connect from y=1.28 on the graph to where x=-3 line drops vertically onto the curve?
    Last edited by wolfhound; October 10th 2010 at 10:26 AM.
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  7. #7
    MHF Contributor Unknown008's Avatar
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    I'm getting f(-3) = -0.684. Check your calculations again.

    You might want to find the equation of the gradient, by using:

    \dfrac{y - y_1}{x - x_1} = m

    You will get the y intercept of the tangent then.

    Then, you connect the y intercept with the point to where the line x = -3 drops vertically on the curve.
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  8. #8
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    You got f(-3) or f'(-3) = -0.684?
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  9. #9
    MHF Contributor Unknown008's Avatar
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    I didn't put any apostrophe in my answer, meaning I got f(-3) = -0.684. f'(-3) = -0.279.

    You need to find the point on the curve first. You were given x = -3. The y coordinate is given by f(-3).

    Now, since you are looking for a tangent at that point, you also need to find f'(3).

    Once you get those, you have your coordinates.

    f(-3) = -0.684
    f'(-3) = -0.279
    x = -3

    Or, (-3, -0.684) with gradient m = -0.279

    Thus, you get the equation of the tangent as being:

    \dfrac{y - (-0.684)}{x - (-3)} = -0.279

    This becomes:

    y = -0.279x - 1.521

    Now that you have the equation of the tangent, you can join the points: (-3, -0.684) and (0, -1.521) to get the tangent.
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