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Math Help - tgnts n norms

  1. #1
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    tgnts n norms

    The equation of a curve is y = 3x^2 - 2x + 1

    find the coordinates of the point R on the curve at which the gradient is -14.

    do i put the result of dy/dx to equal -14? doing that will not find me 2 points though..
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by x-disturbed-x
    The equation of a curve is y = 3x^2 - 2x + 1

    find the coordinates of the point R on the curve at which the gradient is -14.

    do i put the result of dy/dx to equal -14? doing that will not find me 2 points though..
    There is only one point at which the gradient is -14 and yes you do solve:

     \frac{dy}{dx}\ =\ -14

    to find the x-coordinate of the point, then substitute that value back
    into the original equation to find the corresponding y.

    RonL
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  3. #3
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    y y y

    i know the gradient and the x coordinate. the original equation before differentiation / simplification was

    y = 3x^2 - 2x + 1

    how do i find the y coordinate?
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  4. #4
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    Quote Originally Posted by x-disturbed-x
    i know the gradient and the x coordinate. the original equation before differentiation / simplification was

    y = 3x^2 - 2x + 1

    how do i find the y coordinate?
    The x-coordinate of the point at which the derivavtive is -14 is
    -2.

    Since the required point is on the curve y\ =\ 3.x^2\ -\ 2.x\ +\ 1, and we
    already know the x-coordinate, we need only substitute
    this into the equation for y will give the y-coordinate.

    RonL
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  5. #5
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    y = 17?
    Last edited by x-disturbed-x; December 4th 2005 at 02:52 PM.
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by x-disturbed-x
    y = 17?
    Yep.

    RonL
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  7. #7
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    Thanks for all your help
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