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Math Help - Graph tangents

  1. #1
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    Graph tangents

    Curve C has equation y= (1/3)x^3 - 4x^2 + 8x + 3

    The point P has coordinates (3,0) and lies on C. Point Q also lies on C. The tangent to C at Q is parallel to the tangent to C at P.

    Find the coordinates of Q-


    I found the tangent to C at P- y = -7x +21
    And that the gradient of the tangent on C at Q will be -7 but I can't think how to get the coordinates.
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  2. #2
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    Hello, greghunter!

    You've worked out the necessary information.


    Curve C has equation: y\:=\:\tfrac{1}{3}x^3 - 4x^2 + 8x + 3

    The point P has coordinates (3,0) and lies on C.
    Point Q also lies on C, and the tangent at Q is parallel to the tangent at P.

    Find the coordinates of Q.

    The derivative is: . y' \:=\:x^2-8x + 8

    At P(3,0)\!:\;\;y' \:=\:3^2-8(3) + 8 \:=\:\text{-}7

    . . The slope of the tangent at P is -7.


    Where else is the slope of the tangent equal to -7 ?

    . . Solve: . x^2-8x + 8 \:=\:-7 \quad\Rightarrow\quad x^2 -8x + 15 \:=\:0

    . . (x-3)(x-5) \:=\:0 \quad\Rightarrow\quad x \:=\:3,\:5


    So the tangent has slope -7 when x = 3
    (we know!) . . . and when x = 5

    When x=5\!:\;y \:=\:\tfrac{1}{3}(5^3) - 4(5^2) + 8(5) + 3 \:=\:\text{-}\frac{46}{3}

    . . Therefore, the coordinates of Q are: . \left(5,\:\text{-}\tfrac{46}{3}\right)

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