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

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

    Find the coordinates of $\displaystyle Q.$

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

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

    . . The slope of the tangent at $\displaystyle P$ is -7.


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

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

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


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

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

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

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