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    I've tried this question out and I've done a and b but can't do part c.

    The curve C has equation y=x^3 - 2x^2 - x + 9 x>0

    The point P has coordinates (2,7)

    a) Show that P lies on C

    worked out to be ;
    7= 2^3 -2(2)^2-2+9
    7 = 8-8-2+9
    7=7

    b) Find the equation of the tangent to C at P, giving your answer in the form y=mx+c, where m and c are constants.

    found to be ;
    dy/dx= 3x^2-4x - 1
    12-8-1
    m = 3
    y-7=3(x-2)
    y = 3x+1
    The point Q also lies on C

    Given that the tangent to C at Q is perpendicular to the tangent to C at P,

    c) show that the x coordinates of Q is 1/3 (2 + square root 6)
    ??
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    (Original post by shaunobrien1)
    I've tried this question out and I've done a and b but can't do part c.

    The curve C has equation y=x^3 - 2x^2 - x + 9 x>0

    The point P has coordinates (2,7)

    a) Show that P lies on C

    worked out to be ;
    7= 2^3 -2(2)^2-2+9
    7 = 8-8-2+9
    7=7

    b) Find the equation of the tangent to C at P, giving your answer in the form y=mx+c, where m and c are constants.

    found to be ;
    dy/dx= 3x^2-4x - 1
    12-8-1
    m = 3
    y-7=3(x-2)
    y = 3x+1
    The point Q also lies on C

    Given that the tangent to C at Q is perpendicular to the tangent to C at P,

    c) show that the x coordinates of Q is 1/3 (2 + square root 6)
    ??
    So first find the perpendicular gradient.
    Set \dfrac{\mathrm d y}{\mathrm d x} = \dfrac{\mathrm d y}{\mathrm d x}
    Where the perpendicular gradient is equal to the gradient function(your 3x^2-4x - 1)
    Then solve for x and you have your answer
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    (Original post by shaunobrien1)
    I've tried this question out and I've done a and b but can't do part c.

    The curve C has equation y=x^3 - 2x^2 - x + 9 x>0

    The point P has coordinates (2,7)

    a) Show that P lies on C

    worked out to be ;
    7= 2^3 -2(2)^2-2+9
    7 = 8-8-2+9
    7=7

    b) Find the equation of the tangent to C at P, giving your answer in the form y=mx+c, where m and c are constants.

    found to be ;
    dy/dx= 3x^2-4x - 1
    12-8-1
    m = 3
    y-7=3(x-2)
    y = 3x+1
    The point Q also lies on C

    Given that the tangent to C at Q is perpendicular to the tangent to C at P,

    c) show that the x coordinates of Q is 1/3 (2 + square root 6)
    ??
    For part C you already know the gradient of the tangent to C at P, it's 3 (assuming you did it correctly). So you want the gradient at point Q to be -1/3 since it's perpendicular to the tangent at P.
    So you're literally just solving  \displaystyle \frac{dy}{dx}=-\frac{1}{3} .
 
 
 
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