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    I had Paper 3 for IGCSE Maths today and it was a bit of a stinker..too many questions on circle theoreom and no simeltaneous [sp] equations!

    Luckily, there wasn't any questions on calculus, so they must be all on the 2nd paper tomorrow..Can anyone help me with this problem?

    A curve has equations y = x^3 - 6x^2 + 9x - 2

    a) Find the coordinates of the point of this curve at which the tangent is parallel to the line y = -3x + 5

    b) Find the coordinates of the two turning points on this curve

    Any help is much appreciated!
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    a) Find dy/dx first. If the tangent is parallel to the line y=-3x+5 it has the same gradient. The gradient of that line is -3, so you need to find the values of x for which dy/dx = -3.
    b) The stationary points can be found by solving the equation dy/dx=0.

    I hope this helps.
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    (Original post by rachelina*)
    A curve has equations y = x^3 - 6x^2 + 9x - 2

    a) Find the coordinates of the point of this curve at which the tangent is parallel to the line y = -3x + 5

    b) Find the coordinates of the two turning points on this curve
    a) dy/dx = 3x^2 - 12x + 9

    The tangent is parallel to the line y = -3x + 5. This line has a gradient of -3, therefore the tangent has a gradient of -3 as well.

    The tangent has the same gradient as the point on the curve that it touches.

    Hence, gradient of the curve at this point is -3, therefore:

    -3 = 3x^2 - 12x + 9
    3x^2 - 12x + 12 = 0
    (3x - 6)(x - 2) = 0

    At the point on the curve where the gradient is -3, x = 2.

    y = x^3 - 6x^2 + 9x - 2 = 2^3 - 6(2^2) + 9(2) - 2 = 8 - 24 + 18 - 2 = 0

    Hence, the coordinates of the point on the curve at which the tangent is parallel to the line y = -3x + 5, are: (2, 0)

    b) At the two turning points, the gradient of the curve will be zero, hence:

    0 = 3x^2 - 12x + 9
    0 = (3x - 3)(x - 3)

    x = 1 and x = 3 at the two turning points.

    For the turning point whose x-coordinate is 1:
    y = x^3 - 6x^2 + 9x - 2 = 1^3 - 6(1^2) + 9(1) - 2 = 1 - 6 + 9 - 2 = 2

    For the turning point whose x - coordinate is 3:
    y = x^3 - 6x^2 + 9x - 2 = 3^3 - 6(3^2) + 9(3) - 2 = 27 - 54 + 27 - 2 = - 2

    Hence, the coordinates of the two turning points on the curve are:

    (1, 2) and (3, -2)
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    (Original post by mikesgt2)
    a) Find dy/dx first. If the tangent is parallel to the line y=-3x+5 it has the same gradient. The gradient of that line is -3, so you need to find the values of x for which dy/dx = -3.
    b) The stationary points can be found by solving the equation dy/dx=0.

    I hope this helps.
    IGCSE Maths must be harder than GCSE Maths in the UK, because we don't do calculus until P1.
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    Cheers! I'll have to read that about 10 times until it computes in my mind but thank you very much!

    Nah, IGCSE isn't THAT bad. Hard calculus is the worst..We get calculus, sets and function notation added in but lots of nasty things like 3D Trig, Standard Deviation, Sampling [ok thats not so nasty, but I was never that keen on it] are taken out
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    what? i did cambridge's IGCSE Maths and there's no calculus (the extended paper -2 and 4) and 1 and 3 are for core people...
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    Well I don't know how that happened! There was simple differentiation on Paper 3 yesterday..I take Paper 3 and 4, the higher tier papers.
 
 
 
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