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    Topic is differentiation.

    The curve with the equation y=x^3-4x^2+3x crosses the x-axis at the points A, B and C.
    a) Find the coordinates of A, B and C.
    b) Find the gradient of the curve at each of the points.

    I know the answers:
    a) x=0, x=3, x=1
    b) 3, 6, -2

    But how do I get to those answers?
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    So you know that points A, B and C are on the x-axis. This means that at these points, y=0. So make the equation:

    0=x^3-4x^2+3x.

    Then you can factor this by removing an x from the equation to get:

    0= x(x^-4x+3) which is equal to 0=x(x-3)(x-1)

    from this, you know that x=0, x=3, x=1. This is the answer to (a).

    For (b), you have to differentiate. So,

    dy/dx = 3x^2-8x+3

    then substitute x=0 and you get an answer of 3 for the gradient, substitute x=3 and you get 6 and substitute x=1 and you get -2.


    Hope that made sense!
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    (Original post by metellaest)
    so you know that points a, b and c are on the x-axis. This means that at these points, y=0. So make the equation:

    0=x^3-4x^2+3x.

    Then you can factor this by removing an x from the equation to get:

    0= x(x^-4x+3) which is equal to 0=x(x-3)(x-1)

    from this, you know that x=0, x=3, x=1. This is the answer to (a).

    For (b), you have to differentiate. So,

    dy/dx = 3x^2-8x+3

    then substitute x=0 and you get an answer of 3 for the gradient, substitute x=3 and you get 6 and substitute x=1 and you get -2.


    Hope that made sense!
    thank you so so much
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    (Original post by mxgx)
    thank you so so much
    No problem

    Is 'Further Maths' the same as Additional Maths (which is what I'm doing), or are they different subjects?
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    Could anyone also help on b for this question?

    A curve has the equation y=x^3+x^2-4x+1.

    a) Find the gradient of the curve at the point P (-1,5)
    ^^ I've already done this and the answer is -3.

    b) Given that the gradient at the point Q on the curve is the same as the gradient at the point P, find, as exact fractions, the coordinates of the point Q.
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    (Original post by metellaest)
    No problem

    Is 'Further Maths' the same as Additional Maths (which is what I'm doing), or are they different subjects?
    I'm not sure? My exam is the AQA Level 2 Further Maths exam.
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    (Original post by mxgx)
    Could anyone also help on b for this question?

    A curve has the equation y=x^3+x^2-4x+1.

    a) Find the gradient of the curve at the point P (-1,5)
    ^^ I've already done this and the answer is -3.

    b) Given that the gradient at the point Q on the curve is the same as the gradient at the point P, find, as exact fractions, the coordinates of the point Q.
    Find dy/dx and equate it to -3, solve for x (should be a quadratic). One of the solutions will be x=-1, the other solution is the point you want. Then plug this x-value back into the equation to get your coordinates.

    PS: Moved to maths.
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    this is the extra maths which some people do beyond GCSE
 
 
 
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