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    ok the question is basically:

    Find the two set of co-ordinates of the turning point on the curve:

    y^3 + 3xy^2 - x^3 = 3

    I know that inorder to find turning points, I must get dy/dx, and then substitute the equation dy/dx to equal 0

    Differentiating the above equation, led me to get:

    dy/dx = (3x^2 - 3y^2)/(3y^2 + 6xy)

    ......Im not sure where to go from here, coz if I change that differentiated equation into 0, I'll have a bit of controversy between x and ys.......
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    y³+3xy²-x³ = 3

    Differentiates to

    3y²(y')+3(2xy(y')+y²)-3x² = 0
    3y²(y')+6xy(y')+3y² = 3x²
    y'(3y²+6xy) = 3x²-3y²
    y' = (3x²-3y²)/(3y²+6xy)

    Equate to 0:

    0 = (3x²-3y²)/(3y²+6xy)

    The denominator must equal 0. Hence,

    0 = 3x²-3y²
    3x² = 3y²
    x² = y²

    Tricky. I'll plug this in to my graphical calculator and see what happens.
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    (Original post by ~Revol~)
    ok the question is basically:

    Find the two set of co-ordinates of the turning point on the curve:

    y^3 + 3xy^2 - x^3 = 3

    I know that inorder to find turning points, I must get dy/dx, and then substitute the equation dy/dx to equal 0

    Differentiating the above equation, led me to get:

    dy/dx = (3x^2 - 3y^2)/(3y^2 + 6xy)

    ......Im not sure where to go from here, coz if I change that differentiated equation into 0, I'll have a bit of controversy between x and ys.......
    Whoops
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    Fudge, I can't figure out how to make y the subject, I've got both y² and y³ to deal with.
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    At a turning point, the top of the dy/dx fraction has to equal 0. So x = y or x = -y.

    Substitute x = y into y^3 + 3xy^2 - x^3 = 3 and solve, then do the same for x = -y. Scroll down if you get stuck.

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    Case 1: y = x.
    y^3 + 3xy^2 - x^3 = 3
    => x^3 + 3x^3 - x^3 = 3
    => 3x^3 = 3
    => x^3 = 1
    => x = 1
    => y = 1.

    Case 2: y = -x.
    y^3 + 3xy^2 - x^3 = 3
    => -x^3 + 3x^3 - x^3 = 3
    => x^3 = 3
    => x = 3^(1/3)
    => y = -3^(1/3).
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    (Original post by Jonny W)
    At a turning point, the top of the dy/dx fraction has to equal 0. So x = y or x = -y.
    Ok, I see how dy/dx=0 when x=y, but i dont understand how it can also be x=-y???
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    (Original post by ~Revol~)
    Ok, I see how dy/dx=0 when x=y, but i dont understand how it can also be x=-y???
    Because x^2 = y^2 = (-y)^2
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    (Original post by ~Revol~)
    Ok, I see how dy/dx=0 when x=y, but i dont understand how it can also be x=-y???
    x² = y²
    =>
    +/-x = +/-y (taking square roots of both sides)
    So x = y or x = -y.
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    I attach a graph.
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    (Original post by Jonny W)
    I attach a graph.
    Doesn't that show three turning points, whereas your answer only has two?
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    (Original post by ZJuwelH)
    Doesn't that show three turning points, whereas your answer only has two?
    No, the thing on the left isn't a turning point.

    Ben
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    (Original post by Ben.S.)
    No, the thing on the left isn't a turning point.

    Ben
    Oh yeah, just thought it through.
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    (Original post by ZJuwelH)
    Doesn't that show three turning points, whereas your answer only has two?
    A turning point is where the tangent is parallel to the x-axis (so that dy/dx = 0), not where the tangent is parallel to the y-axes (so that dy/dx is undefined).
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    I'm offering rep for the first person who finds, numerically, the equation of each of the graph's three asymptotic lines.
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    Asymptotes at x=-1.5 x = 1 y = 1

    Is that right?
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    (Original post by Jonny W)
    I'm offering rep for the first person who finds, numerically, the equation of each of the graph's three asymptotic lines.
    Roughly ...

    y = -2.89x
    y = -0.65x
    y = 0.53x
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    (Original post by shiny)
    Roughly ...

    y = -2.89x
    y = -0.65x
    y = 0.53x
    Correct!

    Dividing y^3 + 3xy^2 - x^3 = 3 by x^3 gives

    (y/x)^3 + 3(y/x)^2 - 1 = 3/x^3,

    which is approximately

    (y/x)^3 + 3(y/x)^2 - 1 = 0

    when |x| is large. Numerically solving the cubic k^3 + 3k^2 - 1 = 0 gives k = 0.532089, -0.652704 and -2.87939.
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