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C2 - Applications of Differentiation: Sketching Curves watch

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    Q: For this curve, find where the curve meets the axes, and the coordinates of any stationary points and determine their nature. State the behaviour of y as x -> +- infinity and then sketch each curve.

    y= (x-3)(x^2+3x+6)

    Well, this is a bit embarrasing but I pretty much know how to do everything the question asks for but what I don't know is how to factorise (x-3)(x^2+3x+6). Any help would really be appreciated...
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    it is not reducible over the reals


    (Original post by TarotOfMagic)
    Q: For this curve, find where the curve meets the axes, and the coordinates of any stationary points and determine their nature. State the behaviour of y as x -> +- infinity and then sketch each curve.

    y= (x-3)(x^2+3x+6)

    Well, this is a bit embarrasing but I pretty much know how to do everything the question asks for but what I don't know is how to factorise (x-3)(x^2+3x+6). Any help would really be appreciated...
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    (Original post by TarotOfMagic)
    Q: For this curve, find where the curve meets the axes, and the coordinates of any stationary points and determine their nature. State the behaviour of y as x -> +- infinity and then sketch each curve.

    y= (x-3)(x^2+3x+6)

    Well, this is a bit embarrasing but I pretty much know how to do everything the question asks for but what I don't know is how to factorise (x-3)(x^2+3x+6). Any help would really be appreciated...
    You cannot factorise the second bracket normally, try completing the square
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    (Original post by TeeEm)
    it is not reducible over the reals
    Sorry?
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    (Original post by TarotOfMagic)
    Sorry?
    b2 - 4ac < 0
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    (Original post by KaylaB)
    You cannot factorise the second bracket normally, try completing the square
    I got (x-3)(x+3/2)^2+15/4

    But I'm kinda stuck on that point. If I let y = 0, it will still be unfactorisable and the question asks to find where the curve meets the axes and the coordinates of the stationary points.
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    (Original post by TarotOfMagic)
    I got (x-3)(x+3/2)^2+15/4

    But I'm kinda stuck on that point. If I let y = 0, it will still be unfactorisable and the question asks to find where the curve meets the axes and the coordinates of the stationary points.
    If you let y=0 then you will get the point x=3, and as you can't solve the quadratic then the graph must only cross the x axis once
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    (Original post by clo-jo11)
    If you let y=0 then you will get the point x=3, and as you can't solve the quadratic then the graph must only cross the x axis once
    So you know from the equation already that the graph crosses the x axis at 3, what else does the equation tell you about the graph?
    y= (x-3)(x^2+3x+6)
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    (Original post by clo-jo11)
    If you let y=0 then you will get the point x=3, and as you can't solve the quadratic then the graph must only cross the x axis once
    I checked the answers and it said that the graph also crosses at (0, -18)?
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    (Original post by TarotOfMagic)
    I got (x-3)(x+3/2)^2+15/4

    But I'm kinda stuck on that point. If I let y = 0, it will still be unfactorisable and the question asks to find where the curve meets the axes and the coordinates of the stationary points.


    This is what the graph looks like, it only has one root. i.e: it only has one solution to y=0.
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    Question solved! Thanks, everyone!
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    (Original post by TarotOfMagic)
    I checked the answers and it said that the graph also crosses at (0, -18)?
    There's a difference between a x-axis crossing, which is what you get when you plug y=0 into your equation solve to get coordinates of the form (x_0, 0), (x_1, 0), \ldots

    And a y-axis crossing which is what you get when you plug x=0 into your equation to get the intercepts of the y-axis in the form (0, y_0).
 
 
 
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