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    hey this is a question for c1 edexcel.

    This is the quesiton (Im having trouble with i and ii ) :



    This my answer



    This is their answer



    -Okay as you can see my cubic curve intersects my quadractic curve at two places.
    - Their intersects at one place. I have the same working as them, so that bit is fine. Its just the actual sketch. How was i supposed to have know that it intersects at one point?.
    - Would i gain any marks ? Because i still have the individual graphs correct its just that ive shown two intersections when there is only one. I think I would loose all marks because this exercise is about intersections.

    What confuses me even more it that they said cubic curves are steeper than quadractic curves so it will intercept:



    I feel quite annoyed since i calcuted the intercepts correctly. yet i still somehow managed to get it incorrect.

    Please help guys
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    Yes, x^3 is always steeper than x^2, after a certain point. Maybe you should ask yourself what 'steeper' means in this case... have you done C3 differentiation yet? That would be easier to explain.

    Another approach to understand how x^3 and x^2 interact is to plug in the points x = 2, 3, 4, 5, 6, etc, into the equations y = (x-2)^3 and y = x(x-4) until you have convinced yourself that the first one is always going to be greater than the second one, which would prove that your sketch of them intersecting twice is impossible.
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    (Original post by jayseanfan)
    What confuses me even more it that they said cubic curves are steeper than quadractic curves so it will intercept.
    This is rubbish. While the first bit is true, 'so it will intercept' is not. Our two situations are different; it all depends on where the cubic and quadratic curves are.
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    Dream Eater's right - the cubic is steeper than the quartic, but because the positive section of the cubic is already "above" the quadratic, it can't intercept again.

    If you have have a question like this in the exam I would do what Dream Eater suggests, and plug some numbers into the equations to get an idea of where the curves are going.
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    (Original post by Schlemmons)
    Dream Eater's right - the cubic is steeper than the quartic, but because the positive section of the cubic is already "above" the quadratic, it can't intercept again.

    If you have have a question like this in the exam I would do what Dream Eater suggests, and plug some numbers into the equations to get an idea of where the curves are going.
    "but because the positive section of the cubic is already "above" the quadratic" - can you explain this a little more
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    (Original post by jayseanfan)
    "but because the positive section of the cubic is already "above" the quadratic" - can you explain this a little more
    Basically, what's causing you confusion is that the two situations where there is an intercept are not quite the same.

    In your graph of the question, looking from the point x=2 onwards, the x^3 starts higher than the x^2 curve, AND it grows faster. So the x^2 curve will never be able catch up.

    In the bit you printed out of a book, the x^3 graph actually has a little part that is below the x^2. Well, as it is going to eventually become much bigger than x^2, there must be a point where they cross. See why it is different?

    If you don't understand this then I recommend you do what I suggested and just plot about 6 or 7 points until you can convince yourself which way the curves are going -- it doesn't take much more than that. For example:

    \text{When }x = 2,   \: (x-2)^3 = 0, \: x(x-4) = -4
    \text{When }x = 3,  \: (x-2)^3 = 1, \: x(x-4) = -3
    \text{When }x = 4,  \: (x-2)^3 = 8, \: x(x-4) =  0
    \text{When }x = 5, \:  (x-2)^3 = 27, \: x(x-4) =  5
    \text{When }x = 6, \:  (x-2)^3 = 64, \: x(x-4) =  12
    \text{When }x = 7,  \: (x-2)^3 = 125, \: x(x-4) = 21

    Is it clear now why one is always going to be much bigger than the other?
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      Thank you for this thread OP. I know I would've done exactly the same mistake.
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      One solution would be to use simultaneous equations on both to therefore find the points of intersection.
     
     
     
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