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How do we know the degree of a root which is also a point of inflexion?

  1. Apr 17, 2012 #1
    Hey guys, I was looking at an exam I did last year and tried to solve a question, which at the time I couldn't do.

    Unfortunately I'm running into the same problem I had during the exam, so hear me out on this one

    Question:
    The graph below has equation y =ax(x-b)(x+c)^d. Write down the values for a, b, c and d.
    whatthefuuu.png


    Okay, so there's an intercept at x=-1, so b=-1. There's another intercept at x=3, so c = -3

    So, y = ax(x+1)(x-3)^d

    Now here's the problem, the answers simply say "d = 3", and then sub in the point (1,-16) and solve for a,

    But at x=3 (where it's a point on inflexion), can't the degree (d) be ANY odd number? ie. 3,5,7,9,11,...,999999?

    Because that's how I learnt polynomials, a point on inflexion at a root means the degree of the root is odd.

    And so that's where I messed up, becausing I couldn't figure what value of 'd' I should use.

    I must obviously be missing some sort of concept, so can someone please help me out?
     
  2. jcsd
  3. Apr 17, 2012 #2

    epenguin

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    Gold Member

    You idea is right - it could be any odd number - so far as you have got.

    You now have to ask yourself the question, one of the five or so in Polya's 'How to solve it' book

    -

    "Am I using all the information I'm given?"
     
  4. Apr 17, 2012 #3
    I'm aware of the the point (1,-16),

    So as you already know that b=-1 and c=-3, sub in the point to get:

    -16 = a(1)(1+1)(1-3)^d
    = 2a(-2)^d

    Two varibles makes this unsolvable but using guess and check you'll find 3 is the only correct solution (which just makes the whole question practically ****** and pointless).

    I was expecting there to be a way to solve the question without guess and check but I guess there isn't any other way...
     
  5. Apr 17, 2012 #4

    epenguin

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    OK, you have used the fact that (1, -16) is a point on the function.

    But I think you are supposed to use also the fact that it is a special kind of point, a stationary point.
     
    Last edited: Apr 17, 2012
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