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Determining the least possible degree of a polynomial function

  1. Oct 8, 2013 #1
    1. The problem statement, all variables and given/known data
    Determine the least possible degree of the function corresponding to the graph shown below. Justify your answer.

    2. Relevant equations
    The graph is attached. I remade the graph using google grapher, but the graph I got in the test have exactly the same x-intercepts (-2 of order 2 and 1 of order 3), y-intercepts, turning points, and end behaviour.

    3. The attempt at a solution
    So according to the x-intercept, the least possible degree is 5. (adding up the order of the x-intercepts, 2+3=5.

    According to the turning points, the least possible degree is 3. (2 turning points, 2+1=3.)

    Because the question asks for the least possible degree, I thought I should use the larger value here, which is 5.

    But then my teacher said the answer is 3, because 3 is less than 5. The class also got 3 as answer. When I asked my math teacher about this she told me that the question asked about the least degree, so since 3 is less than 5, the answer should have been 3. she said "5 is the exact degree, not the least degree." Some of my classmates also told me that since her question was regarding the "least" degree, I should've put 3 instead of 5.

    And this just completely baffles me-the question clearly says the least possible degree. So for it to be the least possible degree, it has to be possible first. The x-intercepts clearly indicate that it would not be even possible for the function to be a 3rd degree function.

    I was told that when my English improves I will understand the question better and then see why the answer is 3. But I have literally thought about the question for a week now and still don't see the flaw in my logic. Please help me understand this question? Sorry and Thanks in advance.
     

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  2. jcsd
  3. Oct 8, 2013 #2

    symbolipoint

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    Maybe what I say here is not precise enough, but every time the curve bends up or down, the degree increases by 1; the degree assumes at degree 1 at the extreme left side, and then you find four bends which change direction, so 1+4=5.
     
  4. Oct 8, 2013 #3

    Dick

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    I agree. I don't think your English is a problem. The problem is that the curve has 2 turning points. If f is your function, that means f'' has two zeros. If the degree were 3 then f'' would be a linear function and could only have one zero. It has to be at least degree 5.
     
  5. Oct 8, 2013 #4

    haruspex

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    ... and an inflection, which counts as another two for this purpose. So yes, the degree must be at least 5. Take a gold star fakecop for getting the right answer when your teacher and the rest of the class are wrong, and another for not taking their word for it!
     
  6. Oct 8, 2013 #5

    Dick

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    Actually I meant 2 "inflection points", not "turning points". In fact, it has 3, but degree three can't even have 2.
     
  7. Oct 8, 2013 #6
    Yeah, I tried to use my calculus knowledge to talk to my teacher about this (I even made a graph for the second derivative) but she just won't listen to me. She said that if I knew nothing about calculus, then I would understand the question the way she intended to.

    I've been very upset because I feel that I have gone the length to learn calculus outside of school and have been punished in return. I don't know what to do now. It seems that no matter how hard I work to prepare for a test I just get the question wrong!

    And in my belief, even if we were allowed to use flex points (a much better criteria than x-intercepts or turning points) to determine the degree, our result still will not be accurate.
    Why? consider the graph attached. Since there is 1 flex point, we may conclude that the least possible degree is 3. But the degree is actually 5. (As you can see the equation is x^5-5x). Even though the 3rd and 5th degree graphs look similar, they just won't be the same for the reason that the 3rd derivative in the 3rd degree will always be constant, where as the 3rd derivative in the 5th degree will not be constant.

    I believe that to truly find the degree, we need to find the least-ordered derivative for the function that stays at a constant value. For example, if the 5th derivative is constant, then the degree is 5. But if we are given only the graph with no equation, then things become exceedingly difficult. It is difficult to pinpoint zeros of any derivative beyond degree 2.

    zeros of f(x) are the zeros in the graph, zeros of f'(x) are the turning points, and zeros of f''(x) are the points that change from concave up to concave down. But how can anyone visually determine points where the third derivative is zero from just a graph? It's not possible.

    So you see my source of confusion when my teacher asked me the question and expects me to base my criteria off only the turning points!
     

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  8. Oct 8, 2013 #7

    Dick

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    I see your frustration. All you can say by looking a graph is possibly to make some statement about a minimum degree of the polynomial. You certainly can't determine it exactly. Your first graph has to have degree at least 5 because it clearly has 3 flex points. If they don't believe you, I don't know what to do about it. You are still correct. For the second one, it has degree at least 3, both from counting the turning points and from having an inflection point. Beyond that, sure, visual inspection won't tell you it really has degree 5. You can make a degree 3 polynomial that has a graph that looks a lot like that and I wouldn't expect anyone to figure that out by just looking at the graph. You are probably more well informed than your teacher. Sometimes that happens.
     
  9. Oct 8, 2013 #8

    symbolipoint

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    Much of what you say there seems consistent with what I remember from College Algebra --- which I have reviewed a few times since graduating. I agree; calculus is not really needed to respond to the question about degree of the graphed polynomial function.
     
  10. Oct 8, 2013 #9

    Dick

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    Are the rules from College Algebra just arbitrary rules, or is there a noncalculus way you derive them? fakecop seems to have a very good intuitive understanding of what's going on based on calculus. Is there another way?
     
  11. Oct 9, 2013 #10
    I agree it's 5.

    We all know it's odd because it heads off towards +inifinity on the positive side and -infinity on the negative side.

    One inflection is obvious. You might question whether there is a second inflection - unless you know what a cubic curve looks like - and that is clearly not cubic. So there are 2 inflection points.

    This wouldn't be the first time that a math teacher didn't know his math.
     
  12. Oct 9, 2013 #11
    I think you can make an argument that does not use any knowledge of calculus or inflection points. Clearly the given function has a root of even multiplicity at ##x=-2## and one of odd multiplicity at ##x=1##. So ##p(x)=(x+2)^2(x-1)q(x)##. If the degree of ##p## were 3, then ##q(x)=c## for some constant ##c##. The graph of ##p## would then look like the graph of ##y=(x+2)^2(x-1)## with a different scale factor on the ##y##-axis. This graph does not look like a scaled version of the graph of ##y=(x+2)^2(x-1)##, and so the degree of ##p## cannot be 3. So 3 is not a possible degree of the polynomial, and therefore it cannot be the least possible degree.
     
    Last edited: Oct 9, 2013
  13. Oct 9, 2013 #12
    Thank you everyone for your replies. But my teacher seems only to be interested in her own interpretation of the question. I did manage to persuade a few classmates but some of them maintain that the answer is 3. Guess I will have to make do with that...
     
  14. Oct 9, 2013 #13

    epenguin

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    x5 - x ?

    I don't think using the roots was mentioned. From the roots 0 and ±1 in your second graph you know you have a factor
    x( x2 - 1 ).
    As mentioned it does not look like a cubic. The odd symmetry f(x) = - f(-x) of the above factor is shared by your graph, therefore any further factor has to be made of even powers of x but no real roots. (x2 + 1) is just the simplest possibility, but it would have been harder to say it couldn't be other things subject to the stated restrictions.
     
    Last edited: Oct 10, 2013
  15. Oct 9, 2013 #14
    When you expand this, you get x^3-x. I attached the graph below.

    This is what I'm talking about-it's impossible for you to determine the actual degree, because one simply cannot distinguish the sign of derivatives beyond third order. In this case the third degree graph and the fifth degree graph look very similar to each other. There is no visual criteria to distinguish them.

    Suppose we had a geometric interpretation for third derivative-call it "jerk". A very smart person might be able to look at the two graphs and distinguish the cubic graph because it has constant "jerk". But we don't have those capabilities-so stating the least possible degree is impossible.
     

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  16. Oct 10, 2013 #15

    epenguin

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    Sorry it was late night and what I wrote wasn't quite what I'd thought. I have corrected it and from my reasoning you can actually get that the minimum degree of a curve like in your #6 thumbnail is 5.

    It depended on the unformalised 'look' of it, but since you have the real roots and can plot, as you have, the cubic with those roots you can recognise it can't be that. OK you might wonder what about that multiplied by a constant. But that wouldn't change the position of the maximum and minimum, so it can't be that either.
     
    Last edited: Oct 10, 2013
  17. Oct 10, 2013 #16

    LCKurtz

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    Challenge them to produce such a cubic.
     
  18. Oct 10, 2013 #17

    pasmith

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    So she would claim that [itex]x^3[/itex] is of least degree 1?

    Your teacher needs a crash course in basic logic: if I am required to choose [itex]x[/itex] to satisfy [itex]x \geq 3[/itex] and [itex]x \geq 5[/itex] then of course I must take [itex]x \geq 5[/itex].
     
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