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Impossible pre cal question Give formula for the graph of the polynominal

  1. Jul 3, 2012 #1
    1. The problem statement, all variables and given/known data

    http://www.math.poly.edu/courses/ma0914/past_exams/MA0922_Final_2000-12-13.pdf [Broken] First problem on the exam

    2. Relevant equations

    I believe the answer should be a 3rd degree polynomial so Ax^3+Bx^2+Cx+D

    3. The attempt at a solution

    Ax^3+Cx
    -8A-2C=0
    8A+2C=0

    -A-C=3
    -4A=C
    -A+4A=3A

    A=1.

    However that is impossible because A has to be negative for the 2nd derivative on the left side of the equation to be positive. Idk where I went wrong.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jul 3, 2012 #2
    This is not correct. It should be a 5th degree polynomial. Are you able to explain why & continue from there?

    There are several possible ways to form the necessary equations to find out the constants.
     
  4. Jul 3, 2012 #3
    I guess I would continue by plugging in the values for the zeros and plug in the values for (-1,3) and (1,-3) and solve the system of linear equations. However I don't know why it's a 5 degree. Technically I'm not supposed to use derivatives.
     
  5. Jul 3, 2012 #4
    Something like that, yeah : ]

    I'm not too familiar with the US (I assume?) education system, so I don't know what is taught in Pre-calculus. If, however, you already know how to take the derivative of polynomials, you might want to use that in order to make the system of equations easier to solve. It shouldn't make a huge difference, though.

    As for why it's 5th degree: You should be able to see it from the shape of the graph. One way to explain it is that a 3rd degree polynomial can only change direction two times (ie. f'(x) has only two roots and is a 2nd degree polynomial). You can see that the graph changes from increasing to decreasing or vice versa four times, so f'(x) has four roots and f(x) is a 5th-degree polynomial. Well, at least that's one possibility, and you were only looking for ONE possible formula.
     
  6. Jul 3, 2012 #5
    Alright I got it. I Went from Algebra 2 and trig to calc 1 and 2 and then all of the way up to complex variables. However to get a job as a math tutor I need to take a test in pre calculus. However contrary to the name pre calculus I noticed I never used any of this stuff in my upper math classes.
     
  7. Jul 3, 2012 #6
    -32A+16B-8C+4D-2E=0
    32A+16B+8C+4D+2E=0
    -A+B-C+D-E=3
    A+B+C+D=-3

    So I get 4 equations and 5 unknowns. However when I try to solve them I get.
    2B+2D=0
    32B+8D=0
    which implies that both B and D are zero does that seem right so far.

    edit my bad.
     
  8. Jul 3, 2012 #7
    Yeah, B=0, D=0.

    EDIT: You should get 5 equations for the 5 unknowns, though.

    Well, six for the six unknowns, but f(0)=0 -> F=0 is pretty damn obvious, so.
     
  9. Jul 3, 2012 #8
    Where would the 5th one come from besides the face that F would equal zero. Because I'm getting something that can have infinitely many solutions.
     
  10. Jul 3, 2012 #9
    You can use f'(2)=0 or f'(-2)=0, for example. There's an exact solution, don't worry.
     
  11. Jul 3, 2012 #10
    I can't use derivatives this pre calc. Is this question possible without calculus?
     
  12. Jul 3, 2012 #11
    Also wouldn't the 2nd derivative be zero not the first at 2. After looking I noticed that the 1st derivative is zero also. Not like it helps.
     
  13. Jul 3, 2012 #12
    Oh. To be honest, I didn't really think about that possibility. Using derivatives just seemed so obvious that I didn't bother to check how to do it without them, sorry.

    I'm not sure about this, but I don't see any way to do this without the derivatives.

    EDIT: No, the first derivative is zero at the (local) maxima and minima. The sign of the second derivative tells whether it's a maximum or a minimum, unless f''(x)=0.
     
  14. Jul 3, 2012 #13
    Well even though my original hunch that it's not possible is correct I see that my reasoning was wrong. Thanks for the help I learned something.
     
  15. Jul 3, 2012 #14

    Filip Larsen

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    Gold Member

    It is a bit easier if you are familiar with the fundamental theorem of algebra, from which it follows that if a n-degree polynomial has the roots (or zeros) x1, x2, ..., xn (some of which may be equal to each other), then the polynomial can be written as g(x) = a(x-x1)(x-x2)...(x-xn), where a root will be repeated if the function has zero derivative (i.e. horizontal tangent) in that root.

    Looking at the graph you can count 3 roots, of which 2 is repeated since they have a horizontal tangent, and 4 horizontal tangents in total. Taken together that means g must be a 5-degree polynomial with the root in -2 and 2 each repeated twice. From the last information, that g(-1) = 3, you can then solve for a and get one particular polynomial that "fits" all the information given.
     
  16. Jul 3, 2012 #15
    I wrote down the g(-1)=3 and got 4 equations with 5 unknowns.
     
  17. Jul 3, 2012 #16

    D H

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    Staff Emeritus
    Science Advisor

    The graph appears to be <fill in the blank> about x=0. Use that.
     
  18. Jul 3, 2012 #17
    symmetric?
     
  19. Jul 3, 2012 #18

    D H

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    Staff Emeritus
    Science Advisor

    Close. There is a symmetry there.

    Hint: How do f(x) and f(-x) appear to be related?
     
  20. Jul 3, 2012 #19
    f(-x)=-f(x) which allow us setting all even degree (include f) to zero. Makes life easier but still need the knowledge of derivative. How this can be done without derivative?

     
  21. Jul 3, 2012 #20
    Filip Larsen (post #14) has suggested all you need to know to solve it. There's no need to bother about whether it's "symmetric" or looking at any derivatives or any simultaneous equations (as the first few posts were suggesting).
     
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