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Creating a non-cubic polynomial

  1. Jun 26, 2006 #1
    Would you please give one or two (or more, sorry hehehe) to help me?

    How can you make up a polynomial that:
    - crosses the x-axis at -2
    - just touches the x-axis at 1 ("touch" as the graph y=x^2 would only touch, not cross, the x-axis)
    - and is above the x-axis between -2 and 1.

    I know there must be equations that are not cubic that satisfy these conditions. Oh and the graph contains no additional contact with the x-axis. How would you create the equation? Please help^^
     
  2. jcsd
  3. Jun 26, 2006 #2

    0rthodontist

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    If the equation can't be cubic, then make it quintic (highest term x^5) so it will have the same end behavior.

    Let's say that you have a polynomial that is cubic with these properties (do you?). Then can you predict what would happen if you multiply it by (for example) (x^2 + 1)?
     
    Last edited: Jun 26, 2006
  4. Jun 26, 2006 #3

    J77

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    So you want a function [tex]f(x)[/tex] such that:

    [tex]\frac{df}{dx}=0[/tex] and [tex]\frac{d^2f}{dx^2}<0[/tex] at [tex]x=1[/tex]

    and

    [tex]f(-2)=0[/tex]...
     
  5. Jun 26, 2006 #4
    well very many thanks I give you. Sorry, im only at precalculus and just at the basic level, so I don't really know what happens when you multiply an equation by (x^2 + 1), but as I tried doing that it worked! The graph of the quintic equation looks very similar to my original cubic equation lol^^

    Oh and J77, I'm afraid I do not yet know calculus. Sorry hehehe...

    EDIT: Oh I just figured out why. Adding the factor (x^2 + 1) does not add additional x-intercepts!
     
    Last edited: Jun 26, 2006
  6. Jun 26, 2006 #5

    HallsofIvy

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    So one factor is (x-(-2))= (x+2)
    Okay, the parabola is tangent to the x-axis and another factor is
    (x-1)2

    (x+2)(x-1)2 itself, at x= 0, is (0+2)(-1)2= 2 which is "above" the x-axis- i.e. positive. Since a polynomial is of one sign between places where it is 0 and (x+2)(x-1)2 is 0 only at -2 and 1, it's positive between -2 and 1: it looks to me like that works.

    Yes, there are "equation" (I assume you mean polygons) that are not cubic yet satisfy these equation. For example, multiply (x+2)(x-1)2 by x2+1 which is always positive and, since x2+ 1 is never 0, does not cross or touch the x-axis at any other point. But why? Your original problem did not state that the polynomial must be not be cubic.
     
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