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Homework Help: Polynomial Graph Behavior

  1. Oct 26, 2008 #1

    jacksonpeeble

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    1. The problem statement, all variables and given/known data
    Sketch the graph of the polynomial function. Make sure your graph shows all intercepts and exhibits the proper behavior. No calculator allowed.

    P(x)=x(x-3)(x+2)


    2. Relevant equations
    P(x)=x(x-3)(x+2)


    3. The attempt at a solution
    P(x)=x(x-3)(x+2)
    0(x-3)(x+2)=0
    x(3-3)(x+2)=0
    x(x-3)(-2+2)=0
    Zeros: 0, 3, -2
    Degree: 1+1+1=3 (Odd)
    Leading Coefficient: 1 (Positive)
    Multiplicities: 1, 1, 1

    Therefore, I have X-Intercepts at all of the zeros. However, I am confused as to how I determine what direction the graph goes at different intervals and whether it bounces or goes through the zeros. Any help is greatly appreciated.
     
  2. jcsd
  3. Oct 26, 2008 #2

    rock.freak667

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    Find the stationary points by putting P'(x)=0
     
  4. Oct 27, 2008 #3
    There's a reason this is the pre-cal thread, no matter how much some like to ignore the title. Although it is a lot easier with calculus,I suggest you find out what lies in between the zeroes (positive values or negative values). In fact,as a good way to guess, pick x values halfway between the zeroes for substitution.
     
  5. Oct 27, 2008 #4

    HallsofIvy

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    If x> 3, then all of x, x-3, and x+2 are positive so x(x-3)(x+ 2) is positive.
    If 0< x< 3, then x and x+2 are positive but x-3 is negative so x(x-3)(x+2) is negative.
    If -2< x< 0, then x+ 2 is positive but x and x-3 are negative so x(x-3)(x+2) is positive.
    If x< -2 then all of x, x-3, and x+2 are negative so x(x-3)(x+2) is negative.

    That should be exactly what you need.
     
  6. Oct 27, 2008 #5

    jacksonpeeble

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    What's that supposed to mean? This work is from my class, called "Honors Trigonometry and Pre-calculus."

    Thank you to HallsOfIvy and everyone else that contributed, that was exactly what I needed!
     
  7. Oct 27, 2008 #6
    P'(x) means the derivative with respect to x, which I am guessing you wouldn't have learned by now because that is in calculus.
     
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