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Rational Function Graphing

  1. Jun 16, 2009 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations
    Slant Asymptote at y=x+3. (X-1) and (X+2) appear to be intercepts in the back of the book. How do I factor the numerator to get that?

    3. The attempt at a solution
    I know this is simple and there is a method to find the zeros of the numerator that I am overlooking- please help.
    Last edited: Jun 16, 2009
  2. jcsd
  3. Jun 17, 2009 #2


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    Re: Rational Function Graphing - AWAKE TILL SOLVED

    A more systematic method is to apply Rational Roots Theorem. You want to find some binomials that can divide the polynomial numerator and leave no remainder. Try dividing by (X - 1) and see what results. Can you factor this result? OR, try dividing the numerator by (X + 2). How is the result? Remainder?

    My guess is you want three linear binomials as a factorization for the numerator, since it has degree of 3. If those other binomials, X-1 and X+2 are factors, then your function would have value of ZERO when X=+1 and when X=-2.
  4. Jun 17, 2009 #3


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    Re: Rational Function Graphing - AWAKE TILL SOLVED

    The only integer factors of 4 are (1)(4) and (2)(2). As symbolipoint said, by the "rational roots theorem" the only rational (in this case, integer) roots must be factors of 4: [itex]\pm 1[/itex], [itex]\pm 2[/itex], [itex]\pm 4[/itex]. It easy to try those in the polynomial and see that x= 1 makes it 0: x-1 is a factor. Trying x= -2 also gives a root so x+2 is also a factor.
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