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Range of P(x)

  1. Oct 2, 2011 #1
    1. The problem statement, all variables and given/known data
    P(x) = √6+5x-x2

    2. Relevant equations

    3. The attempt at a solution
    P(x) = √6+5x-x2

    -x2+5x+6 >/= 0
    x2-5x-6 </= 0
    (x-6)(x+1) </= 0
  2. jcsd
  3. Oct 2, 2011 #2


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    Do you mean [itex]P(x) =\sqrt{6+5x-x^2}\,?[/itex]

    If so, then use parentheses to make things clear.

    P(x) = √(6+5x-x2).

    Otherwise it looks like you have [itex]P(x) =(\sqrt{6})+5x-x^2\,?[/itex]

    What you have done will allow you to find the domain of P(x).

    To find the range, find the vertex of the parabola of f(x) = 6+5x-x2 . Then proceed from there.
  4. Oct 2, 2011 #3
    In vertex form it is y=(x-2.5)2-12.25
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