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Changing equation to standard form

  1. Sep 6, 2009 #1
    1. The problem statement, all variables and given/known data

    Sketch the region in the xy-plane that is bounded between the graphs of the given functions. Find the points of intersection of the graphs.

    1) y=x^2+2x+2

    2)y=-x^2-2x+2


    3. The attempt at a solution

    I already completed the square for equation 1):
    y=(x+1)^2+1

    Im having trouble completing the square for the second equation because of the negative values. I tried factoring out the negative sign:

    -(x^2+2x-2)

    but that just makes the 2 negative.

    How should I go about converting the second equation?
     
  2. jcsd
  3. Sep 6, 2009 #2
    nevermind, i got it
     
  4. Sep 6, 2009 #3
    Oh well from what i can see,-x^2-2x+2 = -(x^2+2x+2) +4. so....-x^2-2x+2=-(y)+4. If you can sketch the graph for the 1st one the second one shouldn't be a problem:biggrin:
     
  5. Sep 7, 2009 #4

    HallsofIvy

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    No, [itex]-(x^2+ 2x- 2)= -x^2- 2+ 2[/itex], not [itex]x^2- 2x+ 2[/itex].
    [itex]x^2- 2x+ 2= (x- 1)^2+ 1[/itex].

     
  6. Sep 7, 2009 #5
    hmm did i make a mistake somewhere? Where did the [itex] x^2- 2x+ 2 [/itex] come from haha::rolleyes:
     
  7. Sep 7, 2009 #6
    it comes out to be -(x+1)^2+3
     
  8. Sep 7, 2009 #7
    I think you misread the second equation
     
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