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Intersection of two graphs

  1. Sep 7, 2011 #1
    1. The problem statement, all variables and given/known data The graphs y=abs(x) and y=(x^2)-6 intersect at x=3 and x= -3
    What is confusing me is when I set them equal to eachother and solve (x^2)-x-6=0 and (x^2)+x-6=0 I get -3,+3,-2,+2
    What is the deal with the negative 2 and pos 2?


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 7, 2011 #2

    wukunlin

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    Gold Member

    abs(x) = x only if x is positive, so when you solve x^2-x-6=0, you need to disregard negative answers, and vice versa
     
  4. Sep 7, 2011 #3

    rock.freak667

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    Homework Helper

    y= |x| is the graph of y=x with the bottom half reflected in the x-axis (so any value of x gives a positive value of y). It looks like a v basically.

    Since both y=|x| and y=x2-6 are symmetrical about the y-axis, you will get 'mirrored' answers, so if you get x=2, you will get x=-2 on the next side of the graphs.

    Draw them out and you will see the symmetry I am talking about.
     
  5. Sep 7, 2011 #4
    Thanks, I did draw them out and the graphs intersect at x=3 and x=-3
    They do not intersect at x=2 and x=-2 even though 2 and -2 solve the equation when I set the two functions equal to eachother.
     
  6. Sep 7, 2011 #5

    rock.freak667

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    Yeah drawing you would see +3 and -3 but it's like wukunlin said.

    When you solved x2-x-6=0 you had the constraint of x>0, you would ignore the x=-2.
     
  7. Sep 7, 2011 #6
    ohhhhhhhh ok I get it now. Thanks guys
     
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