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A rather straight forward function question.

  1. Jan 22, 2013 #1
    The problem statement:
    A square centered at the origin has its vertices on the x- & y- axes.
    The graph of the function f(x)=ax2-4 , a>0
    Passes through three of the square's vertices.
    You must find what 'a' makes this statement true.
    Other things to know:
    Well I think there are multiple solutions to the problem. But I only need one. It would really help if you could show work and explain how you got your answer.
    Thanks in advance to anyone who assists.




    Solved:
    f(x)=(0.25)X2-4

    'a'=0.25
     
    Last edited: Jan 22, 2013
  2. jcsd
  3. Jan 22, 2013 #2

    jbunniii

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    If you plug in x = 0, you get f(0) = -4. That gives you one of the vertices, right?
     
  4. Jan 22, 2013 #3
    Yeah so you are saying that because it is a square, it has to have vertices at
    (4,0),(-4,0),(0,4), and (0,-4)? Hahah genius. So now it's much simpler to find the formula for what 'a' must be.
     
  5. Jan 22, 2013 #4

    jbunniii

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    Yes, that's right.
     
  6. Jan 22, 2013 #5
    Solved:
    f(x)=(0.25)X2-4

    'a'=0.25
     
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