A rather straight forward function question.

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Dangshnizzle
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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:
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jbunniii said:
If you plug in x = 0, you get f(0) = -4. That gives you one of the vertices, right?
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.
 
Dangshnizzle said:
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.
Yes, that's right.
 
jbunniii said:
Yes, that's right.
Solved:
f(x)=(0.25)X2-4

'a'=0.25