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April 25, 2014

April 25, 2014

Posted by **Anonymous** on Thursday, December 27, 2012 at 4:21pm.

Is this correct? y=16/75(x-7.5)^2

- Algebra 2 -
**Steve**, Thursday, December 27, 2012 at 4:26pmIf the vertex is at (0,0), you know that when x=±15/2, y=12

so, if y = kx^2

12 = k(15/2)^2, so

k = 12*4/225 = 48/225 = 16/75

y = 16/75 x^2

Now, if you don't want to consider negative values for x, then it would make sense to translate the vertex to (7.5,0), so you are correct.

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