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September 3, 2014

September 3, 2014

Posted by **Emily** on Sunday, August 22, 2010 at 8:28pm.

I've got : 8=a(6-h)^2+15, but i don't know what to do next.

- Math -
**Anonymous**, Sunday, August 22, 2010 at 8:40pmI will assume that your equation is such that the vertex is on the y-axis

Then your equation will be

y = ax^2 + 15 , (the h of your equation will be zero)

then (6,8) must lie on the equation

8 = a(6)^2 + 15

a = -7/36

equation: y = (-7/36)x^2 + 15

- Math -
**bobpursley**, Sunday, August 22, 2010 at 8:44pmthe peak is at 3,15, it turns downward so that at h=8 me the difference in x is 6.

h=-a(x-3)^+15

then

8=-a(6-3)^2+15

8=-9a+15

a=7/9

so, here is one that satisfies the statement

h=-7/9 (x-3)^2+15

- Math -
**Emily**, Sunday, August 22, 2010 at 9:24pmBobpursley, how do you know that the peak is at (3,15)?

- Math -
**Anonymous**, Sunday, February 13, 2011 at 3:27pmi dont know howto solve this either someone HELP

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