Question 687718
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http://www.scribblar.com/mg9yqg9

Basically, we are dealing with a quadratic function.  Generally, that will involve using the "vertex form"
{{{y=a(x-h)^2+b}}}
so, we know that there is a maximum of y=35.  what, in the formula, is the y value of the vertex of the parabola?  (b)
Since the y values are the same on the given points, then the axis of symmetry is in the middle of them (2+10)/2  or 6.  that is the h value of the vertex.  If they were not the same y values, then we could still find the h value, but it would be more complicated:
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***alternative way to find h value Start***
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so we can insert that into the formula:  y=a(x-h)^2+35
we also know two points that work.  enter the first point and you get 3=a(2-h)^2+35 
foil (2-h)^2 to get h^2-4h+4, 
solve for "a"
a=-32/(h^2-4h+4)
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now plug the other point into y=a(x-h)^2+35
and you get 3=a(10-h)^2+35
subtract 35 from both sides
-32=a(10-h)^2
substitute -32/(h^2-4h+4) from the first equation in for a in the second equation.
You get:
-32=(-32(10-h)^2)/(h^2-4h+4)
reduce the -32's and move the h^2-4h+4 to the left
reduce to find the h value.
h=6
***alternative way to find h value Stop***
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so now you have y=a(x-6)^2+35
use one of the points for the x and y value to find "a"
3=a(4)^2+35
-32/16=a
a=-2
so 
y=-2(x-6)^2+35