Question 217107
I am having difficulty with a question. The problems is as follows: A parabola has an x-intercept at 2, its axis of symmetry is the l ine x=4, and the y-coordinate of its vertex is 6. Determone the equation of the parabola. 
I am setting up my equation - y = ax^2 + bx + c and plugging in the numbers as follows: y = 4a^2 + 2b + 8. Am I on the right track? 
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y = ax^2 + bx + c  is the correct format.
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Since the axis of symmetry is x=4, the other x-intercept will be at 6.
So the zeroes are 2 and 6.
The vertex at y = +6 and 2 x-intercepts means it opens downward.
The vertex is at (4,6), it has to be on the axis of symmetry.
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There's a way to find a parabola given 3 points (or a circle) using determinants.  I'll send info on that to anyone interested.
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The x value at the vertex is -b/2a, so -b/2a = 4.
f(4) = 6 (vertex)
f(2) = 0
f(6) = 0
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At x = 2:
4a + 2b + c = 0
At x = 6:
36a + 6b + c = 0
At x = 4
16a + 4b + c = 6
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Solving 3 eqns in a, b and c gives:
a = -1.5
b = 12
c = -18
--> y = -1.5x^2 + 12x - 18