SOLUTION: I am trying to find the quadratic function of a parabola with a vertex of 5.3 and X intercepts -5.25 and 5.25

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Question 975497: I am trying to find the quadratic function of a parabola with a vertex of 5.3 and X intercepts -5.25 and 5.25
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Ordered pair for the vertex is (0,5.3) because its x coordinate is in the middle of the zeros' values.

y=a%28x-0%29%5E2%2B5.3, full standard form
y=ax%5E2%2B5.3
-
Solve for a
y-5.3=ax%5E2
%28y-5.3%29%2Fx%5E2=a
-
Substitute either of the given zeros to have an expression for a
a=%280-5.3%29%2F%285.25%29%5E2
a=5.3%2F5.25%5E2
a=0.193333cross%283repeating%29
which could really be expressed as a rational number, but if your 5.3 was a measurement
then a good choice could well be more like a=0.193.

highlight%28y=%280.193%29x%5E2%2B5.3%29

a might in fact NOT be a repeating decimal in the way I thought first, but using 1=16%2F16, you should be able to find a=848%2F4410.