Algebra 2

An archer releases an arrow from a shoulder height of 1.39 m. When
the arrow hits the target 18 m away, it hits point A. When the target is
removed, the arrow lands 45 m away. Find the maximum height of the arrow along its parabolic path.
My points are
(0,1.39)
(18,1.5)
(45,0)
I'm supposed to use this formula: y = ax^2 + bx + c
I get C = 1.39 after that I get lost

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  1. ax^2+bx+c=y

    0a+0b+c=1.39
    324a+18b+c=1.5
    2025a+45b+c=0

    a=-.00137
    b=0.0308
    c=1.39

    y = -.00137x^2 + 0.031x + 1.39

    max height on the parabola is at x = -b/2a, so at
    x = .031/.00274 = 11.31
    y = 1.565

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