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1 June, 10:16

In the xy-coordinate plane, the graph of the equation y=2x^2-12x-32 has zeros at x=d and x=e, where d>e. The graph has a minimum at (f,-50) what are the vaules of d, e, and f?

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  1. 1 June, 12:38
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    To find the zeroes solve fro x:-

    2 (x^2 - 6x - 16) = 0

    2 (x - 8) (x + 2) = 0

    x = {8, - 2

    so d = 8 and e = - 2

    the x coordinate of the minimum value = - b / 2a = 12 / 2*2 = 3

    f = 3
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