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What is the quadratic function that is created with roots - 10 and - 6 and a vertex at (-8, - 8) ?

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  1. 4 May, 09:42
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    f (x) = x^2 + 16x + 56.

    Step-by-step explanation:

    The vertex form is:

    y = a (x + 8) ^2 - 8 where a is a constant.

    When x = - 6, y = 0 so:

    0 = a (-6+8) ^2 - 8

    4^2a = 8

    a = 1

    So the function is y = (x + 8) ^2 - 8

    in standard form this is

    y = x^2 + 16x + 64 - 8

    y = x^2 + 16x + 56.
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