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8 August, 05:03

What is the vertex of the quadratic finction f (x) = (x-6) (x+2)

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  1. 8 August, 07:58
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    The zeros are where the factors are zero, at x=-2 and x=6. The line of symmetry and the x-value of the vertex is halfway between these at (-2+6) / 2 = 2. Then the y-value of the vertex is

    ... f (2) = (2-6) (2+2) = - 16

    The vertex is (x, y) = (2, - 16).
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