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22 January, 20:37

What is the equation of the quadratic function with a vertex at (2,-25), and an x-intercept at (7,0)

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  1. 22 January, 20:55
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    f (x) = x^2 - 4x - 21.

    Step-by-step explanation:

    Since the vertex is at (2, - 25) we can write the vertex form as:

    y = f (x) = a (x - 2) ^2 - 25

    Given the x-intercept (7, 0), x = 7 when y = 0, so substituting:

    0 = a (7 - 2) ^2 - 25

    0 = 25a - 25

    giving a = 1.

    So our equation is f (x) = (x - 2) ^2 - 25.

    Converting to standard form

    f (x) = x^2 - 4x + 4 - 25

    f (x) = x^2 - 4x - 21.
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