Create the equation of a quadratic function, in standard form, that has one zero of -3 and a turning point at (-1,-16)

Question

Create the equation of a quadratic function, in standard form, that has one zero of -3 and a turning point at (-1,-16)

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Gabriella 2 weeks 2022-01-15T17:21:12+00:00 1 Answer 0 views 0

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    2022-01-15T17:22:28+00:00

    Answer:

    y=4x^2+8x-12

    Step-by-step explanation:

    we know that

    The quadratic equation in vertex form is equal to

    y=a(x-h)^2+k

    where

    a is the leading coefficient

    (h,k) is the vertex of the quadratic equation

    Remember that

    In a quadratic equation the turning point is the same that the vertex

    so

    (h,k)=(-1,-16)

    substitute

    y=a(x+1)^2-16

    we have one zero at (-3,0)

    substitute and solve for a

    0=a(-3+1)^2-16

    0=4a-16\\4a=16\\a=4

    substitute

    y=4(x+1)^2-16

    Convert to standard form

    y=4(x^2+2x+1)-16\\y=4x^2+8x+4-16\\y=4x^2+8x-12

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