Find the quadratic equation whose parabola has vertex (3,-2) and y-intercept (0, 16). Give your answer in vertex form.

Question

Find the quadratic equation whose parabola has vertex (3,-2) and y-intercept (0, 16). Give your
answer in vertex form.

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Thanh Hà 5 years 2021-08-12T04:57:09+00:00 1 Answers 53 views 0

Answers ( )

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    2021-08-12T04:58:46+00:00

    Answer:

    y = 2*(x – 3)^2 – 2

    Step-by-step explanation:

    Remember that a quadratic equation of vertex (h, k) is written as:

    y = a*(x – h)^2 + k

    Where a is the leading coefficient.

    So, if we know that the vertex is at (3, – 2)

    we have:

    y = a*(x – 3)^2 + (-2)

    And we want the y-intercept to be (0, 16)

    This means that, when we take x = 0, we must have y = 16

    if we replace these in the above equation we get:

    16 = a*(0 – 3)^2 – 2

    now we can solve this for a

    16 = a*(-3)^2 – 2

    16 = a*9 – 2

    16 + 2 = a*9

    18 = a*9

    18/9 = a

    2 = a

    Then the quadratic equation is:

    y = 2*(x – 3)^2 – 2

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )