Find the coordinates of the turning point for the equation of the graph below y=x^2+14x+45

Question

Find the coordinates of the turning point for the equation of the graph below y=x^2+14x+45

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Dâu 2 weeks 2021-09-04T21:14:58+00:00 1 Answers 0 views 0

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    2021-09-04T21:16:28+00:00

    Answer:

    (- 7, – 4 )

    Step-by-step explanation:

    Given a quadratic in standard form

    y = ax² + bx + c ( a ≠ 0 )

    Then the x- coordinate of the turning point is

    x = – \frac{b}{2a}

    y = x² + 14x + 45 ← is in standard form

    with a = 1, b = 14 , then

    x = – \frac{14}{2} = – 7

    Substitute x = – 7 into the equation and evaluate for y

    y = (- 7)² + 14(- 7) + 45 = 49 – 98 + 45 = – 4

    coordinates of turning point = (- 7, – 4 )

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