(Easy) For the following quadratic function, find the axis of symmetry, the vertex and the y-intercept. y = x^2 + 12x + 32

Question

(Easy) For the following quadratic function, find the axis of symmetry, the vertex and the y-intercept. y = x^2 + 12x + 32

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Adela 3 years 2021-08-23T09:02:04+00:00 2 Answers 33 views 0

Answers ( )

    0
    2021-08-23T09:03:42+00:00

    Answer:

    y  ( \times  + 6) 2 - 4

    0
    2021-08-23T09:03:55+00:00

    Answer:

    see explanation

    Step-by-step explanation:

    Given a quadratic in standard form , y = ax² + bx + c ( a ≠ 0 ), then

    The x- coordinate of the vertex, which is also the equation of the axis of symmetry is

    x_{vertex} = – \frac{b}{2a}

    y = x² + 12x + 32 ← is in standard form

    with a = 1, b = 12 , then

    x_{vertex} = – \frac{12}{2} = – 6

    Substitute x = – 6 into y for corresponding y- coordinate

    y = (- 6)² + 12(- 6) + 32 = 36 – 72 + 32 = – 4

    Thus

    equation of axis of symmetry is x = – 6

    vertex = (- 6, – 4 )

    To find the y- intercept , let x = 0

    y = 0² + 12(0) + 32 = 32 ← y- intercept

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