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16 October, 14:00

Y=-2 (x-1) ^2+8 how do I factor this parabola? I need to find the zeros, vertex, and axis of symmetry

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  1. 16 October, 14:35
    0
    Y = - 2 (x - 1) ² + 8

    y = - 2 ((x - 1) (x - 1)) + 8

    y = - 2 (x² - x - x + 1) + 8

    y = - 2 (x² - 2x + 1) + 8

    y = - 2 (x²) + 2 (2x) - 2 (1) + 8

    y = - 2x² + 4x - 2 + 8

    y = 2x² + 4x + 6

    2x² + 4x + 6 = 0

    x = - 4 + / - √ (4² - 4 (2) (6))

    2 (2)

    x = - 4 + / - √ (16 - 48)

    4

    x = - 4 + / - √ (-32)

    4

    x = - 4 + / - 5.6568i

    4

    x = - 4 + / - 1.4142i

    x = - 4 + 1.4142i x = - 4 - 1.4142

    y = - 2x² + 4x + 6

    y = - 2 (-4 + 1.4142i) ² + 4 (-4 + 1.4142i) + 6

    y = - 2 ((-4 + 1.4142i) (-4 + 4.4142i) - 16 + 5.6568i + 6

    y = - 2 (16 - 5.6568i - 5.6568i + 1.99996164i²) - 16 + 5.6568i + 6

    y = - 2 (16 - 11.3136i + 1.9996164) - 16 + 5.6568i + 6

    y = - 32 + 22.6272i - 3.99992328 - 16 + 5.6568i + 6

    y = - 32 - 3.99992328 - 16 + 6 + 22.6272i + 5.6568i

    y = - 35.99992328 - 16 + 6 + 28.284i

    y = - 51.99992328 + 6 + 28.284i

    y = - 45.99992328 + 28.284i

    (x, y) = (-4 + 1.4142i, - 45.99992328 + 28.284i)

    y = - 2x² + 4x + 6

    y = - 2 (-4 - 1.4142i) ² + 4 (-4 - 1.4142i) + 6

    y = - 2 ((-4 - 1.4142i) (-4 - 1.4142i)) - 16 - 5.6568i + 6

    y = - 2 (16 + 5.6568i + 5.6568i + 1.99996164i²) - 16 - 5.6568i + 6

    y = - 2 (16 + 11.3136i + 1.99996164) - 16 - 5.6568i + 6

    y = - 32 - 22.6272i - 3.99992328 - 16 - 5.6568i + 6

    y = - 32 - 3.99992328 - 16 + 6 - 22.6272i - 5.6568i

    y = - 35.99992328 - 16 + 6 - 28.284i

    y = - 51.99992328 + 6 - 28.284i

    y = - 45.99992328 - 28.284i

    (x, y) = (-4 - 1.4142i, - 45.99992328 - 28.284i)

    zeros: - 4 + 1.4142i or - 4 - 1.4142i

    vertex: (-4 + 1.4142i, - 45.99992328 + 28.284i) and (-4 - 1.4142i, - 45.99992328 - 28.284i)

    axis of symmetry: 0 + 2.8284i
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