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A.P asked in Science & MathematicsMathematics · 2 months ago

Precal Help!! How can I solve this? Thank you in advance!!?

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2 Answers

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  • atsuo
    Lv 6
    2 months ago

    Let the function be y = f(x).

    f(x) has three zeros, at x = -4, -3 and 1/2.

    So it must be

    f(x) = k(x + 4)(x + 3)(x - 1/2)

    = k(x^3 + (13/2)x^2 + (17/2)x - 6)

    And the y-intercept is -2, so f(0) = -2. Therefore,

    -2 = k(-6)

    k = 1/3

    So the answer is

    y = (1/3)(x^3 + (13/2)x^2 + (17/2)x - 6)

    = (1/3)x^3 + (13/6)x^2 + (17/6)x - 2

  • 2 months ago

    The graph has zeros at (1/2,0), (-4,0) and (-3,0).

    The graph has a y-intercept at (0,-2).

    y = ax^3 + bx^2 + cx + d

    Plugging in our four points gives 4 linear equations in a,b,c,d:

    0 = (1/8)a + (1/4)b + (1/2)c + d

    0 = -64a + 16b - 4c + d

    0 = -27a + 9b -3c + d

    -2 = d

    Solving these simultaneous equations gives a = 1/3, b = 13/6, c = 17/6, d = -2

    y(x) = (1/3)x^3 + (13/6)x^2 + (17/6)x - 2

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