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# Precal Help!! How can I solve this? Thank you in advance!!? ### 2 Answers

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• 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

• 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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