Question 936364: Write a cubic function that passes through the following points: (-2,0) (3,0) (-1,0) and (1,2). Explain how you got the answer
Answer by TimothyLamb(4379)   (Show Source):
You can put this solution on YOUR website! the general standard cubic:
ax^3 + bx^2 + cx + d = 0
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from the first 3 points (-2,0) (3,0) (-1,0), we have 3 roots at:
(x + 2)(x - 3)(x + 1)
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so the equation is as follows, where k is an unknown constant:
(x + 2)(x - 3)(x + 1) + k = 0
(xx - x - 6)(x + 1) + k = 0
xxx + xx - xx - x - 6x - 6 + k = 0
xxx - 7x - 6 + k = 0
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but from the 4th point (1,2) we know that:
xxx - 7x - 6 + k = 0
1*1*1 - 7*1 - 6 + k = 0
-12 + k = 0
k = 12
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combining the constants, the equation is:
xxx - 7x - 6 + k = 0
xxx - 7x - 6 + 12 = 0
xxx - 7x + 6 = 0
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answer:
x^3 - 7x + 6 = 0
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