Question 196682
please help me solve: the cubic function whose graph is shown. 
(-3,0)(0,-7) (1,0) (3,0)
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Form: y = ax^3 + bx^2 + cx + d
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Substitute for x/y and solve for a,b,c,d
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(-3,0): (-3)^3a + (-3)^2b + (-3)c + d = 0
(0,-7):  0a     +   0b    +  0c   + d = -7
(1,0) :  1a     +   1b    +  1c   + d = 0
(3,0) : (3)^3   + (3)^2b  + (3)c  + d = 0
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Solve the system of 4 equations with 4 unknowns by any method you know
to get:
a = -7/9
b =  7/9
c =  7
d =  -7
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Equation:
y = (-7/9)x^3 + (7/9)x^2 + 7x - 7
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Cheers,
Stan H.