I'm trying to get a cubic function from the points (0, 7), (2, 10), (4, 5), and (6, 0). I know the equation is y = ax^3 + bx^2 + cx + d, and
d = 7
After plugging in the other points I got 3 equations:
8a + 4b + 2c = 3
64a + 16b + 4c = -2
216a + 36b + 6c = -7
My substitution and elimination between three unknowns is more than shabby, could someone please help out?
Third way! If you don't like fractions, Multiply 8a+ 4b+ 2c= 3 by 2 to get 16a+ 8b+ 4c= 6 and subtract from the second equation, 64a+ 16b+ 4c= -2, to get 48a+ 8b= -8, eliminating c. Similarly, multiply 8a+ 4b+ 2c= 3 by 3 to get 24a+ 12b+ 6c= 9 and subtract from the third equation, 216a+ 36b+ 6c= -7 to get 192a+ 24b= -16, again eliminating c.
Now we have 48a+ 8b= -8 and 192a+ 24b= -16. Multiply 48a+ 8b= -8 by 3 to get 144a+ 24b= -24 and subtract from 192a+ 24b= -16 to get 48a= 8.
as Soroban got.
(Of course, since a is a fraction, you can't avoid a fraction here.)
Now put that back into 48a+ 8b= -8, which is the same as 6a+ b= -1 to get 1+ b= -1 so b= -2.
Put a= 1/6 and b= -2 into any of the original three equations to solve for c.