Find an expression for a cubic function f if f(5) = 300 and f(-5) = f(0) = f(6) = 0.
Start with the general cubic:
f(x) = A·x³ + B·x² + C·x + D
Use f(5) = 300 and f(-5) = f(0) = f(6) = 0 to form four equations:
f(5) = A·(5)³ + B·(5)² + C·(5) + D = 300
f(-5) = A·(-5)³ + B·(-5)² + C·(-5) + D = 0
f(0) = A·(0)³ + B·(0)² + C·(0) + D = 0
f(6) = A·(6)³ + B·(6)² + C·(6) + D = 0
We simplify them to this system of four equations in four unknowns
125A + 25B + 5C + D = 300
-125A + 25B - 5C + D = 0
D = 0
216A + 36B + 6C + D = 0
And since the third equation tells us that D=0 we can simplify the
other equations and just have three equations in three unknowns:
125A + 25B + 5C = 300
-125A + 25B - 5C = 0
216A + 36B + 6C = 0
That's pretty easy to solve.
A=-6, B=6, C=180, and we already had D=0
So
f(x) = A·x³ + B·x² + C·x + D
becomes:
f(x) = -6x³ + 6x² + 180x
Edwin