Let's solve for the variable coefficients a, b, and c by using the given three points. We'll establish three equations by plugging our x-values into our function, and solve this system of equations. Here are the equations:
1. For f(1)=-12, we substitute x=1 into the function f(x)=ax^(2)+bx+c giving us
a(1)^2 + b(1) + c = -12
This simplifies to:
a + b + c = -12
2. For f(-3)=8, we substitute x=-3 into the function f(x)=ax^(2)+bx+c giving us
a(-3)^2 + b(-3) + c = 8
This simplifies to:
9a - 3b + c = 8
3. For f(3)=-46, we substitute x=3 into the function f(x)=ax^(2)+bx+c giving us
a(3)^2 + b(3) + c = -46
This simplifies to:
9a + 3b + c = -46
Now we have a system of 3 equations, and we can solve it by using linear algebra methods. We create a matrix A for the coefficients of a, b and c, and a matrix B for numbers on the right side of the equations.
After solving the system of linear equations, we get the values a = -2.0, b = -9.0, and c = -1.0 (approximately).
Thus, the final quadratic function is given by:
f(x) = -2.0x^2 + -9.0x + -1.0.