Answer:
To fit a quadratic function of the form f(t) = c0 + c1t + c2t^2 to the given data points using the least squares method, we need to find the values of c0, c1, and c2 that minimize the sum of squared errors between the predicted values and the actual data points. The sum of squared errors (SSE) is given by:
SSE = Σ [f(t_i) - y_i]^2
where (t_i, y_i) are the data points, and f(t_i) is the value predicted by the quadratic function at t_i.
Let's calculate the values of c0, c1, and c2 that minimize SSE for the given data points.
Data points: (0, 4), (1, 7), (2, -12), (3, -13)
Step 1: Set up the equations using the data points:
For (0, 4):
4 = c0 + c1*0 + c2*0^2 = c0
For (1, 7):
7 = c0 + c1*1 + c2*1^2 = c0 + c1 + c2
For (2, -12):
-12 = c0 + c1*2 + c2*2^2 = c0 + 2*c1 + 4*c2
For (3, -13):
-13 = c0 + c1*3 + c2*3^2 = c0 + 3*c1 + 9*c2
Step 2: Solve the system of equations:
c0 = 4 (from the first equation)
c0 + c1 + c2 = 7 (from the second equation)
c0 + 2*c1 + 4*c2 = -12 (from the third equation)
c0 + 3*c1 + 9*c2 = -13 (from the fourth equation)
Substitute the value of c0 into the second, third, and fourth equations:
4 + c1 + c2 = 7
4 + 2*c1 + 4*c2 = -12
4 + 3*c1 + 9*c2 = -13
Step 3: Solve the system of equations to find c1 and c2:
c1 + c2 = 3
2*c1 + 4*c2 = -16
3*c1 + 9*c2 = -17
Now, we can use any suitable method to solve this system of equations. Let's use the substitution method.
From the first equation, we get c1 = 3 - c2. Substitute this value into the second equation:
2*(3 - c2) + 4*c2 = -16
6 - 2*c2 + 4*c2 = -16
2*c2 = -16 - 6
2*c2 = -22
c2 = -11
Now, substitute the value of c2 back into c1 = 3 - c2:
c1 = 3 - (-11) = 3 + 11 = 14
So, we have c1 = 14 and c2 = -11. The quadratic function is:
f(t) = c0 + c1*t + c2*t^2
f(t) = 4 + 14*t - 11*t^2
Therefore, the quadratic function that fits the given data points using the least squares method is f(t) = 4 + 14*t - 11*t^2.
Explanation: