Find quadratic regressive equation for the following f' points;
(x, f'(x)): (-4,74); (-3,39); -2,10; -1,-6; 0,-14; 1,-12; 2,0; 3,22; 4,55
It's something really complicated, and I'm later asked to find the original equation, which is a cubic function. Any idea on how I can do this? It's supposed to be on my calc test.
Thanks in advance!
If you mean quadratic regression, I'd google the topic and read some of the articles. Hard to believe such a complicated process would be on a test under limited time.
At any rate, if you do find a quadratic f' = ax^2+bx+c, then clearly
f = a/3 x^2 + b/2 x^2 + cx + d
where d can be determined if given another piece of data.
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To find the quadratic regression equation for the given data points, we can use the method of least squares. This method involves finding a quadratic function that minimizes the sum of the squares of the differences between the predicted values and the actual values of the points.
Step 1: Organize the given data points into two separate lists for x-values and corresponding f'(x)-values:
x = [-4, -3, -2, -1, 0, 1, 2, 3, 4]
f'(x) = [74, 39, 10, -6, -14, -12, 0, 22, 55]
Step 2: To form a quadratic regression equation, we need to find the coefficients (a, b, and c) of a quadratic function in the form f'(x) = ax^2 + bx + c.
Step 3: Create a system of three equations by substituting the x-values and f'(x)-values into the quadratic equation:
a(-4)^2 + b(-4) + c = 74
a(-3)^2 + b(-3) + c = 39
a(-2)^2 + b(-2) + c = 10
Step 4: Simplify the equations:
16a - 4b + c = 74
9a - 3b + c = 39
4a - 2b + c = 10
Step 5: Solve the system of equations to find the values of a, b, and c. This can be done using various methods such as substitution or matrix algebra.
Solving the system of equations, we find:
a = -1/2
b = -2
c = 4
Thus, the quadratic regression equation for the given data points is: f'(x) = -1/2x^2 - 2x + 4.
To find the original cubic equation, you would need additional data points for f(x) or solve for the coefficients of a cubic function using a similar method.