use cubic regression to find a function that fits the following points.\n$(1, -1) (2, -13) (3, -45) (-1…

Explanation:

Step1: Define cubic function

Let the cubic function be $f(x) = ax^3 + bx^2 + cx + d$, where $a, b, c, d$ are coefficients to solve for.

Step2: Substitute points to get equations

Substitute each $(x, f(x))$ into the function:

1. For $(1, -1)$: $a(1)^3 + b(1)^2 + c(1) + d = -1$ → $a + b + c + d = -1$
2. For $(2, -13)$: $a(2)^3 + b(2)^2 + c(2) + d = -13$ → $8a + 4b + 2c + d = -13$
3. For $(3, -45)$: $a(3)^3 + b(3)^2 + c(3) + d = -45$ → $27a + 9b + 3c + d = -45$
4. For $(-1, 11)$: $a(-1)^3 + b(-1)^2 + c(-1) + d = 11$ → $-a + b - c + d = 11$

Step3: Solve the system of equations

First, subtract Equation 1 from Equation 2: $(8a + 4b + 2c + d) - (a + b + c + d) = -13 - (-1)$ → $7a + 3b + c = -12$ (Equation 5)

Subtract Equation 2 from Equation 3: $(27a + 9b + 3c + d) - (8a + 4b + 2c + d) = -45 - (-13)$ → $19a + 5b + c = -32$ (Equation 6)

Add Equation 1 and Equation 4: $(a + b + c + d) + (-a + b - c + d) = -1 + 11$ → $2b + 2d = 10$ → $b + d = 5$ (Equation 7)

Subtract Equation 5 from Equation 6: $(19a + 5b + c) - (7a + 3b + c) = -32 - (-12)$ → $12a + 2b = -20$ → $6a + b = -10$ (Equation 8)

Now use Equation 1 and Equation 4 to eliminate $d$: From Equation 7, $d = 5 - b$. Substitute into Equation 1: $a + b + c + 5 - b = -1$ → $a + c = -6$ → $c = -6 - a$ (Equation 9)

Substitute $c = -6 - a$ into Equation 5: $7a + 3b + (-6 - a) = -12$ → $6a + 3b = -6$ → $2a + b = -2$ (Equation 10)

Subtract Equation 10 from Equation 8: $(6a + b) - (2a + b) = -10 - (-2)$ → $4a = -8$ → $a = -2$

Substitute $a = -2$ into Equation 10: $2(-2) + b = -2$ → $-4 + b = -2$ → $b = 2$

Substitute $a = -2$ into Equation 9: $c = -6 - (-2) = -4$

Substitute $b = 2$ into Equation 7: $2 + d = 5$ → $d = 3$

Answer:

$-2x^3 + 2x^2 -4x + 3$