use cubic regression to find a function that fits the following points. (1, -9) (2, -15) (3, -29) (-1, 3)…

Explanation:

Step1: Assume the cubic function

Let the cubic function be $y = ax^{3}+bx^{2}+cx + d$.

Step2: Substitute the points

For point $(1,-9)$: $a\times1^{3}+b\times1^{2}+c\times1 + d=-9$, i.e., $a + b + c + d=-9$. For point $(2,-15)$: $a\times2^{3}+b\times2^{2}+c\times2 + d=-15$, i.e., $8a + 4b+2c + d=-15$. For point $(3,-29)$: $a\times3^{3}+b\times3^{2}+c\times3 + d=-29$, i.e., $27a+9b + 3c + d=-29$. For point $(-1,3)$: $a\times(-1)^{3}+b\times(-1)^{2}+c\times(-1)+d = 3$, i.e., $-a + b - c + d=3$.

Step3: Set up a system of linear - equations

We have the system of equations: $\begin{cases}a + b + c + d=-9\8a + 4b+2c + d=-15\27a+9b + 3c + d=-29\-a + b - c + d=3\end{cases}$ Subtract the first equation from the second: $(8a + 4b+2c + d)-(a + b + c + d)=-15+9$, $7a + 3b + c=-6$. Subtract the second equation from the third: $(27a+9b + 3c + d)-(8a + 4b+2c + d)=-29 + 15$, $19a+5b + c=-14$. Subtract the first equation from the fourth: $(-a + b - c + d)-(a + b + c + d)=3 + 9$, $-2a-2c = 12$, or $a + c=-6$, so $c=-6 - a$.

Step4: Substitute $c=-6 - a$ into $7a + 3b + c=-6$ and $19a+5b + c=-14$

$7a + 3b-6 - a=-6$, $6a+3b=0$, $2a + b=0$, so $b=-2a$. $19a+5b-6 - a=-14$, $18a+5b=-8$. Substitute $b = - 2a$ into $18a+5b=-8$: $18a+5\times(-2a)=-8$, $18a-10a=-8$, $8a=-8$, $a=-1$.

Step5: Find $b$, $c$, and $d$

Since $a=-1$, then $b=-2a = 2$. Since $a=-1$, then $c=-6 - a=-6+1=-5$. Substitute $a=-1$, $b = 2$, $c=-5$ into $a + b + c + d=-9$: $-1+2-5 + d=-9$, $d=-5$.

Answer:

$-1x^{3}+2x^{2}-5x - 5$