Question

use cubic regression to find a function that fits the following points. (1,4)(2,9)(3,34)(-2,-11) $x^{3}+x^{2}+x+$

Explanation:

Step1: Set up the cubic function

Let the cubic function be \(y = ax^{3}+bx^{2}+cx + d\).
Substitute the points \((x,y)\) into the function:
For \((1,4)\): \(a(1)^{3}+b(1)^{2}+c(1)+d=4\), so \(a + b + c + d=4\).
For \((2,9)\): \(a(2)^{3}+b(2)^{2}+c(2)+d=9\), so \(8a + 4b+2c + d=9\).
For \((3,34)\): \(a(3)^{3}+b(3)^{2}+c(3)+d=34\), so \(27a+9b + 3c + d=34\).
For \((-2,-11)\): \(a(-2)^{3}+b(-2)^{2}+c(-2)+d=-11\), so \(-8a + 4b-2c + d=-11\).

Step2: Solve the system of equations

Subtract the first equation from the second:
\((8a + 4b+2c + d)-(a + b + c + d)=9 - 4\)
\(7a+3b + c=5\).
Subtract the second equation from the third:
\((27a+9b + 3c + d)-(8a + 4b+2c + d)=34 - 9\)
\(19a+5b + c=25\).
Subtract the second equation from the fourth:
\((-8a + 4b-2c + d)-(8a + 4b+2c + d)=-11 - 9\)
\(-16a-4c=-20\), simplify to \(4a + c = 5\).

Step3: Solve for \(a\), \(b\), \(c\), \(d\)

Subtract \(7a+3b + c=5\) from \(19a+5b + c=25\):
\((19a+5b + c)-(7a+3b + c)=25 - 5\)
\(12a+2b=20\), simplify to \(6a + b=10\).
From \(4a + c = 5\), we have \(c=5 - 4a\).
Substitute \(c = 5 - 4a\) into \(7a+3b + c=5\):
\(7a+3b+(5 - 4a)=5\)
\(3a+3b=0\), so \(b=-a\).
Substitute \(b=-a\) into \(6a + b=10\):
\(6a-a=10\), \(5a=10\), \(a = 2\).
If \(a = 2\), then \(b=-2\), \(c=5-4\times2=-3\), \(d=4-(2-2 - 3)=7\).

Answer:

\(2x^{3}-2x^{2}-3x + 7\)