Use cubic regression to find a function that fits the following points. \[ \begin{array}{l}(1,4)(2,9)(3,34)(-2,-11) \\ \left.[?] x^{3}+\square x^{2}+\square x+\square\right]\end{array} \]

Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}a+b+c+d=4\\8a+4b+2c+d=9\\27a+9b+3c+d=34\\-8a+4b-2c+d=-11\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}a=4-b-c-d\\8a+4b+2c+d=9\\27a+9b+3c+d=34\\-8a+4b-2c+d=-11\end{array}\right.\) - step2: Substitute the value of \(a:\) \(\left\{ \begin{array}{l}8\left(4-b-c-d\right)+4b+2c+d=9\\27\left(4-b-c-d\right)+9b+3c+d=34\\-8\left(4-b-c-d\right)+4b-2c+d=-11\end{array}\right.\) - step3: Simplify: \(\left\{ \begin{array}{l}32-4b-6c-7d=9\\108-18b-24c-26d=34\\-32+12b+6c+9d=-11\end{array}\right.\) - step4: Solve the equation: \(\left\{ \begin{array}{l}b=\frac{23-6c-7d}{4}\\108-18b-24c-26d=34\\-32+12b+6c+9d=-11\end{array}\right.\) - step5: Substitute the value of \(b:\) \(\left\{ \begin{array}{l}108-18\times \frac{23-6c-7d}{4}-24c-26d=34\\-32+12\times \frac{23-6c-7d}{4}+6c+9d=-11\end{array}\right.\) - step6: Simplify: \(\left\{ \begin{array}{l}108-\frac{9\left(23-6c-7d\right)}{2}-24c-26d=34\\37-12c-12d=-11\end{array}\right.\) - step7: Solve the equation: \(\left\{ \begin{array}{l}c=\frac{59-11d}{6}\\37-12c-12d=-11\end{array}\right.\) - step8: Substitute the value of \(c:\) \(37-12\times \frac{59-11d}{6}-12d=-11\) - step9: Simplify: \(-81+10d=-11\) - step10: Move the constant to the right side: \(10d=-11+81\) - step11: Add the numbers: \(10d=70\) - step12: Divide both sides: \(\frac{10d}{10}=\frac{70}{10}\) - step13: Divide the numbers: \(d=7\) - step14: Substitute the value of \(d:\) \(c=\frac{59-11\times 7}{6}\) - step15: Calculate: \(c=-3\) - step16: Substitute the values of \(c,d:\) \(b=\frac{23-6\left(-3\right)-7\times 7}{4}\) - step17: Calculate: \(b=-2\) - step18: Substitute the values of \(b,c,d:\) \(a=4-\left(-2\right)-\left(-3\right)-7\) - step19: Substitute back: \(a=4+2+3-7\) - step20: Calculate: \(a=2\) - step21: Calculate: \(\left\{ \begin{array}{l}a=2\\b=-2\\c=-3\\d=7\end{array}\right.\) - step22: Check the solution: \(\left\{ \begin{array}{l}a=2\\b=-2\\c=-3\\d=7\end{array}\right.\) - step23: Rewrite: \(\left(a,b,c,d\right) = \left(2,-2,-3,7\right)\) To find a cubic function that fits the given points \((1,4)\), \((2,9)\), \((3,34)\), and \((-2,-11)\), we can express the cubic function in the form: \[ f(x) = ax^3 + bx^2 + cx + d \] We will set up a system of equations based on the provided points: 1. For the point \((1, 4)\): \[ a(1)^3 + b(1)^2 + c(1) + d = 4 \quad \Rightarrow \quad a + b + c + d = 4 \] 2. For the point \((2, 9)\): \[ a(2)^3 + b(2)^2 + c(2) + d = 9 \quad \Rightarrow \quad 8a + 4b + 2c + d = 9 \] 3. For the point \((3, 34)\): \[ a(3)^3 + b(3)^2 + c(3) + d = 34 \quad \Rightarrow \quad 27a + 9b + 3c + d = 34 \] 4. For the point \((-2, -11)\): \[ a(-2)^3 + b(-2)^2 + c(-2) + d = -11 \quad \Rightarrow \quad -8a + 4b - 2c + d = -11 \] Now we have the following system of equations: \[ \begin{align*} 1. & \quad a + b + c + d = 4 \\ 2. & \quad 8a + 4b + 2c + d = 9 \\ 3. & \quad 27a + 9b + 3c + d = 34 \\ 4. & \quad -8a + 4b - 2c + d = -11 \end{align*} \] Solving this system, we find: \[ (a, b, c, d) = (2, -2, -3, 7) \] Thus, the cubic function that fits the given points is: \[ f(x) = 2x^3 - 2x^2 - 3x + 7 \]