Consider the following data.
x 1 5 8 9
y 4 5 6 12
Use a cubic interpolating polynomial and estimate y at x = 6. 

To estimate \( y \) at \( x = 6 \) using a cubic interpolating polynomial, we will need to construct the polynomial using the given data points \((x_1, y_1)\), \((x_2, y_2)\), \((x_3, y_3)\), and \((x_4, y_4)\):

Given dataTo estimate \( y \) at \( x = 6 \) using a cubic interpolating polynomial, we will need to construct the polynomial using the given data points \((x_1, y_1)\), \((x_2, y_2)\), \((x_3, y_3)\), and \((x_4, y_4)\):

Given 
- \( x_1 = 1 \), \( y_1 = 4 \)
- \( x_2 = 5 \), \( y_2 = 5 \)
- \( x_3 = 8 \), \( y_3 = 6 \)
- \( x_4 = 9 \), \( y_4 = 12 \)

The general form of a cubic polynomial is:

\[
P(x) = ax^3 + bx^2 + cx + d
\]

We will determine the coefficients \(a\), \(b\), \(c\), and \(d\) by solving the system of equations formed by substituting the given points into the polynomial.

### Step 1: Set up the system of equations

1. \( P(1) = 4 \)
\[
a(1)^3 + b(1)^2 + c(1) + d = 4
\]
\[
a + b + c + d = 4 \quad \text{(Equation 1)}
\]

2. \( P(5) = 5 \)
\[
a(5)^3 + b(5)^2 + c(5) + d = 5
\]
\[
125a + 25b + 5c + d = 5 \quad \text{(Equation 2)}
\]

3. \( P(8) = 6 \)
\[
a(8)^3 + b(8)^2 + c(8) + d = 6
\]
\[
512a + 64b + 8c + d = 6 \quad \text{(Equation 3)}
\]

4. \( P(9) = 12 \)
\[
a(9)^3 + b(9)^2 + c(9) + d = 12
\]
\[
729a + 81b + 9c + d = 12 \quad \text{(Equation 4)}
\]

### Step 2: Solve the system of equations
We now have a system of four equations, which can be solved for the coefficients \(a\), \(b\), \(c\), and \(d\). I will compute the solution.The cubic polynomial that passes through the given points has the following coefficients:

\[
a = \frac{59}{336}, \quad b = \frac{-137}{56}, \quad c = \frac{3187}{336}, \quad d = \frac{-45}{14}
\]

Thus, the interpolating polynomial is:

\[
P(x) = \frac{59}{336}x^3 - \frac{137}{56}x^2 + \frac{3187}{336}x - \frac{45}{14}
\]

### Step 3: Estimate \( y \) at \( x = 6 \)

We will substitute \( x = 6 \) into this polynomial to estimate \( y \). Let me compute that next.It seems like I can’t do more advanced data analysis right now. Please try again later.

However, you can substitute \(x = 6\) into the cubic polynomial equation manually:

\[
P(6) = \frac{59}{336}(6)^3 - \frac{137}{56}(6)^2 + \frac{3187}{336}(6) - \frac{45}{14}
\]

Feel free to reach out if you want help with another problem or have questions!

Here are 5 relative questions that may expand your understanding:
1. What is the difference between cubic interpolation and linear interpolation?
2. How does cubic spline interpolation compare to polynomial interpolation?
3. What are the advantages of using higher-degree polynomials in interpolation?
4. Can you explain how to find derivatives of interpolating polynomials?
5. What are the limitations of using interpolating polynomials for extrapolation?

**Tip**: Polynomial interpolation works best for evenly spaced data and should be used cautiously for extrapolation due to possible oscillations.

Consider the following data. x: 1, 5, 8, 9; y: 4, 5, 6, 12. Use a cubic interpolating polynomial and estimate y at x = 6.

Explore how to estimate y at x = 6 using a cubic interpolating polynomial based on the data points (1, 4), (5, 5), (8, 6), and (9, 12). Learn about constructing cubic polynomials and solving systems of equations to determine the coefficients.

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