2b- Solved problems for quadratic interpolation

Last Updated on September 24, 2026 by Maged kamel

Solved problems for quadratic interpolation using the Vandermonde Matrix.

Two solved problems for quadratic interpolation are introduced using the given three points and their corresponding y-values to obtain the coefficients a0, a1, and a2. Then we can write the quadratic polynomial as P(x) = a0 + a1*x + a2*x^2.

The last step is to multiply the inverse Matrix V-1 by the X-X Matrix to find the factor column vector.

We ultimately obtained these values, as shown in the last slide image.

These equations can be written in Matrix form. Use the form V*X=Y, where V is the Vandermonde Matrix, and X is the column vector of the coefficients a₀, a₁, and a2. Y is the column vector of y values for the three points.

These are the final coefficient values. For more details, please refer to the previous Post to learn how to derive this Expression. Matrix B is 3×3, and the factors for each Row and column are shown on the next slide.

Solved 1/2 of the problems for quadratic interpolation.

Derive an Expression for the quadratic polynomial for a given three-point x and y Value, and check the P(x) Value at x = 2.70.

The values of b₁₁, b₁₂, and b₁₃ are the first Row of Matrix B.

The first step is to find the coefficients a0, a1, and a2 by substituting the corresponding elements of the B Matrix, with x0 = 1, y0 = 3; x1 = 2, y1 = 5; x2 = 3, y2 = 8. Each Row has different x0, x1, x2, and fraction values.

The next slide shows the values in the first Row of Matrix B: data for items b11, b12, and b13.

The values of b₂₁, b₂₂, and b₂₃ in the second Row of Matrix B.

The next slide shows the values in the second Row of Matrix B: data for items b21, b22, and b23.

The values of b₃₁, b₃₂, and b33 in the third Row of Matrix B.

The next slide image explains the values in the third Row of Matrix B: data for items b31, b32, and b33.

The values of a₀,a₁, and a₂ for solved example 1.

Once we have written the three rows of Matrix B, we will multiply Matrix B by the vector (y0, y1, y2). The Product will yield the corresponding coefficient values: a0 = 2, a1 = 1/2, and a2 = 1/2.

The final Expression for P(x) and the Value of P(2.707).

The quadratic polynomial can be written as P(x) = a0 + a1x + a2*x^2. The last step is to find the Point P with coordinate x = 2.70. P(2.707) will be 6.995. Please check the next slide image for more details.

Solved Problem 2/2 for quadratic interpolation.

Derive an Expression for the quadratic polynomial using quadratic interpolation for a given three-point x and y Value, and check the P(x) Value at x = PI/12.

The function is f(x)=(sin x+cos x). the given x values are x₀=10 degrees, x₁=20 degrees and x₂=30 degrees. The first step is to find the values of coefficients a₀,a₁, and a₂ by substituting the corresponding elements of the B Matrix.

After converting the values in degrees to radians, consider x₀=0.1745, y₀=1.1585, x₁=0.3491, y₁=1.2817, and x₂=0.5236, y₂=1.3660.

The values of b₁₁, b₂₁, and b₃₁ in the first column of Matrix B.

The next slide shows the values in the first column of Matrix B and the values of the different elements.

The values of b₁₂, b₂₂, and b₂₃ in the second column of Matrix B.

The next slide shows the values in the second column of Matrix B and the values of the different elements.

The values of b₁₃, b₂₃, and b₃₃ in the third column of Matrix B.

The next slide shows the values in the third column of Matrix B and the values of the different elements.

The values of a₀,a₁, and a₂.

Once we have written the three rows or columns, as we did for Matrix B in this example, we multiply Matrix B by the vector (y0, y1, y₂). The Product will give the corresponding values of the coefficients as follows:

a₀=0.9968, a₁=1.0176 and a₂=-0.6009.

The final Expression for the quadratic interpolation is P(x), and we need to evaluate P(PI/12).

The quadratic polynomial can be written as P(x) = a0 + a1x + a2x^2. The last step is to find P(pi/12), which will be=1.222.

The original function’s valueat pii/12 is obtained by writing si (pi/12 )+ cos(pi/12 ) =1.2247, whichdifferss little from the quadratic polynomial Value.

For the PDF data for this Post, you can review and download it from the following document.

This is a Link to Post 9, titled “How to Use a Matrix for the Quadratic Function?“

The previous Post is 2a– Easy introduction to quadratic interpolation.

The next Post is an introduction to Newton-divided difference interpolation.

This is a Link to Holistic Numerical Methods-Newton Divided Differences.