{"id":4572,"date":"2026-09-14T12:25:12","date_gmt":"2026-09-14T12:25:12","guid":{"rendered":"https:\/\/www.mathros.net.ua\/en\/?p=4572"},"modified":"2026-09-14T14:25:47","modified_gmt":"2026-09-14T14:25:47","slug":"quadratic-interpolation","status":"publish","type":"post","link":"https:\/\/www.mathros.net.ua\/en\/quadratic-interpolation.html","title":{"rendered":"Quadratic Interpolation: Function Approximation Using Three Nodes"},"content":{"rendered":"

Quadratic interpolation is a numerical method that makes it possible to approximate intermediate function values using three known nodes. Unlike linear interpolation<\/a>, this approach uses a quadratic polynomial and can better represent the nonlinear behavior of a function between given points. Let us look at how such a polynomial is constructed, how its coefficients are found, and how the method can be used for functions given in tabular form.<\/p>\n

Quadratic Interpolation: From Three Nodes to a Parabola<\/h2>\n

Suppose a function is given in tabular form, and three of its values are known at the nodes<\/p>\n

\\[
\n(x_{i-1},y_{i-1}),\\qquad (x_i,y_i),\\qquad (x_{i+1},y_{i+1}).
\n\\]<\/p>\n

To approximate function values between the two outer nodes, a parabola is drawn through these three points. It is described by the quadratic interpolation polynomial<\/p>\n

\\[
\nP_2(x)=a\\cdot x^2+b\\cdot x+c.
\n\\]<\/p>\n

This polynomial contains three unknown coefficients \\( a \\), \\( b \\), and \\( c \\). To determine them uniquely, three independent conditions are required. These conditions come from the requirement that the value of the constructed polynomial at each of the three nodes must match the corresponding tabulated function value:<\/p>\n

\\[
\nP_2(x_{i-1})=y_{i-1},\\qquad P_2(x_i)=y_i,\\qquad P_2(x_{i+1})=y_{i+1}.
\n\\]<\/p>\n

Thus, the interpolation<\/a> nodes provide the known function values, while the quadratic polynomial is constructed from these data and used to approximate intermediate values.<\/p>\n

\"Quadratic<\/p>\n

The interpolation nodes do not have to be equally spaced – it is enough for their \\( x \\)-coordinates to be different. Therefore, quadratic interpolation can be used for both equally spaced and unequally spaced tabular data.<\/p>\n

If the three nodes lie on the same straight line, this does not affect the validity of the method. In this case, the coefficient \\( a \\) will be equal to zero, so the resulting polynomial will effectively become linear.<\/p>\n

From Interpolation Conditions to a System: Determining the Unknown Coefficients<\/h2>\n

To find the coefficients \\( a \\), \\( b \\), and \\( c \\), substitute the coordinates of each node into the quadratic polynomial. This gives a system of three linear equations:<\/p>\n

\\[
\n\\begin{cases}
\na\\cdot x_{i-1}^2+b\\cdot x_{i-1}+c=y_{i-1},\\\\
\na\\cdot x_i^2+b\\cdot x_i+c=y_i,\\\\
\na\\cdot x_{i+1}^2+b\\cdot x_{i+1}+c=y_{i+1}.
\n\\end{cases}
\n\\]<\/p>\n

The unknowns in this system are the coefficients \\( a \\), \\( b \\), and \\( c \\). The system has a unique solution if the \\( x \\)-coordinates of the three nodes are pairwise distinct:<\/p>\n

\\[
\nx_{i-1}\\ne x_i,\\qquad x_i\\ne x_{i+1},\\qquad x_{i-1}\\ne x_{i+1}.
\n\\]<\/p>\n

Under this condition, the determinant<\/a> of the coefficient matrix is nonzero, so the coefficients of the quadratic interpolation polynomial are determined uniquely.<\/p>\n

The resulting system can be solved using any standard method. To find the coefficients, we will use Cramer’s rule<\/a>:<\/p>\n

\\[
\na=\\frac{\\det(A_1)}{\\det(A)},\\qquad b=\\frac{\\det(A_2)}{\\det(A)},\\qquad c=\\frac{\\det(A_3)}{\\det(A)}.
\n\\]<\/p>\n

The determinant of the coefficient matrix is<\/p>\n

\\[
\n\\det(A)=
\n\\begin{vmatrix}
\nx_{i-1}^2 & x_{i-1} & 1\\\\
\nx_i^2 & x_i & 1\\\\
\nx_{i+1}^2 & x_{i+1} & 1
\n\\end{vmatrix}.
\n\\]<\/p>\n

To find the coefficient \\( a \\), replace the first column of the coefficient matrix with the column of tabulated function values:<\/p>\n

\\[
\n\\det(A_1)=
\n\\begin{vmatrix}
\ny_{i-1} & x_{i-1} & 1\\\\
\ny_i & x_i & 1\\\\
\ny_{i+1} & x_{i+1} & 1
\n\\end{vmatrix}.
\n\\]<\/p>\n

To find the coefficient \\( b \\), replace the second column with the tabulated function values:<\/p>\n

\\[
\n\\det(A_2)=
\n\\begin{vmatrix}
\nx_{i-1}^2 & y_{i-1} & 1\\\\
\nx_i^2 & y_i & 1\\\\
\nx_{i+1}^2 & y_{i+1} & 1
\n\\end{vmatrix}.
\n\\]<\/p>\n

To find the coefficient \\( c \\), make the corresponding replacement in the third column:<\/p>\n

\\[
\n\\det(A_3)=
\n\\begin{vmatrix}
\nx_{i-1}^2 & x_{i-1} & y_{i-1}\\\\
\nx_i^2 & x_i & y_i\\\\
\nx_{i+1}^2 & x_{i+1} & y_{i+1}
\n\\end{vmatrix}.
\n\\]<\/p>\n

After calculating the determinants and finding the coefficients \\( a \\), \\( b \\), and \\( c \\), the quadratic interpolation polynomial is completely determined.<\/p>\n

From the Constructed Polynomial to the Result: Approximating a Function Value<\/h2>\n

Once the coefficients have been determined, the constructed polynomial can be used to find function values at points where tabulated data are not available. If \\( x \\) lies between the two outer nodes, the approximate function value is found from<\/p>\n

\\[
\nf(x)\\approx P_2(x).
\n\\]<\/p>\n

Thus, once the polynomial has been constructed, the interpolation problem reduces to directly evaluating it at the required point.<\/p>\n

If the table contains more than three nodes, three nodes close to the given value of \\( x \\) are selected for a local approximation so that the point at which we want to find the function value lies between the two outer nodes. A quadratic polynomial is then constructed from the selected nodes and used over the corresponding part of the interval. Applying this approach to different parts of the table forms the basis of piecewise quadratic interpolation.<\/p>\n

\"Selecting<\/p>\n

The accuracy of the approximation depends on the behavior of the function and the positions of the selected nodes. In general, it is best to use nodes located as close as possible to the point at which we want to approximate the function value.<\/p>\n

Quadratic Interpolation: Step-by-Step Examples<\/h2>\n

Let us look at the practical use of quadratic interpolation for functions given in tabular form. In each case, we will select the required nodes, set up a system for the coefficients of the quadratic polynomial, find the coefficients using Cramer’s rule, and calculate the approximate function value.<\/p>\n

Example 1. Using the given table of function values, find the approximate value of \\( f(2.1) \\)<\/h3>\n\n\n\n\n\n\n\n
\\( i \\)<\/th>\n\\( x_i \\)<\/th>\n\\( y_i \\)<\/th>\n<\/tr>\n<\/thead>\n
\\( 0 \\)<\/td>\n\\( 0 \\)<\/td>\n\\( 2 \\)<\/td>\n<\/tr>\n
\\( 1 \\)<\/td>\n\\( 1.5 \\)<\/td>\n\\( 5.75 \\)<\/td>\n<\/tr>\n
\\( 2 \\)<\/td>\n\\( 3 \\)<\/td>\n\\( 14 \\)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

To construct the quadratic interpolation polynomial, we use the three given nodes:<\/p>\n

\\[
\n(x_0,y_0)=(0,2),\\qquad (x_1,y_1)=(1.5,5.75),\\qquad (x_2,y_2)=(3,14).
\n\\]<\/p>\n

Write the quadratic polynomial:<\/p>\n

\\[
\nP_2(x)=a\\cdot x^2+b\\cdot x+c.
\n\\]<\/p>\n

Substitute the coordinates of the nodes:<\/p>\n

\\[
\n\\begin{cases}
\nc=2,\\\\
\n2.25\\cdot a+1.5\\cdot b+c=5.75,\\\\
\n9\\cdot a+3\\cdot b+c=14.
\n\\end{cases}
\n\\]<\/p>\n

Calculate the determinant of the coefficient matrix:<\/p>\n

\\[
\n\\det(A)=
\n\\begin{vmatrix}
\n0 & 0 & 1\\\\
\n2.25 & 1.5 & 1\\\\
\n9 & 3 & 1
\n\\end{vmatrix}.
\n\\]<\/p>\n

In this case,<\/p>\n

\\[
\n\\det(A)=0\\cdot1.5\\cdot1+0\\cdot1\\cdot9+1\\cdot2.25\\cdot3-1\\cdot1.5\\cdot9-0\\cdot1\\cdot3-0\\cdot2.25\\cdot1=-6.75.
\n\\]<\/p>\n

Since<\/p>\n

\\[
\n\\det(A)\\ne0,
\n\\]<\/p>\n

the system has a unique solution.<\/p>\n

To find the coefficient \\( a \\), calculate<\/p>\n

\\[
\n\\det(A_1)=
\n\\begin{vmatrix}
\n2 & 0 & 1\\\\
\n5.75 & 1.5 & 1\\\\
\n14 & 3 & 1
\n\\end{vmatrix}
\n=-6.75.
\n\\]<\/p>\n

Then<\/p>\n

\\[
\na=\\frac{\\det(A_1)}{\\det(A)}=\\frac{-6.75}{-6.75}=1.
\n\\]<\/p>\n

To find the coefficient \\( b \\), calculate<\/p>\n

\\[
\n\\det(A_2)=
\n\\begin{vmatrix}
\n0 & 2 & 1\\\\
\n2.25 & 5.75 & 1\\\\
\n9 & 14 & 1
\n\\end{vmatrix}
\n=-6.75.
\n\\]<\/p>\n

Therefore,<\/p>\n

\\[
\nb=\\frac{\\det(A_2)}{\\det(A)}=\\frac{-6.75}{-6.75}=1.
\n\\]<\/p>\n

To find the coefficient \\( c \\), calculate<\/p>\n

\\[
\n\\det(A_3)=
\n\\begin{vmatrix}
\n0 & 0 & 2\\\\
\n2.25 & 1.5 & 5.75\\\\
\n9 & 3 & 14
\n\\end{vmatrix}
\n=-13.5.
\n\\]<\/p>\n

Thus,<\/p>\n

\\[
\nc=\\frac{\\det(A_3)}{\\det(A)}=\\frac{-13.5}{-6.75}=2.
\n\\]<\/p>\n

Therefore,<\/p>\n

\\[
\na=1,\\qquad b=1,\\qquad c=2.
\n\\]<\/p>\n

The quadratic interpolation polynomial is<\/p>\n

\\[
\nP_2(x)=x^2+x+2.
\n\\]<\/p>\n

For \\( x=2.1 \\), we obtain<\/p>\n

\\[
\nP_2(2.1)=2.1^2+2.1+2.
\n\\]<\/p>\n

Calculate:<\/p>\n

\\[
\nP_2(2.1)=4.41+2.1+2=8.51.
\n\\]<\/p>\n

Therefore,<\/p>\n

\\[
\nf(2.1)\\approx8.51.
\n\\]<\/p>\n

Example 2. Using the given table of function values, find the approximate value of \\( f(5.2) \\)<\/h3>\n\n\n\n\n\n\n\n\n
\\( i \\)<\/th>\n\\( x_i \\)<\/th>\n\\( y_i \\)<\/th>\n<\/tr>\n<\/thead>\n
\\( 0 \\)<\/td>\n\\( 1 \\)<\/td>\n\\( 4.5 \\)<\/td>\n<\/tr>\n
\\( 1 \\)<\/td>\n\\( 2.4 \\)<\/td>\n\\( 7.72 \\)<\/td>\n<\/tr>\n
\\( 2 \\)<\/td>\n\\( 4 \\)<\/td>\n\\( 9 \\)<\/td>\n<\/tr>\n
\\( 3 \\)<\/td>\n\\( 7 \\)<\/td>\n\\( 4.5 \\)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

The value \\( x=5.2 \\) lies between the nodes with \\( x \\)-coordinates \\( 4 \\) and \\( 7 \\):<\/p>\n

\\[
\n4<5.2<7.
\n\\]<\/p>\n

To construct the local quadratic polynomial, we use the nodes<\/p>\n

\\[
\n(x_1,y_1)=(2.4,7.72),\\qquad (x_2,y_2)=(4,9),\\qquad (x_3,y_3)=(7,4.5).
\n\\]<\/p>\n

For the selected nodes, set up a system for the coefficients \\( a \\), \\( b \\), and \\( c \\):<\/p>\n

\\[
\n\\begin{cases}
\n5.76\\cdot a+2.4\\cdot b+c=7.72,\\\\
\n16\\cdot a+4\\cdot b+c=9,\\\\
\n49\\cdot a+7\\cdot b+c=4.5.
\n\\end{cases}
\n\\]<\/p>\n

Calculate the determinant of the coefficient matrix:<\/p>\n

\\[
\n\\det(A)=
\n\\begin{vmatrix}
\n5.76 & 2.4 & 1\\\\
\n16 & 4 & 1\\\\
\n49 & 7 & 1
\n\\end{vmatrix}
\n=-22.08.
\n\\]<\/p>\n

To find the coefficient \\( a \\), calculate<\/p>\n

\\[
\n\\det(A_1)=
\n\\begin{vmatrix}
\n7.72 & 2.4 & 1\\\\
\n9 & 4 & 1\\\\
\n4.5 & 7 & 1
\n\\end{vmatrix}
\n=11.04.
\n\\]<\/p>\n

Then<\/p>\n

\\[
\na=\\frac{\\det(A_1)}{\\det(A)}=\\frac{11.04}{-22.08}=-0.5.
\n\\]<\/p>\n

To find the coefficient \\( b \\), calculate<\/p>\n

\\[
\n\\det(A_2)=
\n\\begin{vmatrix}
\n5.76 & 7.72 & 1\\\\
\n16 & 9 & 1\\\\
\n49 & 4.5 & 1
\n\\end{vmatrix}
\n=-88.32.
\n\\]<\/p>\n

Thus,<\/p>\n

\\[
\nb=\\frac{\\det(A_2)}{\\det(A)}=\\frac{-88.32}{-22.08}=4.
\n\\]<\/p>\n

To find the coefficient \\( c \\), calculate<\/p>\n

\\[
\n\\det(A_3)=
\n\\begin{vmatrix}
\n5.76 & 2.4 & 7.72\\\\
\n16 & 4 & 9\\\\
\n49 & 7 & 4.5
\n\\end{vmatrix}
\n=-22.08.
\n\\]<\/p>\n

Therefore,<\/p>\n

\\[
\nc=\\frac{\\det(A_3)}{\\det(A)}=\\frac{-22.08}{-22.08}=1.
\n\\]<\/p>\n

Thus,<\/p>\n

\\[
\na=-0.5,\\qquad b=4,\\qquad c=1.
\n\\]<\/p>\n

We obtain the quadratic interpolation polynomial<\/p>\n

\\[
\nP_2(x)=-0.5\\cdot x^2+4\\cdot x+1.
\n\\]<\/p>\n

For \\( x=5.2 \\), we have<\/p>\n

\\[
\nP_2(5.2)=-0.5\\cdot5.2^2+4\\cdot5.2+1.
\n\\]<\/p>\n

Since<\/p>\n

\\[
\n5.2^2=27.04,
\n\\]<\/p>\n

we obtain<\/p>\n

\\[
\nP_2(5.2)=-0.5\\cdot27.04+20.8+1.
\n\\]<\/p>\n

Then<\/p>\n

\\[
\nP_2(5.2)=-13.52+20.8+1=8.28.
\n\\]<\/p>\n

Therefore,<\/p>\n

\\[
\nf(5.2)\\approx8.28.
\n\\]<\/p>\n

Example 3. Using the given table of function values, find the approximate value of \\( f(1.2) \\)<\/h3>\n\n\n\n\n\n\n\n\n\n
\\( i \\)<\/th>\n\\( x_i \\)<\/th>\n\\( y_i \\)<\/th>\n<\/tr>\n<\/thead>\n
\\( 0 \\)<\/td>\n\\( -3 \\)<\/td>\n\\( 11.5 \\)<\/td>\n<\/tr>\n
\\( 1 \\)<\/td>\n\\( -1 \\)<\/td>\n\\( 4.3 \\)<\/td>\n<\/tr>\n
\\( 2 \\)<\/td>\n\\( 0.5 \\)<\/td>\n\\( 2.05 \\)<\/td>\n<\/tr>\n
\\( 3 \\)<\/td>\n\\( 2 \\)<\/td>\n\\( 2.5 \\)<\/td>\n<\/tr>\n
\\( 4 \\)<\/td>\n\\( 4.5 \\)<\/td>\n\\( 9.25 \\)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

The value \\( x=1.2 \\) lies between the nodes with \\( x \\)-coordinates \\( 0.5 \\) and \\( 2 \\):<\/p>\n

\\[
\n0.5<1.2<2.
\n\\]<\/p>\n

To construct the local quadratic polynomial, we use the nodes<\/p>\n

\\[
\n(x_1,y_1)=(-1,4.3),\\qquad (x_2,y_2)=(0.5,2.05),\\qquad (x_3,y_3)=(2,2.5).
\n\\]<\/p>\n

For the selected nodes, set up the system:<\/p>\n

\\[
\n\\begin{cases}
\na-b+c=4.3,\\\\
\n0.25\\cdot a+0.5\\cdot b+c=2.05,\\\\
\n4\\cdot a+2\\cdot b+c=2.5.
\n\\end{cases}
\n\\]<\/p>\n

Calculate the determinant of the coefficient matrix:<\/p>\n

\\[
\n\\det(A)=
\n\\begin{vmatrix}
\n1 & -1 & 1\\\\
\n0.25 & 0.5 & 1\\\\
\n4 & 2 & 1
\n\\end{vmatrix}
\n=-6.75.
\n\\]<\/p>\n

To find the coefficient \\( a \\), calculate<\/p>\n

\\[
\n\\det(A_1)=
\n\\begin{vmatrix}
\n4.3 & -1 & 1\\\\
\n2.05 & 0.5 & 1\\\\
\n2.5 & 2 & 1
\n\\end{vmatrix}
\n=-4.05.
\n\\]<\/p>\n

Then<\/p>\n

\\[
\na=\\frac{\\det(A_1)}{\\det(A)}=\\frac{-4.05}{-6.75}=0.6.
\n\\]<\/p>\n

To find the coefficient \\( b \\), calculate<\/p>\n

\\[
\n\\det(A_2)=
\n\\begin{vmatrix}
\n1 & 4.3 & 1\\\\
\n0.25 & 2.05 & 1\\\\
\n4 & 2.5 & 1
\n\\end{vmatrix}
\n=8.1.
\n\\]<\/p>\n

Thus,<\/p>\n

\\[
\nb=\\frac{\\det(A_2)}{\\det(A)}=\\frac{8.1}{-6.75}=-1.2.
\n\\]<\/p>\n

To find the coefficient \\( c \\), calculate<\/p>\n

\\[
\n\\det(A_3)=
\n\\begin{vmatrix}
\n1 & -1 & 4.3\\\\
\n0.25 & 0.5 & 2.05\\\\
\n4 & 2 & 2.5
\n\\end{vmatrix}
\n=-16.875.
\n\\]<\/p>\n

Therefore,<\/p>\n

\\[
\nc=\\frac{\\det(A_3)}{\\det(A)}=\\frac{-16.875}{-6.75}=2.5.
\n\\]<\/p>\n

Thus,<\/p>\n

\\[
\na=0.6,\\qquad b=-1.2,\\qquad c=2.5.
\n\\]<\/p>\n

The quadratic interpolation polynomial is<\/p>\n

\\[
\nP_2(x)=0.6\\cdot x^2-1.2\\cdot x+2.5.
\n\\]<\/p>\n

For \\( x=1.2 \\), we obtain<\/p>\n

\\[
\nP_2(1.2)=0.6\\cdot1.2^2-1.2\\cdot1.2+2.5.
\n\\]<\/p>\n

Calculate:<\/p>\n

\\[
\n1.2^2=1.44.
\n\\]<\/p>\n

Then<\/p>\n

\\[
\nP_2(1.2)=0.6\\cdot1.44-1.44+2.5.
\n\\]<\/p>\n

We obtain<\/p>\n

\\[
\nP_2(1.2)=0.864-1.44+2.5=1.924.
\n\\]<\/p>\n

Therefore,<\/p>\n

\\[
\nf(1.2)\\approx1.924.
\n\\]<\/p>\n

The Next Step in Interpolation: Methods for Further Study<\/h2>\n

Quadratic interpolation clearly shows how intermediate function values can be approximated from a few known nodes. Next, it is worth exploring other interpolation methods designed for smooth, periodic, and equally spaced data.<\/p>\n

    \n
  1. Cubic Spline Interpolation: Smooth Approximation of Tabular Data<\/a> \u2014 Learn how cubic splines connect neighboring nodes with smooth curves and help determine intermediate values of a function given in tabular form.<\/li>\n
  2. Trigonometric Interpolation: Approximation of Periodic Functions<\/a> \u2014 Learn how sines and cosines are used to construct interpolation polynomials for approximating functions with periodic behavior.<\/li>\n
  3. Gauss Interpolation Formulas: Calculations Near Central Nodes<\/a> \u2014 Explore the first and second Gauss interpolation formulas and learn how they help determine intermediate values for equally spaced nodes.<\/li>\n<\/ol>\n

    Quadratic Interpolation: From the Algorithm to Your Own Program<\/h2>\n

    If you enjoy programming, try turning what you have learned about quadratic interpolation into your own program. The flowchart below shows an algorithm that takes tabulated values, selects an appropriate group of nodes, constructs the matrices for Cramer’s rule, finds the coefficients of the quadratic polynomial, and calculates the approximate function value at a given point. Implement this algorithm in your favorite programming language and test it with different sets of input data.<\/p>\n

    \"Flowchart<\/p>\n","protected":false},"excerpt":{"rendered":"

    Quadratic interpolation is a numerical method that makes it possible to approximate intermediate function values using three known nodes. Unlike<\/p>\n","protected":false},"author":1,"featured_media":4594,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"template-centered.php","format":"standard","meta":{"footnotes":""},"categories":[574],"tags":[577,576,134,595,594],"class_list":["post-4572","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-function-approximation-methods","tag-function-interpolation","tag-interpolation-polynomial","tag-numerical-methods","tag-piecewise-quadratic-interpolation","tag-quadratic-interpolation"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/posts\/4572","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/comments?post=4572"}],"version-history":[{"count":20,"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/posts\/4572\/revisions"}],"predecessor-version":[{"id":4611,"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/posts\/4572\/revisions\/4611"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/media\/4594"}],"wp:attachment":[{"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/media?parent=4572"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/categories?post=4572"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.mathros.net.ua\/en\/wp-json\/wp\/v2\/tags?post=4572"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}