In the article Quadratic Functions, we infer the properties of quadratic functions based on given equations of the form f(x) = ax^{2} + bx + c where a ≠ 0. In the opposite direction, this article describes how to find the equations of quadratic functions with some given properties. The techniques vary depending on the properties given.

**Case 1: Given the coordinates of the vertex and another point through which the parabola passes**

Suppose that the coordinates of the vertex of a parabola are (p,q) and the parabola passes through (x_{1},y_{1}) where p ≠ x_{1} and q ≠ y_{1}. Then, the equation of the parabola can be expressed as:

…………………………………………………………………………………………………………………………………………….. (1)

**Example 1**

A parabola passes through (5,2) and its vertex is at (3,7). Find the equation of the parabola.

**Answer**

Substituting p = 3, q = 7, x_{1} = 5 and y_{1} = 2 into (1), we get:

**Case 2: Given the coordinates of the point(s) of intersection with the x-axis and another point through which the parabola passes**

Suppose that a parabola has the following properties: 1) It intersects the x-axis at (x_{1},0) and (x_{2},0) where x_{1} ≠ x_{2}, and 2) It passes through (x_{0},y_{0}) where x_{0} ≠ x_{1} and x_{0} ≠ x_{2}. Then, the equation of the parabola can be expressed as:

……………………………………………………………………………………………………………………… (2)

**Example 2**

Find the quadratic function whose graph passes through (1,-3) and intersects the x-axis at the points (-2,0) and (6.0).

**Answer**

Substitute x_{1} = -2, x_{2} = 6, x_{0} = 1, and y_{0} = -3 into (2). This results in:

**Example 3**

Find the quadratic function whose graph passes through (-2.5) and touches the x-axis at (1.0).

**Answer**

In this example, it can be concluded that the parabola intersects the x-axis at exactly one point, namely at (1,0). Thus, in this case x_{1} = x_{2} = 1, x_{0} = -2 and y_{0} = 5. Substituting these values into (2), we get:

**Case 3: Given the coordinates of three points through which the parabola passes**

In this case, the search for the equation starts with assuming that the equation of the parabola is y = ax^{2} + bx + c with a ≠ 0. Then, substitute the abscissae and ordinates of the coordinates of the three points into the equation. The values of a, b, and c can determined by solving the system of linear equations as a result of these substitutions.

**Example 4**

Find the quadratic function whose graph passes through the points A(1,1), B(-2,10), and C(-1,3).

**Answer**

Let the equation of the parabola be y = ax^{2} + bx + c where a ≠ 0. Substitute the abscissae and ordinates of the coordinates of A, B, and C into the equation. This yields.

A(1,1): 1 = a⋅1^{2} + b⋅1 + c ⇔ a + b + c = 1 …………………………………………………….. (*)

B(-2,10): 10 = a⋅(-2)^{2} + b⋅(-2) + c ⇔ 4a – 2b + c = 10 …………………………………………………….. (**)

C(-1,3): 3 = a⋅(-1)^{2} + b⋅(-1) + c ⇔ a – b + c = 3 …………………………………………………….. (***)

It is easy to check that a = 2, b = -1, and c = 0 satisfy (*), (**), and (***) simultaneously. Therefore, the desired quadratic function has the equation f(x) = 2x^{2} – x.