In the article Quadratic Functions, we infer the properties of quadratic functions based on given equations of the form f(x) = ax2 + 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 (x1,y1) where p ≠ x1 and q ≠ y1. Then, the equation of the parabola can be expressed as:

\frac{y-q}{y_1 - q} = (\frac{x-p}{x_1 - p})^2 …………………………………………………………………………………………………………………………………………….. (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, x1 = 5 and y1 = 2 into (1), we get:

\frac{y-7}{2-7} = (\frac{x-3}{5-3})^2

\frac{y-7}{-5} = \frac{(x-3)^2}{4}

y = - \frac{5}{4} (x - 3)^2 + 7

y = - \frac{5}{4} x^2 + \frac{15}{2} x - \frac{17}{4}

 

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 (x1,0) and (x2,0) where x1 ≠ x2, and 2) It passes through (x0,y0) where x0 ≠ x1 and x0 ≠ x2. Then, the equation of the parabola can be expressed as:

\frac{y}{y_0} = \frac{(x - x_1)(x - x_2)}{(x_0 - x_1 )(x_0 - x_2)} ……………………………………………………………………………………………………………………… (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 x1 = -2, x2 = 6, x0 = 1, and y0 = -3 into (2). This results in:

\frac{y}{-3} = \frac{(x - (-2))(x - 6)}{(1 - (-2))(1 - 6)}

\frac{y}{-3} = \frac{(x + 2)(x - 6)}{3 \cdot (-5)}

y = \frac{1}{5} (x^2 - 4x - 12)

y = \frac{1}{5} x^2 - \frac{4}{5} x - \frac{12}{5}

 

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 x1 = x2 = 1, x0 = -2 and y0 = 5. Substituting these values ​​into (2), we get:

\frac{y}{5} = \frac{(x - 1)^2}{(-2 - 1)^2}

\frac{y}{5} = \frac{(x - 1)^2}{9}

y = \frac{5}{9} (x - 1)^2

y = \frac{5}{9} x^2 - \frac{10}{9} x + \frac{5}{9}

 

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 = ax2 + 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 = ax2 + 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⋅12 + 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) = 2x2 – x.

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