Quadratic Function Modelling: Solving Real-World Problems Mathematically | 二次函数建模：现实问题的数学求解

📚 Quadratic Function Modelling: Solving Real-World Problems Mathematically | 二次函数建模：现实问题的数学求解

Quadratic functions appear everywhere in nature, business and engineering. A quadratic model can describe the path of a thrown ball, the shape of a suspension bridge, or the profit of a company. In this article, we will explore how to translate a real situation into a quadratic equation, interpret its key features, and solve practical problems step by step.

二次函数在自然、商业和工程中无处不在。一个二次模型可以描述投掷球的轨迹、悬索桥的形状，或者一家公司的利润。在本文中，我们将学习如何把现实情境转化为二次方程，并解读其关键特征，从而一步步解决实际问题。

1. What Is Quadratic Modelling? | 什么是二次函数建模？

Quadratic modelling is the process of representing a relationship between two variables with a quadratic equation of the form y = ax² + bx + c, where a, b and c are constants and a ≠ 0. The graph of such an equation is a parabola, which has a distinctive curved shape that either opens upward (a > 0) or downward (a < 0).

二次函数建模，是指用形如 y = ax² + bx + c 的二次方程来表示两个变量之间的关系。其中 a、b、c 是常数，且 a ≠ 0。这种方程的图像是抛物线，具有明显的弯曲形状：当 a > 0 时开口向上，当 a < 0 时开口向下。

The simplest quadratic model is a perfect square such as y = x², but real models always involve coefficients that reflect the units and scale of the problem. For instance, the height h of a ball at time t may be h = −4.9t² + v₀t + h₀, where the coefficient −4.9 comes from half the gravitational acceleration.

最简单的二次模型如 y = x²，但现实模型中的系数总是反映问题的单位和尺度。例如，球在时间 t 的高度可写成 h = −4.9t² + v₀t + h₀，其中系数 −4.9 来自重力加速度的一半。

2. Key Features of a Parabola | 抛物线的重要特征

To use a quadratic model effectively, you must know its four key features: the axis of symmetry, the vertex, the roots, and the y-intercept.

要有效使用二次模型，你必须掌握它的四个关键特征：对称轴、顶点、根（零点）和 y 轴截距。

For a function y = ax² + bx + c, the axis of symmetry is given by x = −b/(2a). The vertex lies on this line, so its x-coordinate is −b/(2a), and its y-coordinate is obtained by substituting this value back into the equation. The roots are the solutions of ax² + bx + c = 0, and the y-intercept is simply c.

对于函数 y = ax² + bx + c，对称轴为 x = −b/(2a)。顶点就在这条直线上，因此其横坐标为 −b/(2a)，纵坐标则把这个值代回原方程得到。根是方程 ax² + bx + c = 0 的解，y 轴截距就是 c。

Vertex: x = −b/(2a), y = f(−b/(2a))

顶点：x = −b/(2a)，y = f(−b/(2a))

The discriminant Δ = b² − 4ac tells us how many real roots exist: if Δ > 0 there are two distinct roots, if Δ = 0 there is one repeated root, and if Δ < 0 there are no real roots.

判别式 Δ = b² − 4ac 告诉我们存在多少个实根：Δ > 0 时有两个不同的根，Δ = 0 时有一个重根，Δ < 0 时没有实根。

3. The Modelling Cycle | 建模的一般流程

Modelling a real problem with a quadratic function usually follows a five-step cycle. This cycle is a standard approach in mathematics and is often tested in exams.

用二次函数给实际问题建模通常遵循五步流程。这个流程是数学中的标准方法，也常常出现在考试中。

• Identify the variables. Decide which quantity is the input x and which is the output y. Make sure the units are consistent.

  确定变量。判断哪个量是输入 x，哪个量是输出 y，并确保单位一致。

• Gather data or conditions. Find fixed values such as a maximum or minimum, or points that the parabola must pass through.

  收集数据或条件。找出定值，例如最大值或最小值，或者抛物线必须经过的某些点。

• Choose the form. Select y = ax² + bx + c, or the vertex form y = a(x − h)² + k, depending on which information is available.

  选择形式。根据给出的信息选择 y = ax² + bx + c，或顶点式 y = a(x − h)² + k。

• Solve for the unknown coefficients. Use the given conditions to set up equations and calculate a, b and c.

  求解未知系数。利用已知条件列方程，算出 a、b 和 c。

• Interpret and validate. Explain the result in the context of the problem and check whether it makes sense.

  解释和验证。在问题的背景下解释结果，并检验它是否合理。

4. Example 1: Projectile Motion | 例1：抛物运动

A ball is thrown vertically upward from a height of 2 m with an initial velocity of 20 m/s. Ignoring air resistance, its height in metres after t seconds is modelled by h = −4.9t² + 20t + 2. Let us find the maximum height and the time at which it occurs.

一个球从离地 2 m 处以 20 m/s 的初速度竖直上抛。忽略空气阻力，t 秒后它的高度（米）可用 h = −4.9t² + 20t + 2 表示。我们来求最大高度及其出现的时间。

Here a = −4.9, b = 20,

Published by TutorHao | Mathematics Revision Series | aleveler.com

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