📚 Modelling with Quadratics | 二次函数建模
Quadratic modelling is one of the most practical topics in the Edexcel A-Level Mathematics specification. It shows how a function of the form y = ax² + bx + c can describe real situations such as projectile motion, area, revenue and profit. In the exam, you are expected to form a quadratic model, interpret its features, and use the vertex or roots to answer contextual questions.
二次函数建模是 Edexcel A-Level 数学大纲中最实用的主题之一。它展示形如 y = ax² + bx + c 的函数如何描述抛体运动、面积、收入和利润等实际情况。在考试中，你需要建立二次模型、解释其特性，并利用顶点或根来回答情境问题。
1. What is quadratic modelling? | 什么是二次函数建模？
A quadratic model uses a function of degree two to represent a real-world relationship between two variables. The general form is:
二次模型使用二次函数来表示现实世界中两个变量之间的关系。一般形式为：
y = ax² + bx + c, a ≠ 0
In modelling, the variable x is often replaced by t for time, n for number of items, or another context letter. The quadratic term can represent acceleration, area, or decreasing returns. The model is a simplified version of reality, so you must always check whether the answers make sense in the original context.
在建模中，变量 x 经常被换成表示时间的 t、表示物品数量的 n 或其他情境字母。二次项可以表示加速度、面积或收益递减。模型是对现实情况的简化，因此你必须始终检查答案在原情境中是否合理。
2. Key features of a quadratic model | 二次模型的关键特征
Every quadratic model has a parabolic graph. The shape opens upwards if a > 0 and opens downwards if a < 0. This tells you whether the model has a minimum or a maximum value.
每个二次模型的图像都是一条抛物线。如果 a > 0，开口向上；如果 a < 0，开口向下。这告诉你模型是有最小值还是最大值。
The line of symmetry passes through the vertex. For y = ax² + bx + c, the x-coordinate of the vertex is given by:
对称轴通过顶点。对于 y = ax² + bx + c，顶点的 x 坐标由下式给出：
x = −b / (2a)
Other important features include the y-intercept at (0, c), the roots where y = 0, and the vertex itself, which represents the maximum or minimum value of the model. In context, the vertex often answers questions about maximum height, maximum profit or minimum cost.
其他重要特征包括 y 轴截距 (0, c)、当 y = 0 时的根，以及顶点本身，它表示模型的最大值或最小值。在情境中，顶点通常用于回答关于最大高度、最大利润或最小成本的问题。
3. Standard forms and converting between them | 标准形式及其转换
A quadratic model can be written in three main forms, each useful for a different purpose. The standard form is y = ax² + bx + c, the factored form is y = a(x − p)(x − q), and the vertex form is y = a(x − h)² + k.
二次模型可以写成三种主要形式，每种形式适用于不同目的。标准形式为 y = ax² + bx + c，因式分解形式为 y = a(x − p)(x − q)，顶点形式为 y = a(x − h)² + k。
Completing the square converts the standard form into the vertex form. For example, y = 2x² − 8x + 5 becomes:
配方法可以将标准形式转换为顶点形式。例如，y = 2x² − 8x + 5 可化为：
y = 2(x − 2)² − 3
From this form, the vertex is immediately visible as (2, −3). The table below summarises the three forms:
从这个形式可以直接看出顶点为 (2, −3)。下表总结了三种形式：
| Form | Equation | Most useful for |
|---|---|---|
| Standard form | 标准形式 | y = ax² + bx + c | Identifying y-intercept and using the discriminant | 识别 y 截距并使用判别式 |
| Factored form | 因式形式 | y = a(x − p)(x − q) | Reading roots directly | 直接读出根 |
| Vertex form | 顶点形式 | y = a(x − h)² + k | Finding maximum or minimum value | 求最大值或最小值 |
4. Finding the vertex and maximum/minimum values | 求顶点与最大/最小值
To find the maximum or minimum output of a quadratic model, you can complete the square or use the vertex formula. The x-coordinate of the vertex is x = −b / (2a), and the corresponding y-value is found by substitution.
要求二次模型的最大或最小输出，你可以使用配方法或顶点公式。顶点的 x 坐标为 x = −b / (2a)，对应的 y 值通过代入求得。
For a projectile height model h = −5t² + 20t + 15, the vertex occurs at:
对于抛体高度模型 h = −5t² + 20t + 15，顶点出现在：
t = −20 / (2 × −5) = 2
Substituting t = 2 gives h = −5(2)² + 20(2) + 15 = 35. Since a = −5 is negative, the parabola opens downwards, so the maximum height is 35 metres at t = 2 seconds.
代入 t = 2 得 h = −5(2)² + 20(2) + 15 = 35。因为 a = −5 为负，抛物线开口向下，所以最大高度为 35 米，发生在 t = 2 秒时。
5. Solving quadratic equations in context | 在情境中解二次方程
Solving a quadratic equation in context gives the input values where the model equals a target value, often zero. For example, setting height equal to zero finds when a projectile hits the ground.
在情境中解二次方程可得到模型等于目标值（通常为零）时的输入值。例如，令高度等于零可求出抛体何时落地。
Using h = −5t² + 20t + 15, set h = 0:
使用 h = −5t² + 20t + 15，令 h = 0：
−5t² + 20t + 15 = 0
Divide by −5 and solve:
除以 −5 并求解：
t² − 4t − 3 = 0
t = [4 ± √(16 + 12)] / 2 = 2 ± √7
This gives t ≈ 4.65 or t ≈ −0.65. Since time cannot be negative, the projectile hits the ground after about 4.65 seconds.
解得 t ≈ 4.65 或 t ≈ −0.65。由于时间不能为负，抛体约在 4.65 秒后落地。
6. The discriminant in modelling | 判别式在建模中的应用
The discriminant Δ = b² − 4ac tells you how many real solutions a quadratic equation has. In modelling, it shows whether a target value is reached and how many times.
判别式 Δ = b² − 4ac 告诉你二次方程有多少个实数解。在建模中，它可以显示目标值是否达到以及达到多少次。
Δ = b² − 4ac
If Δ > 0, the model reaches the target twice. If Δ = 0, it reaches the target exactly once. If Δ < 0, the model never reaches that value.
如果 Δ > 0，模型两次达到目标值。如果 Δ = 0，模型恰好一次达到目标值。如果 Δ < 0，模型从未达到该值。
For example, does the projectile reach 40 m? Set h = 40:
例如，抛体能否达到 40 米？令 h = 40：
−5t² + 20t + 15 = 40
−5t² + 20t − 25 = 0
t² − 4t + 5 = 0
Here Δ = (−4)² − 4(1)(5) = 16 − 20 = −4. Since Δ < 0, the projectile never reaches 40 metres.
这里 Δ = (−4)² − 4(1)(5) = 16 − 20 = −4。由于 Δ < 0，抛体从未达到 40 米。
7. Projectile motion and height models | 抛体运动与高度模型
Projectile motion under constant gravitational acceleration produces a quadratic height-time model. A common form is:
在恒定重力加速度下的抛体运动产生二次的高度-时间模型。常见形式为：
h(t) = −½gt² + ut + h₀
Here g is the acceleration due to gravity, u is the initial vertical velocity, and h₀ is the initial height. In Edexcel questions, you are often given a simplified equation such as h = −5t² + 20t + 15, where g is taken as 10 m/s².
这里 g 是重力加速度，u 是初始竖直速度，h₀ 是初始高度。在 Edexcel 题目中，通常会给出简化后的方程，如 h = −5t² + 20t + 15，其中 g 取 10 m/s²。
Remember that the t-intercept h = 0 gives the time at ground level, while the vertex gives the maximum height. The domain t ≥ 0 must always be considered because negative time is not physically meaningful.
请记住，t 轴截距 h = 0 给出落地时间，而顶点给出最大高度。必须始终考虑定义域 t ≥ 0，因为负时间在物理上没有意义。
8. Revenue, cost and profit models | 收入、成本与利润模型
Quadratic models also arise in economics. If the price per item decreases linearly with quantity, revenue is a quadratic function of the number of items sold.
二次模型也出现在经济学中。如果每件商品的价格随数量线性下降，收入就是销售数量的二次函数。
Suppose the demand price is p = 40 − 2x, where x is the number of items sold. Revenue is:
假设需求价格为 p = 40 − 2x，其中 x 是销售数量。收入为：
R = xp = x(40 − 2x) = 40x − 2x²
The revenue-maximising quantity is found from the vertex:
使收入最大化的数量由顶点求得：
x = −40 / (2 × −2) = 10
Substituting x = 10 gives maximum revenue R = 40(10) − 2(10)² = 200. If a cost function C = 4x + 50 is introduced, profit becomes P = R − C = −2x² + 36x − 50, which is also quadratic and can be maximised at x = 9.
代入 x = 10 得最大收入 R = 40(10) − 2(10)² = 200。如果引入成本函数 C = 4x + 50，利润变为 P = R − C = −2x² + 36x − 50，它也是二次函数，并在 x = 9 处取得最大值。
9. Geometric modelling with quadratics | 几何建模
Geometric problems with fixed perimeter often lead to quadratic area models. Suppose a rectangular enclosure uses 100 m of fencing for all four sides. If one side is x, the other is (100 − 2x) / 2 = 50 − x.
固定周长的几何问题常常引出二次面积模型。假设一个矩形围栏用 100 米围栏围成四条边。如果一条边为 x，另一边为 (100 − 2x) / 2 = 50 − x。
The area is:
面积为：
A = x(50 − x) = 50x − x²
This is a downward-opening quadratic, so the maximum area occurs at x = −50 / (2 × −1) = 25. The maximum area is A = 25 × 25 = 625 m². In such problems, the domain is usually 0 < x < 50, because both side lengths must be positive.
这是一个开口向下的二次函数，因此最大面积出现在 x = −50 / (2 × −1) = 25。最大面积为 A = 25 × 25 = 625 平方米。在此类问题中，定义域通常为 0 < x < 50，因为两条边长都必须为正。
10. Setting up equations from real-life constraints | 从实际约束建立方程
To build a quadratic model from a word problem, follow these steps:
从文字题建立二次模型，可按以下步骤进行：
- Identify the quantity to be maximised or minimised, such as area, height, revenue or profit.
- 明确需要最大化或最小化的量，例如面积、高度、收入或利润。
- Choose a variable for the unknown input and express all other quantities in terms of that variable.
- 为未知输入选择一个变量，并用该变量表示所有其他量。
- Use a diagram, formula or given relationship to form the quadratic equation.
- 使用图形、公式或已知关系建立二次方程。
- State the domain of the model based on physical or economic constraints.
- 根据物理或经济约束说明模型的定义域。
- Find the vertex or roots as required, then interpret the results in context.
- 按需要求顶点或根，然后在情境中解释结果。
11. Interpreting roots and domain restrictions | 解释根与定义域限制
Not every solution of a quadratic is valid in the original context. You must reject values that are impossible, such as negative time, negative length, or quantities greater than the feasible range.
并非二次方程的每个解在原情境中都有效。你必须排除不可能的值，例如负时间、负长度或大于可行范围的量。
For example, if a profit model P = −2x² + 36x − 50 has two positive roots, only the first positive root represents the break-even point before profit starts to fall. The other root may still be positive but occurs after the maximum and may be outside the practical range of production.
例如，如果利润模型 P = −2x² + 36x − 50 有两个正根，则只有第一个正根代表利润开始下降前的盈亏平衡点。另一个根可能仍为正，但出现在最大值之后，并且可能超出实际生产范围。
Always restate the domain clearly, such as t ≥ 0, 0 ≤ x ≤ 50, or x being an integer number of items. Domain restrictions often change which solutions are acceptable.
始终清楚地重新说明定义域，例如 t ≥ 0、0 ≤ x ≤ 50 或 x 为整数件数。定义域限制通常会改变哪些解是可接受的。
12. Common mistakes and exam tips | 常见错误与考试技巧
Common mistakes in quadratic modelling include forgetting to reject negative answers, losing units in the final statement, and using the vertex formula with the wrong sign. Another frequent error is solving for x but not substituting back to find the actual maximum or minimum value.
二次建模中的常见错误包括忘记排除负答案、在最终结论中遗漏单位，以及顶点公式符号使用错误。另一个常见错误是求出了 x，但没有代回求出实际的最大值或最小值。
Exam tips:
考试技巧：
- Write down the model before calculating, and label the variables clearly.
- 在计算前先写下模型，并清楚地标注变量。
- Always state whether the vertex is a maximum or minimum by referring to the sign of a.
- 始终通过说明 a 的符号来指出顶点是最大值还是最小值。
- Check the discriminant if the question asks whether a certain value is reached.
- 如果题目问是否达到某个值，请检查判别
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