📚 Modelling with Quadratics | 二次函数建模
Modelling with quadratic functions is a core topic in Edexcel A-Level Mathematics. It applies parabolic curves to represent real-world phenomena such as projectile paths, areas, and economic profit. Mastering this area requires understanding the different forms of quadratics, solving equations, interpreting the discriminant, and using the vertex to optimise quantities. This article provides a comprehensive revision guide with bilingual explanations to help you excel in modelling problems.
二次函数建模是 Edexcel A-Level 数学的核心主题之一。它利用抛物线来表示现实世界中的现象，如抛体轨迹、面积和经济利润。掌握这一领域需要理解二次函数的各种形式、解方程、解读判别式，并利用顶点进行最优化。本文提供全面的复习指南，并配以双语解释，帮助你攻克建模类题目。
1. Quadratic Functions and Their Forms | 二次函数及其形式
ach form reveals different properties directly. The standard form gives the y-intercept c; the vertex form gives the turning point (h, k); the factored form gives the roots p and q if a ≠ 0. The coefficient a determines the direction of opening: if a > 0 the parabola opens upwards (∪), if a < 0 it opens downwards (∩).
二次函数主要有三种形式：一般式 y = ax² + bx + c、顶点式 y = a(x − h)² + k 和因式分解式 y = a(x − p)(x − q)。每种形式直接展示不同的性质。一般式给出 y 轴截距 c；顶点式给出顶点 (h, k)；因式分解式给出根 p 和 q（a ≠ 0）。系数 a 决定开口方向：a > 0 时抛物线开口向上 (∪)，a < 0 时开口向下 (∩)。
| Form | Expression | 直接可见特征 |
|---|---|---|
| Standard / 一般式 | y = ax² + bx + c | y-intercept (0, c) |
| Vertex / 顶点式 | y = a(x − h)² + k | Vertex (h, k) |
| Factored / 因式分解式 | y = a(x − p)(x − q) | Roots p, q |
his reveals the vertex (2, −3) and the minimum value −3 directly. Choosing the right form simplifies the modelling process.
将一般式化为顶点式可进行配方。例如 y = 2x² − 8x + 5 可化为 y = 2(x − 2)² − 3，直接得到顶点 (2, −3) 和最小值 −3。选择合适的形式将简化建模过程。
2. Key Features of a Parabola | 抛物线的关键特征
pt, and the roots (x-intercepts) if they exist. The vertex can be found by completing the square or using the formula x = −b/(2a) and then substituting back. The axis of symmetry is the vertical line through the vertex. The discriminant Δ = b² − 4ac tells whether the parabola crosses the x-axis.
每个二次模型都具有抛物线图形。关键特征包括：顶点（最大值或最小值点）、对称轴 x = h、y 轴截距，以及可能存在的根（x 轴截距）。顶点可通过配方法或使用公式 x = −b/(2a) 并代回求得。对称轴是经过顶点的垂直线。判别式 Δ = b² − 4ac 表明抛物线是否与 x 轴相交。
the domain is often restricted (e.g. x ≥ 0), so some mathematical features might be irrelevant. Always check which part of the parabola is valid for the real-world situation.
确定 y 轴截距很简单：令 x = 0 得 y = c。若根存在，则通过解 ax² + bx + c = 0 求得。在建模背景中，定义域常受限（如 x ≥ 0），因此某些数学特征可能与实际无关。务必检查抛物线的哪一部分对现实情况有效。
3. Solving Quadratic Equations | 解二次方程
s: factorising (if the expression factorises nicely), completing the square, and the quadratic formula. The formula is given by:
为求二次模型等于特定值（通常为零）的位置，我们需要解 ax² + bx + c = 0。主要有三种方法：因式分解（如果表达式易于分解）、配方法以及求根公式。公式如下：
x = (−b ± √b² − 4ac) / (2a)
ble, only t ≥ 0 is acceptable. If a problem asks for the time when a ball hits the ground, solve h(t) = 0, take the positive root. Similarly, when finding dimensions, reject negative lengths.
在建模中，常常需要结合实际背景来解读解，舍弃负数或不合理的值。例如，若变量是时间 t，则只有 t ≥ 0 可接受。若题目要求小球落地的时间，解 h(t) = 0，取正根。类似地，求尺寸时需舍去负长度。
4. The Discriminant and Modelling Constraints | 判别式与建模约束
angent), Δ < 0 → no real roots. In modelling, the discriminant helps check feasibility. For example, if a projectile' certain height, the object never reaches that height. This gives a constraint on the model's parameters.
判别式 Δ = b² − 4ac 决定根的性质：Δ > 0 有两个不等实根，Δ = 0 有一个重根（相切），Δ < 0 无实根。在建模中，判别式可用于检验可行性。例如，若某抛体高度模型对某目标高度算得 Δ < 0，则物体永远达不到该高度，从而给模型参数带来约束。
Exam questions may ask: 'Prove that this object never reaches 15 m.' You would substitute the target height into the equation, rearrange to a quadratic, and show the discriminant is negative. This technique is especially common in connected particles or mechanics problems involving quadratics.
考试题目可能要求：'证明该物体永远达不到 15 米。' 将目标高度代入方程，整理成二次方程，并证明其判别式为负。这类技巧在涉及二次函数的连接体或力学问题中尤为常见。
5. Maximising and Minimising in Models | 建模中的最大化与最小化
m; if a < 0, it has a maximum. For y = ax² + bx + c, the x-coordinate of the vertex is x = −b/(2a) and the extreme value is f(−b/2a). Completing the square to the form y = a(x − h)² + k directly gives the optimum value k. This is essential in area, revenue and cost models.
许多建模问题要求求二次函数的最大值或最小值。最值出现在顶点处。若 a > 0，函数有最小值；若 a < 0，函数有最大值。对 y = ax² + bx + c，顶点横坐标为 x = −b/(2a)，极值为 f(−b/2a)。配成 y = a(x − h)² + k 的形式可直接给出最值 k。这在面积、收益和成本模型中至关重要。
When finding the maximum profit or minimum cost, differentiate if you have studied calculus, but completing the square is a purely algebraic method that always works. Always verify whether the vertex lies within the practical domain. For instance, if x represents items produced, x must be a non-negative integer, so round sensibly.
求最大利润或最小成本时，若已学微积分可求导，但配方法是纯代数方法，始终可行。务必验证顶点是否在实际定义域内。例如，若 x 代表产量，x 必须为非负整数，因此要合理取整。
6. Modelling Projectile Motion | 抛体运动建模
e to gravity (approx. 9.8 m/s²), u is initial vertical velocity, and h₀ is initial height. The negative t² term reflects downward curvature. Typical questions: maximum height, time of flight, time to reach a certain height. Solve h(t)=0 for landing time.
经典应用之一是抛体运动。高度 h 作为时间 t 的函数可建模为 h(t) = −½ g t² + u t + h₀，其中 g 为重力加速度（约 9.8 m/s²），u 是初速度的竖直分量，h₀ 是初始高度。负的 t² 项反映开口向下。常见问题：最大高度、飞行时间、达到某高度的时间。解 h(t)=0 得到落地时间。
t − 5t² (taking g = 10). This simplifies to h = −5(t² − 4t) = −5(t − 2)² + 20, giving max height 20 m at t = 2 s. Always state units clearly.
求最大高度时，用 t = −u / (2 × (−½ g)) = u/g，代回即可。此外，题目可能给出如 h = 20t − 5t² 的高度函数（取 g = 10）。可化为 h = −5(t² − 4t) = −5(t − 2)² + 20，得最大高度 20 米，发生在 t = 2 秒时。务必清晰注明单位。
7. Area Optimisation Problems | 面积优化问题
the wall be y and the other two sides be x. The total fence used is y + 2x = L, so y = L − 2x. The area A = x·y = x(L − 2x) = −2x² + Lx. This is a quadratic in x with negative a, giving a maximum area at x = L/4. Such problems require expressing one variable in terms of another and forming a quadratic expression.
典型的建模问题：一位农夫有固定长度的围栏，想靠墙围出一块矩形区域。设与墙平行的边长为 y，另外两边长为 x。所用围栏总长为 y + 2x = L，故 y = L − 2x。面积 A = x·y = x(L − 2x) = −2x² + Lx。这是关于 x 的二次函数，a 为负，最大值在 x = L/4 处取得。此类问题需使用一个变量表示另一个，并构造二次表达式。
at both dimensions are positive. This method extends to paddocks divided into sections, often leading to quadratic models with multiple variables that reduce to one.
求得最优 x 后，计算 y = L − 2(L/4) = L/2，最大面积为 A = (L/4)(L/2) = L²/8。需检查两维数均为正。此方法可推广至分区围场，常涉及多个变量但可简化为单一变量的二次模型。
8. Revenue, Cost and Profit Modelling | 收入、成本与利润建模
ion p = a − bx, revenue R(x) = xp = ax − bx² is a quadratic. Fixed and variable costs yield cost functions C(x), and profit P(x) = R(x) − C(x) often is quadratic. Finding the maximum profit involves locating the vertex of this quadratic.
在经济学中，收入 R(x) 和利润 P(x) 可由二次函数建模。例如，若单价由线性需求函数 p = a − bx 给出，则收入 R(x) = xp = ax − bx² 为二次函数。固定和可变成本构成成本函数 C(x)，利润 P(x) = R(x) − C(x) 通常为二次。求最大利润需要找到此二次函数的顶点。
For instance, a company sells x items at price £ + 200) = −0.5x² + 35x − 200. Completing the square gives maximum profit at x = 35 and P = £412.50. Such problems test algebraic manipulation and interpretation.
例如，某公司以单价 £(40 − 0.5x) 销售 x 件产品，收入 R = 40x − 0.5x²。若总成本为 C = 5x + 200，则利润 P = −0.5x² + 35x − 200。配方可得 x = 35 时利润最大，为 £412.50。此类问题考查代数处理能力和解读。
9. Finding Unknown Parameters from Data | 根据数据求未知参数
Often a quadratic model is given with
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