📚 Quadratic Function Modelling: Real-World Problems and Parabolas | 二次函数建模：实际问题与抛物线
Quadratic functions are among the most flexible tools in A-Level mathematics. Whenever a real quantity has a maximum or minimum, or when changes follow a symmetric curved pattern, a quadratic model can describe it usefully.
二次函数是 A-Level 数学中适应性最强的工具之一。当某个实际问题存在最大值或最小值，或变化规律呈现对称的曲线形态时，二次函数常常能成为一个简洁而有效的数学模型。
1. Why Quadratic Functions Model Real-World Situations | 为什么二次函数能建模现实问题
Many physical and economic situations involve a quantity that initially increases or decreases and then turns around. The graph of a quadratic function is a parabola, and a parabola has exactly one turning point, making it ideal for describing phenomena with a single maximum or minimum.
许多物理与经济情境都涉及一个先上升或先下降、随后发生转折的量。二次函数的图像是抛物线，而抛物线恰好只有一个转向点，因此非常适合描述带有单一最大值或最小值的现象。
For example, the height of a thrown object under gravity follows a quadratic expression in time. Similarly, profit, revenue, and costs can sometimes be written as quadratic functions of the selling price.
例如，物体在重力作用下被抛出时，其高度是关于时间的二次表达式。类似地，利润、收入与成本有时也能写成关于售价的二次函数。
The mathematical power comes from the fact that a quadratic function is simple enough to solve by algebra, yet rich enough to model acceleration, curvature, and symmetric shapes.
二次函数的数学优势在于：它足够简单，可以用代数方法方便求解；又足够丰富，可以描述加速度、弯曲程度和对称图形。
2. Key Features: Standard Form, Completed Square, Factorised Form | 关键特征：一般式、配方法与因式分解式
Every quadratic function can be written in at least three useful forms. The choice of form depends on the question being asked.
每一个二次函数至少可以写成三种有用的形式，具体采用哪种形式取决于题目要求我们求什么。
| Form | Equation | Most Useful For |
| General form | y = ax² + bx + c | reading y-intercept c |
| Completed square | y = a(x − p)² + q | identifying vertex (p, q) |
| Factorised form | y = a(x − r₁)(x − r₂) | reading x-intercepts r₁ and r₂ |
The completed square form is especially important for modelling because the turning point can be read directly. If a > 0, the parabola opens upwards and q is the minimum value.
配方法形式在建模中尤其重要，因为转向点可以直接读出。若 a > 0，抛物线开口向上，q 就是最小值。
If a < 0, the parabola opens downwards and q is the maximum value. In real problems we then interpret q as the largest height, maximum profit, or greatest possible output.
若 a < 0，抛物线开口向下，q 就是最大值。在实际问题中，我们可把 q 理解为最大高度、最大利润或最大产出。
3. Projectile Motion: Modelling the Path of a Thrown Ball | 抛体运动：为扔出的球建立抛物线模型
When air resistance is ignored, the height h metres of a ball thrown vertically with initial speed u m/s from height s metres is modelled by
在忽略空气阻力时，以初速度 u m/s 从高度 s 米竖直上抛的球，其高度 h 米可由下式建模：
h(t) = −5t² + ut + s
where t is time in seconds and the coefficient −5 is half of the approximate value of g = 10 m/s².
其中 t 为以秒为单位的时间，系数 −5 是 g ≈ 10 m/s² 的一半。
For example, if a ball is thrown upwards from 1.5 m with speed 12 m/s, the model is
例如，若一个球从 1.5 m 的高度以 12 m/s 的速度竖直上抛，则模型为
h(t) = −5t² + 12t + 1.5
The equation is quadratic in t, so its graph is a downward-opening parabola. This tells us that the ball first rises, reaches one highest point, and then falls back down.
该式关于 t 是二次的，因此其图像是一条开口向下的抛物线。这说明球先上升，达到唯一最高点后，再下落。
4. Finding Maximum Height and Time of Flight | 求最大高度与飞行时间
For the model h(t) = −5t² + 12t + 1.5, we can complete the square to find when the maximum height occurs.
对于模型 h(t) = −5t² + 12t + 1.5，我们可以通过配方法求出最大高度出现的时刻。
h(t) = −5(t − 1.2)² + 8.7
Therefore the maximum height is 8.7 m, reached at t = 1.2 seconds.
因此最大高度为 8.7 m，在 t = 1.2 秒时达到。
To find the time of flight, solve h(t) = 0. Applying the quadratic formula gives
若要求飞行总时间，则解 h(t) = 0。使用求根公式可得
t = (−12 ± √(144 + 30)) ÷ (−10) = (−12 ± √174) ÷ (−10)
The positive root is approximately t ≈ 2.52 seconds. The negative root has no physical meaning and is rejected.
正根约为 t ≈ 2.52 秒。负根没有物理意义，应舍去。
This example shows that a quadratic model does more than draw a curve; it answers practical questions such as “how high?” and “how long?”
这个例子说明二次模型不仅能画出一条曲线，还能回答”有多高”和”有多久”等实际问题。
5. Parabolic Arches and Engineering Structures | 抛物线拱桥与工程结构中的二次模型
Arch bridges and suspension cables often have shapes close to parabolas under idealised load conditions. An engineer can describe the arch by a quadratic equation and then calculate heights at different positions.
拱桥和悬索在理想荷载条件下，形状常接近抛物线。工程师可以用二次方程描述拱形，再计算不同位置处的高度。
Suppose an arch is 6 m high and 12 m wide, with its vertex at the origin. If its two ends meet the ground at x = −6 and x = 6, one possible model is
假设一座拱桥高 6 m、宽 12 m，顶点在原点。若两端分别落在地面 x = −6 和 x = 6 处，一个可行的模型是
y = 6 − (x² ÷ 6)
At x = 3, the height is y = 6 − 9/6 = 4.5 m. This is exactly the kind of calculation a structural engineer might use.
在 x = 3 处，高度为 y = 6 − 9/6 = 4.5 m。这正是结构工程师可能使用的计算类型。
Because the parabola is symmetric, checking values on one side automatically gives values on the other side. This symmetry is why quadratic models are so convenient in design work.
由于抛物线具有对称性，检查一侧的数值即可自动得到另一侧的数值。这种对称性正是二次模型在设计中如此方便的原因。
6. Fitting a Quadratic Model from Experimental Data | 由实验数据拟合二次模型
Sometimes we do not have a formula from theory; we only have data points. If the data clearly change direction once, we can fit a quadratic of the form y = ax² + bx + c.
有时我们并没有理论公式，只有若干数据点。如果数据明显只转折一次，我们便可以拟合形如 y = ax² + bx + c 的二次函数。
For example, suppose the points (0, 1), (1, 2) and (2, 7) are known. Substituting them gives three equations:
例如，已知点 (0, 1)、(1, 2) 和 (2, 7)。代入后得到三个方程：
c = 1, a + b + c = 2, 4a + 2b + c = 7
Solving these simultaneous equations gives a = 2, b = −1 and c = 1, so the model is
解这个方程组得到 a = 2，b = −1，c = 1，因此模型为
y = 2x² − x + 1
In Edexcel A-Level problems, the data are often chosen to give integer coefficients, but the same method works for any three distinct x-values.
在 Edexcel A-Level 的题目中，数据通常被设计成可以得到整数系数，但无论取哪三个不同的 x 值，方法都是一样的。
7. Profit, Revenue and Optimisation Problems | 利润、收入与优化问题
Business problems are a common source of quadratic models. Revenue is price × quantity; if quantity demanded depends linearly on price, then revenue becomes quadratic.
商业问题是二次模型的常见来源。收入 = 价格 × 数量；若需求量与价格呈线性关系，那么收入就会变成二次函数。
Suppose demand is q = 300 − 5p, where p is the price per item. Revenue is
设需求量为 q = 300 − 5p，其中 p 为单价。则收入为
R(p) = p(300 − 5p) = 300p − 5p²
Completing the square gives
配方得到
R(p) = −5(p − 30)² + 4500
So maximum revenue is 4500, reached when the price is 30. The quadratic tells us immediately that charging too much lowers revenue because demand falls.
因此最大收入为 4500，在价格为 30 时取得。二次函数立刻告诉我们：定价过高会使需求量下降，从而减少收入。
This kind of optimisation is central to economics and business studies, and the vertex is the key answer the examiner expects.
这类最优化问题是经济学与商科的核心内容，而顶点正是阅卷者期望考生给出的关键答案。
8. Stopping Distance: Speed vs Distance | 停车距离：速度与距离的二次关系
Typical highway research shows that stopping distance can be modelled as a quadratic function of speed. A simplified model is
典型公路研究表明，停车距离可以表示为车速的二次函数。一个简化模型为
d(v) = 0.05v² + 0.2v
where v is speed in m/s and d is distance in metres.
其中 v 的单位是 m/s，d 的单位是米。
At v = 20 m/s, the stopping distance is
当 v = 20 m/s 时，停车距离为
d(20) = 0.05 × 400 + 0.2 × 20 = 24 m
The quadratic term becomes more important as speed increases. Doubling speed does not simply double the distance; it more than doubles it. This is a crucial real-world insight from the quadratic model.
当速度增加时，二次项变得更加重要。速度加倍并不会让距离刚好加倍，而是会使其增加更多。这是二次模型带来的一个重要现实结论。
9. Using the Discriminant to Test Whether the Model Can Reach a Value | 用判别式判断模型是否能达到某个数值
In a practical problem, we often ask whether a model can reach a particular value, such as a height of 25 m. This is equivalent to asking whether the related quadratic equation has real roots.
在实际问题中，我们常问模型能否达到某个特定值，例如能否达到高度 25 m。这等价于判断相应二次方程是否有实数根。
For h(t) = −5t² + 20t + 2, to test whether the ball can reach 20 m, solve
对于 h(t) = −5t² + 20t + 2，若要检验球能否达到 20 m，解方程
−5t² + 20t + 2 = 20 ⇒ −5t² + 20t − 18 = 0
The discriminant is 20² − 4 × 5 × 18 = 400 − 360 = 40, which is positive. Therefore the ball reaches 20 m at two different times.
判别式为 20² − 4 × 5 × 18 = 400 − 360 = 40，大于零。因此球会在两个不同时刻到达 20 m。
If we test 25 m, the equation becomes −5t² + 20t − 23 = 0 and the discriminant is 400 − 460 = −60, which is negative. The model says 25 m is physically impossible.
若检验 25 m，方程变为 −5t² + 20t − 23 = 0，判别式为 400 − 460 = −60，小于零。该模型说明 25 m 在物理上不可能达到。
This shows how algebra connects to reality: a negative discriminant means the curve never reaches that height.
这说明代数与现实之间的连接：判别式为负，意味着曲线永远不会触及该高度。
10. Domain Restrictions and Model Validity | 定义域限制与模型的有效范围
A quadratic function is defined for all real x, but the real-world situation usually restricts the domain. Time, length, price and distance are normally non-negative.
二次函数对一切实数 x 都有定义，但现实情境通常会限制定义域。时间、长度、价格和距离一般都不能取负数。
For a ball, the model h(t) is only meaningful for t between 0 and the time when it hits the ground. For an arch, x is only meaningful between the two supports.
对于抛出的球，模型 h(t) 只在 t 从 0 到落地时刻之间有意义。对于拱桥，x 只在两个支座之间有实际意义。
Examiners reward students who state this explicitly. Writing “the domain is t ≥ 0” may seem small, but it shows mathematical maturity.
阅卷者会奖励那些明确写出这一点的考生。写上”定义域为 t ≥ 0″看似简单，却能体现数学成熟度。
Extrapolation beyond the valid domain can also be misleading. A quadratic trend observed in one range should not be trusted far outside that range.
在有效范围之外进行外推也可能产生误导。在某个范围内观察到的二次趋势，不应在远离该范围处轻易相信。
11. Identifying Quadratic Relationships from Tables | 从数据表中识别二次关系
When a table of values is given, a constant second difference signals a quadratic relationship. Consider the values of y = x² for x = 0, 1, 2, 3:
当给出一组数值表时，二阶差分恒定就表明存在二次关系。考虑 y = x² 在 x = 0、1、2、3 处的取值：
| x | 0 | 1 | 2 | 3 |
| y | 0 | 1 | 4 | 9 |
| First difference | − | 1 | 3 | 5 |
| Second difference | − | − | 2 | 2 |
The first differences are 1, 3, 5, and the second difference is always 2. A constant second difference tells us the data are perfectly quadratic.
一阶差分为 1、3、5，二阶差分恒为 2。二阶差分恒定说明数据是完全二次的。
In real data, small errors will make the second differences almost constant. Recognising this pattern is a useful diagnostic skill.
在真实数据中，小误差会让二阶差分几乎恒定。识别这种模式是一项实用的判断技能。
12. Common Pitfalls and Exam Tips | 常见错误与考试心得
The most common error is using the wrong sign in the completed square form. For y = −5t² + 12t + 1.5, students sometimes write +5(t − 1.2)² and lose the maximum value.
最常见的错误是配方法中符号出错。例如 y = −5t² + 12t + 1.5，有些学生会误写成 +5(t − 1.2)²，从而失去最大值。
Another common mistake is forgetting that a < 0 means a maximum, while a > 0 means a minimum. Always check the coefficient of x² before interpreting the vertex.
另一个常见错误是忘记 a < 0 对应最大值，而 a > 0 对应最小值。在解读顶点之前，务必先检查 x² 的系数。
Always answer the practical question with units and a contextual sentence. If t is time, write t = 2.52 seconds, not just 2.52.
回答实际问题时，要带上单位并结合情境作解释。如果 t 是时间，应写 t = 2.52 秒，而不是只写 2.52。
When solving equations that come from a model, check whether negative roots should be rejected. A negative length, time or price usually has no physical meaning.
当解出由模型得到的方程时，要检查负根是否应舍去。负的长度、时间或价格通常没有物理意义。
Finally, practise converting between the three forms quickly. In an Edexcel exam, completing the square is often the shortest route to the maximum or minimum value.
最后，要练习在三种形式之间快速转换。在 Edexcel 考试中，配方法通常是求最大值或最小值的最短路径。
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