📚 Exercise 6D.3: Quadratic Function Modelling | 二次函数建模练习
In IB Mathematics, Exercise 6D.3 typically challenges students to model real-world situations using quadratic functions, extract key features like the vertex, intercepts, and interpret the results within a given context. Mastering such exercises builds a strong bridge between abstract algebraic manipulation and practical problem solving.
在 IB 数学中，练习 6D.3 通常会让学生运用二次函数对现实情境进行建模，提取顶点、截距等关键特征，并在给定语境下解释结果。掌握这类练习能在抽象的代数运算与实际解决问题之间架起坚实的桥梁。
1. Understanding the Problem Context | 理解问题背景
Before writing any equation, read the wording carefully to identify what the variables represent. In many quadratic modelling questions, the independent variable (often x) stands for time, horizontal distance, or quantity produced, while the dependent variable (y) represents height, profit, revenue, or cost. Understanding the context helps you decide whether you need a maximum, minimum, or a specific value of the function.
在写下任何方程之前，仔细阅读题干，确定变量所代表的含义。在许多二次建模问题中，自变量（通常是 x）表示时间、水平距离或产量，而因变量（y）表示高度、利润、收入或成本。理解背景有助于你判断是需要最大值、最小值还是函数的某个特定值。
- n object over time, with a < 0 indicating a downward opening parabola and a maximum point.
- 如果问题涉及抛射体，y = ax² + bx + c 可能表示物体随时间变化的高度，此时 a < 0 表明抛物线开口向下，存在最大值。
- If the problem relates to profit or cost, the quadratic may model the profit as a function of the number of items sold; you will likely be asked to find the maximum profit or the break‑even points.
- 如果问题与利润或成本有关，二次函数可能表示利润与销售数量的关系；通常会要求找出最大利润或盈亏平衡点。
2. Identifying the Quadratic Model | 识别二次模型
construct it from a table of values or from a description involving initial conditions and rates of change. Look for clues such as "the height is given by h = -5t² + 20t + 1" or "the profit P(x) = -2x² + 120x - 800".
二次模型的一般形式是 y = ax² + bx + c，其中 a、b、c 为常数且 a ≠ 0。有些题目直接给出方程；另一些则需要你根据数值表格或包含初始条件及变化率的描述来建立模型。注意寻找诸如"高度由 h = -5t² + 20t + 1 给出"或"利润 P(x) = -2x² + 120x - 800"等线索。
(2, -1) and (5, 8), substitute each point into y = ax² + bx + c and solve simultaneously.
如果给出三个点，你可以建立三个方程构成的方程组来求解 a、b、c。例如，当抛物线经过 (0, 3)、(2, -1) 和 (5, 8) 时，将每个点代入 y = ax² + bx + c 并联立求解。
3. Finding the Vertex (Maximum or Minimum) | 求顶点（最大值或最小值）
The vertex of a quadratic function y = ax² + bx + c occurs at x = -b/(2a). This x‑ a > 0). Once you have x, substitute it back into the equation to find the corresponding y‑e. In contextual problems, this vertex often answers the most critical practical question.
二次函数 y = ax² + bx + c 的顶点出现在 x = -b/(2a) 处。该 x 坐标给出了最大值（当 a < 0）或最小值（当 a > 0）的位置。求得 x 后，将其代回方程即可求出相应的 y 值。在情境问题中，这个顶点通常回答了最关键的实践问题。
Vertex: x = -b / (2a), y = a(-b/(2a))² + b(-b/(2a)) + c
For example, if a company' )² + 120(30) - 800 = 1000. Always state the units, e.g., 30 items give a maximum profit of $1000.
例如，若某公司的利润模型为 P(x) = -2x² + 120x - 800，那么 a = -2，b = 120。最大利润出现在 x = -120 / (2 × -2) = 30 处，最大利润为 P(30) = -2(30)² + 120(30) - 800 = 1000。务必注明单位，例如 30 件产品带来最大利润 1000 美元。
4. Solving for x-intercepts (Roots/Zeros) | 求解 x 轴截距（根/零点）
X-intercepts are the points where y = 0. In real‑ mes when a projectile hits the ground or the production levels where profit is zero (break‑ and solve using factorisation, completing the square, or the quadratic formula: x = [-b ± √b² - 4ac] / (2a).
X 轴截距是使 y = 0 的点。在现实模型中，它们可以表示抛射体落地的时刻或利润为零（盈亏平衡点）的产量。将二次函数设为零，然后通过因式分解、配方法或求根公式求解：x = [-b ± √b² - 4ac] / (2a)。
x = [-b ± √b² - 4ac] / (2a)
Be careful: sometimes the negative root may not make sense in context (e.g., negative time) and should be discarded, but you should still compute it and explain why you reject it. If the discriminant b² - 4ac is negative, there are no real x-intercepts, meaning the model never reaches zero within the given domain.
注意：有时负根在实际情境中无意义（例如时间为负），应予以舍弃，但你仍需计算出该根并说明舍弃的理由。如果判别式 b² - 4ac 为负，则不存在实根，这意味着模型在给定定义域内永远不会达到零。
5. Solving for the y‑intercept | 求解 y 轴截距
The y‑ epresents the initial value: the starting height, initial cost, or profit when no items are sold. Always interpret the y‑intercept with a short sentence that ties back to the scenario.
y 轴截距出现在 x = 0 处。它就是方程 y = ax² + bx + c 中的常数项 c。在情境中，它通常表示初始值：初始高度、初始成本或未售出任何产品时的利润。始终用一句简短的话将 y 轴截距与情境联系起来。
For example, if a bridge arch is modelled by h = -0.02x² + 1.2x + 3, the y‑ s 3, meaning the arch starts at a height of 3 metres above the reference point when x = 0.
例如，若桥拱的模型为 h = -0.02x² + 1.2x + 3，则 y 轴截距为 3，表示当 x = 0 时，拱在参考点上方 3 米处开始。
6. Axis of Symmetry and Its Practical Meaning | 对称轴及其实际意义
l line x = -b/(2a) that passes through the vertex. It divides the parabola into two mirror‑ ymmetric about t = 2, so the time to go up equals the time to come down to the same level.
对称轴是通过顶点且垂直于 x 轴的直线 x = -b/(2a)，它将抛物线分成镜像对称的两半。在许多建模问题中，对称轴可以帮助你找到缺失的坐标，或验证某一情境在时间或距离上是否对称。例如，如果球在 t = 2 秒时达到最大高度，则抛物线关于 t = 2 对称，因此上升和下降到同一高度的时间相等。
You can use the axis of symmetry to find a second x‑ the axis is x = 4, the other root is x = 7, because they are equally spaced around x = 4.
你可以利用对称轴在已知一个根的情况下求出另一个 x 轴截距。如果一根为 x = 1，对称轴为 x = 4，则另一根为 x = 7，因为它们关于 x = 4 对称排列。
7. Domain and Range in Context | 情境中的定义域与值域
s the ground. For a profit function, x (number of items) cannot be negative. Identifying the sensible domain allows you to state a relevant range. The range then describes the possible output values (height, profit, etc.) within the scenario.
尽管二次函数通常的定义域是所有实数，但实际模型往往会限制定义域。对于抛射体，定义域可能是从 t = 0 到它落地的时刻。对于利润函数，x（产品数量）不能为负。确定合理的定义域后，你就可以给出相应的值域。值域描述了在该情境下可能的输出值（高度、利润等）。
the initial height to the maximum height (21 metres) and then down to 0, i.e., 0 ≤ h ≤ 21.
例如，若 h = -5t² + 20t + 1 表示球在 t = 0 到落地时间 t = 4.05 之间的高度，定义域为 0 ≤ t ≤ 4.05，值域从初始高度到最大高度（21 米）再降到 0，即 0 ≤ h ≤ 21。
8. Sketching the Quadratic Graph | 绘制二次函数图像
A clear sketch is a powerful tool for checking your work and communicating results. Start by marking the y‑intercept, then plot the x‑intercepts (if they exist) and the vertex. Draw a smooth curve through these points. Always label the axes with the appropriate variables and units, and indicate the vertex coordinates. Use a dashed line for the axis of symmetry if it helps visualisation.
一个清晰的草图是检查计算和传达结果的有力工具。首先标出 y 轴截距，然后描出 x 轴截距（如果存在）和顶点。用光滑的曲线连接各点。始终用适当的变量和单位标注坐标轴，并标出顶点坐标。如果有助于可视化，可以用虚线画出对称轴。
Many marks in IB are awarded for correct shape: if a < 0, the parabola must open downward; if a > 0, it opens upward. Do not rely on a perfect scale, but make sure the relative positions of the vertex and intercepts are plausible. A quick sketch also helps you spot silly mistakes, like a negative root where a positive one is required.
IB 评分中许多分数都对应正确的形状：如果 a < 0，抛物线必须开口向下；如果 a > 0，则开口向上。不必追求精确的比例，但要确保顶点和截距的相对位置合理。快速的草图还有助于你发现低级错误，比如在需要正根的地方出现了负根。
9. Interpreting Results in Context | 在上下文中解释结果
Finding a numerical answer is only half the battle; you must also explain what that number means in the given situation. If you calculate a maximum, state it as "the maximum height is 21 metres, reached after 2 seconds". If you find x‑intercepts, explain them as "the company breaks even when 10 and 40 units are sold". Always answer in full sentences that refer back to the scenario's variables.
求出数值只是成功的一半，你还必须解释这个数字在给定情境中的含义。如果你算出了最大值，要表述为"最大高度为 21 米，在 2 秒后达到"。如果求出了 x 轴截距，要解释为"公司在售出 10 件和 40 件产品时达到盈亏平衡"。始终用完整的句子作答，并回扣情境中的变量。
If the question asks "Is it possible to reach a profit of $1200?" he resulting quadratic has no real roots, state that the target cannot be reached, and perhaps use the vertex to confirm the maximum possible profit.
如果题目问"是否可能达到 1200 美元的利润？"，你可以设 P(x) = 1200 并求解。如果由此得到的二次方程无实根，则说明无法达到该目标，并可借助顶点来确认最大可能利润。
10. Using the Discriminant to Predict Outcomes | 利用判别式预测结果
w many real solutions a quadratic equation has. In modelling, it can answer questions like "Does the projectile reach a height of 10 m?" Δ = 0, the height is reached exactly at the vertex. If Δ < 0, the height is never reached.
判别式 Δ = b² - 4ac 可以告诉你二次方程有多少个实数解。在建模中，它可以在不完全解方程的情况下回答诸如"抛射体能达到 10 米的高度吗？"之类的问题。如果 Δ > 0，则有两个实数解，表明该高度达到了两次（一次上升、一次下降）。如果 Δ = 0，则该高度恰好在顶点处达到。如果 Δ < 0，则永远达不到该高度。
so the ball never reaches 25 m. Such reasoning saves time and shows deeper understanding.
这种方法极其高效。例如，已知 h = -5t² + 20t + 2，要检查球是否能达到 25 米，可设 -5t² + 20t + 2 = 25，整理得 -5t² + 20t - 23 = 0。判别式为 (20)² - 4(-5)(-23) = 400 - 460 = -60 < 0，因此球永远达不到 25 米。这样的推理既节省时间，又能展示更深层次的理解。
11. Transforming Between Quadratic Forms | 不同二次型之间的转换
² + k, where (h, k) is the vertex. Completing the square transforms standard form into vertex form. In Exercise 6D.3, you might be asked to write a given quadratic in this form to read off the maximum or minimum directly. This is particularly helpful when the coefficients do not yield neat factorisation.
将模型转换为顶点式 y = a(x - h)² + k（其中 (h, k) 为顶点）后，分析往往更加简便。通过配方法可将标准式化为顶点式。在练习 6D.3 中，你可能会被要求将给定的二次函数写成此形式，以便直接读出最大值或最小值。当系数无法进行简洁的因式分解时，这一方法尤为有用。
5 = 2(x - 2)² - 8 + 5 = 2(x - 2)² - 3. The vertex is (2, -3) and the minimum value is -3.
示例：将 y = 2x² - 8x + 5 转换为顶点式。从含 x 的项中提取 2：y = 2(x² - 4x) + 5。在括号内配平方：x² - 4x 化为 (x - 2)² - 4。于是 y = 2[(x - 2)² - 4] + 5 = 2(x - 2)² - 8 + 5 = 2(x - 2)² - 3。顶点为 (2, -3)，最小值为 -3。
12. Worked Example with Step‑by‑Step Solution | 分步解答的完整示例
Let us work through a typical question similar to those in Exercise 6D.3: " P(x) = -x² + 18x - 72, where x is the number of hundreds of units produced. Find the break‑n production levels, the maximum profit, and the range of production that yields a profit."
让我们完整解答一道类似于练习 6D.3 的典型题目："某公司发现其每周利润 P（千美元）由 P(x) = -x² + 18x - 72 给出，其中 x 是生产的百件数。求盈亏平衡产量、最大利润以及能产生利润的产量范围。"
Step 1 – Break‑even: Set P = 0: -x² + 18x - 72 = 0 → Multiply by -1: x² - 18x + 72 = 0. Solve: x = [18 ± √324 - 288] / 2 = [18 ± √36 = 12 or x = 6. Interpretation: The company breaks even when producing 600 and 1200 units.
第 1 步 – 盈亏平衡：令 P = 0：-x² + 18x - 72 = 0 → 两边乘以 -1：x² - 18x + 72 = 0。解得 x = [18 ± √324 - 288] / 2 = [18 ± √36] / 2 = [18 ± 6] / 2。因此 x = 12 或 x = 6。解释：公司生产 600 件和 1200 件产品时达到盈亏平衡。
Step 2 – Maximum profit: -18 / (2 × -1) = 9. P(9) = -(9)² + 18(9) - 72 = -81 + 162 - 72 = 9. The maximum profit is $9000 when 900 units are produced.
第 2 步 – 最大利润：顶点在 x = -b/(2a) = -18 / (2 × -1) = 9 处。P(9) = -(9)² + 18(9) - 72 = -81 + 162 - 72 = 9。当生产 900 件产品时，最大利润为 9000 美元。
Step 3 – Range of production for a profit: Profit is positive when P(x) > 0, which occurs between the roots: 6 < x < 12. So the company must produce between 600 and 1200 units to make a positive profit. Always state the domain: x is in hundreds of units, so 600 < units < 1200.
第 3 步 – 产生利润的产量范围：当 P(x) > 0 时利润为正，这发生在两根之间：6 < x < 12。因此公司必须生产 600 到 1200 件产品才能获得正利润。务必注明定义域：x 代表百件数，所以 600 < 件数 < 1200。
This structured approach, with clear reasoning and contextual interpretation, mirrors what IB examiners expect in questions linked to Exercise 6D.3. Practise similar problems until the steps become automatic.
这种结构化的解题方法，配合清晰的推理和情境解释，正是 IB 考官在与练习 6D.3 相关的题目中所期望的。反复练习类似问题，直到这些步骤变得自动。
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