📚 Mastering Example 1.6.1: A Step-by-Step Approach to A-Level Maths Problems | 掌握例题1.6.1：A-Level数学问题分步解决指南
In A-Level Pure Mathematics, worked examples act as miniature roadmaps. Example 1.6.1 is typically designed to test your understanding of quadratic functions, including substitution, factorisation, completing the square, and curve sketching. In this article, we will dissect a representative problem of this type from start to finish, ensuring you know not just the answer, but the reasoning behind every single step.
在A-Level纯数中，例题就像微型路线图。例题1.6.1通常旨在考查你对二次函数的理解，包括代入求值、因式分解、配方和曲线草图。在本文中，我们将从头到尾剖析一道此类代表性题目，确保你不仅知道答案，更理解每一步背后的逻辑。
1. Understanding the Problem Statement | 理解题目要求
Let us consider the classic Example 1.6.1 style question: Given the quadratic function f(x) = 3x² − 5x + 2, find f(2), solve f(x) = 0, express f(x) in completed square form, and determine the minimum value of f(x). This single question combines several fundamental skills that appear repeatedly across AQA A-Level papers.
我们来看一道典型的例题1.6.1风格问题：已知二次函数 f(x) = 3x² − 5x + 2，求 f(2)，解方程 f(x) = 0，将 f(x) 表示为完全平方形式，并确定 f(x) 的最小值。这一问题综合了多项基本技能，这些技能在AQA A-Level试卷中反复出现。
The first and most important step is to identify exactly what the question asks. Notice that there are four distinct tasks. If you miss one, you lose marks even if the other three are perfect. Always underline key verbs such as “solve”, “find”, “express”, and “determine”.
第一步也是最关键的一步，是明确题目究竟问了什么。注意这里有四项不同的任务。只要你漏掉其中一项，即使另外三项全部正确也会失分。务必圈出“求”“解”“表示”“确定”等关键动词。
2. Direct Substitution: Evaluating f(2) | 直接代入：计算 f(2)
For the first task, we substitute x = 2 into the function. This is called direct substitution. Replacing every occurrence of x with 2 gives f(2) = 3(2)² − 5(2) + 2. We must follow the order of operations: brackets first, then powers, then multiplication, then addition and subtraction.
对于第一项任务，我们将 x = 2 代入函数。这叫作直接代入。把每个 x 都替换为 2，得到 f(2) = 3(2)² − 5(2) + 2。我们必须遵循运算顺序：先算括号，再算幂，然后乘除，最后加减。
f(2) = 3 × 4 − 10 + 2 = 12 − 10 + 2 = 4. Therefore the point (2, 4) lies on the curve. This simple calculation is often used as a warm-up question in exams, but it also establishes a foundation for checking more complex results later.
f(2) = 3 × 4 − 10 + 2 = 12 − 10 + 2 = 4。因此点 (2, 4) 位于这条曲线上。这个简单计算在考试中常作为热身题，但它也为之后检验更复杂的结果奠定了基础。
3. Solving f(x) = 0 by Factorisation | 用因式分解求解 f(x) = 0
To solve f(x) = 0, we factorise the quadratic. We need two numbers that multiply to give 3 × 2 = 6 and add to give −5. These numbers are −2 and −3. Therefore we split the middle term: 3x² − 3x − 2x + 2, which allows us to group and factorise the expression as (3x − 2)(x − 1).
为求解 f(x) = 0，我们对二次式进行因式分解。我们需要找到两个数：相乘为 3 × 2 = 6，相加为 −5。这两个数是 −2 和 −3。因此我们拆分中间项：3x² − 3x − 2x + 2，从而可以进行分组并因式分解为 (3x − 2)(x − 1)。
Setting each factor equal to zero gives 3x − 2 = 0 or x − 1 = 0. Hence x = 2/3 or x = 1. These are the roots of the equation, meaning the curve crosses the x-axis at exactly two points: (2/3, 0) and (1, 0).
令每个因式等于零，得到 3x − 2 = 0 或 x − 1 = 0。因此 x = 2/3 或 x = 1。这就是方程的两个根，意味着曲线与 x 轴恰好相交于两点：(2/3, 0) 和 (1, 0)。
4. Completing the Square | 配方法
Completing the square is a powerful algebraic technique. For f(x) = 3x² − 5x + 2, we first factor out the coefficient of x² from the first two terms: f(x) = 3[x² − (5/3)x] + 2. Inside the bracket, we halve the coefficient of x, which is −5/6, and then square it to obtain 25/36.
配方是一种强大的代数技巧。对于 f(x) = 3x² − 5x + 2，我们首先从前两项中提取 x² 的系数：f(x) = 3[x² − (5/3)x] + 2。在括号内，我们将 x 的系数取半，得到 −5/6，然后平方得到 25/36。
Now we rewrite the bracket as a perfect square minus the square term: f(x) = 3[(x − 5/6)² − 25/36] + 2. Expanding the outer bracket gives f(x) = 3(x − 5/6)² − 75/36 + 2. Simplifying, we obtain f(x) = 3(x − 5/6)² − 25/12 + 24/12 = 3(x − 5/6)² − 1/12.
现在我们把括号改写为“完全平方减去平方项”：f(x) = 3[(x − 5/6)² − 25/36] + 2。展开外层括号得到 f(x) = 3(x − 5/6)² − 75/36 + 2。化简后得到 f(x) = 3(x − 5/6)² − 25/12 + 24/12 = 3(x − 5/6)² − 1/12。
f(x) = 3(x − 5/6)² − 1/12
5. Determining the Minimum Value | 确定最小值
Once the function is in completed square form, the minimum value is easy to read. Since (x − 5/6)² is always non-negative, the smallest possible value of 3(x − 5/6)² is 0, which occurs when x = 5/6. Substituting this into the completed square form gives f(5/6) = 3(0)² − 1/12 = −1/12.
一旦函数写成完全平方形式，最小值就一目了然。由于 (x − 5/6)² 永远非负，因此 3(x − 5/6)² 的最小可能值是 0，此时 x = 5/6。把它代入完全平方形式，得到 f(5/6) = 3(0)² − 1/12 = −1/12。
Thus the minimum point of the quadratic is (5/6, −1/12). Because the coefficient of x² is positive, the parabola opens upwards, confirming that this critical point is indeed a minimum rather than a maximum.
因此该二次函数的最小值点是 (5/6, −1/12)。由于 x² 的系数为正，抛物线开口向上，从而确认这个临界点确实是极小值点而非极大值点。
6. Sketching the Curve | 绘制函数图像
A good sketch of a quadratic function requires four key features: the roots, the y-intercept, the vertex, and the direction of opening. We already have the roots x = 2/3 and x = 1. The y-intercept is found by setting x = 0, which gives f(0) = 2. The vertex is at (5/6, −1/12). Since the leading coefficient is positive, the curve opens upward.
绘制二次函数图像的要点包括四个关键特征：根、y截距、顶点和开口方向。我们已经得到根 x = 2/3 和 x = 1。y截距通过令 x = 0 求得，f(0) = 2。顶点位于 (5/6, −1/12)。由于首项系数为正，曲线开口向上。
When sketching, accuracy matters. The vertex sits between the two roots, which is consistent because the axis of symmetry is x = 5/6, exactly halfway between 2/3 and 1. Label all intercepts clearly and indicate the minimum point. In AQA mark schemes, unlabelled intercepts often lose valuable marks.
绘图时准确性很重要。顶点位于两个根之间，这与对称轴 x = 5/6 恰好位于 2/3 和 1 的正中间这一事实相符。请清楚标出所有截距并指出最小值点。在AQA评分标准中，未标注的截距常常导致宝贵分数丢失。
7. Exploring Inequalities with the Roots | 利用根探究不等式
The roots also allow us to solve inequalities such as f(x) < 0. Since the parabola opens upward, the curve lies below the x-axis between the two roots. Therefore the solution to f(x) < 0 is the interval 2/3 < x < 1.
根还可以帮助我们求解 f(x) < 0 这类不等式。由于抛物线开口向上，曲线在两个根之间位于 x 轴下方。因此 f(x) < 0 的解集是区间 2/3 < x < 1。
Similarly, f(x) ≥ 0 is satisfied when x ≤ 2/3 or x ≥ 1. Many A-Level questions ask for the range of values of x that make a quadratic positive or negative, so this skill should be practised until it becomes automatic.
类似地，f(x) ≥ 0 在 x ≤ 2/3 或 x ≥ 1 时成立。许多A-Level题目会要求求出使二次函数为正或为负的 x 取值范围，因此这项技能必须练习到自动化的程度。
8. Domain and Range | 定义域与值域
In pure mathematics, the domain of this quadratic function is all real numbers, since there are no denominators, square roots, or logarithms that impose restrictions. We write the domain as x ∈ ℝ, meaning x can be any real number.
在纯数中，这个二次函数的定义域是所有实数，因为表达式中没有分母、平方根或对数带来限制。我们把定义域写作 x ∈ ℝ，意思是 x 可以是任意实数。
The range, however, is restricted by the minimum value. Since the lowest output is −1/12 and the curve rises indefinitely, the range is f(x) ≥ −1/12. Written in set notation, this is { f(x) : f(x) ≥ −1/12 }.
但值域受到最小值的限制。由于最低输出值为 −1/12，且曲线无限上升，因此值域为 f(x) ≥ −1/12。用集合记号表示为 { f(x) : f(x) ≥ −1/12 }。
9. Verifying with the Discriminant | 用判别式进行验证
For a quadratic ax² + bx + c, the discriminant is Δ = b² − 4ac. In our case, a = 3, b = −5, and c = 2, so Δ = (−5)² − 4 × 3 × 2 = 25 − 24 = 1. Because Δ > 0, the equation has two distinct real roots, which matches our factorisation result exactly.
对于二次式 ax² + bx + c，判别式为 Δ = b² − 4ac。在本例中，a = 3，b = −5，c = 2，因此 Δ = (−5)² − 4 × 3 × 2 = 25 − 24 = 1。因为 Δ > 0，方程有两个不同的实数根，这与我们的因式分解结果完全一致。
Scholars often make the mistake of thinking that a perfect square discriminant is needed. In fact, any positive discriminant, even 0.0001, guarantees two distinct roots. Here Δ = 1, which is small but positive, explaining why the two roots 2/3 and 1 are very close together on the x-axis.
学生们常常误以为判别式必须是完全平方数。事实上，任何正的判别式，即使只有0.0001，也能保证有两个不同的根。这里 Δ = 1，虽然小但为正，这解释了为什么两个根 2/3 和 1 在 x 轴上距离很近。
10. Connecting Substitution and Graph | 联系代入法与图像
The earlier substitution f(2) = 4 can now be interpreted graphically. The point (2, 4) lies above the x-axis and to the right of the larger root x = 1, which is consistent with our inequality result that f(x) > 0 for x > 1.
之前的代入结果 f(2) = 4 现在可以从图像上理解。点 (2, 4) 位于 x 轴上方，并且处于较大根 x = 1 的右侧，这与我们关于 x > 1 时 f(x) > 0 的不等式结论一致。
This shows the beauty of pure mathematics: algebra and geometry tell the same story. By checking a single point like x = 2, you can confirm whether your sketch and inequality solutions are correct. Always perform this simple verification when you have time at the end of an exam.
这体现了纯数之美：代数与几何讲述着同一个故事。通过检查像 x = 2 这样的一个点，你就能确认图像和不等式解是否正确。在考试结束前如有时间，务必进行这种简单的验证。
11. Common Mistakes to Avoid | 需要避免的常见错误
One frequent error in completing the square is forgetting to multiply the square term by the coefficient 3. If you write 3(x − 5/6)² − 25/36, you have effectively ignored the factor of 3 outside the bracket. Always expand your completed square form to check that it matches the original quadratic.
配方中最常见的错误是忘记将平方项乘以系数 3。如果你写成 3(x − 5/6)² − 25/36，实际上就忽略了括号外的系数 3。始终展开你的完全平方形式，检查它是否与原二次式一致。
Another common error is sign confusion when substituting negative values into the quadratic. Remember that (−5)² is 25, not −25. Also, when solving f(x) = 0, never divide both sides by the factor (x − 1) unless you first note that x = 1 is a valid root; otherwise you may lose the other root 2/3.
另一个常见错误是代入负值时出现符号混淆。记住 (−5)² 等于 25，而不是 −25。另外，在解 f(x) = 0 时，不要直接两边同除以因式 (x − 1)，除非你首先注明 x = 1 是有效根；否则你可能会丢失另一个根 2/3。
12. Wider Problem-Solving Strategies | 更广泛的解题策略
Example 1.6.1 is not just about one quadratic function. It demonstrates a general four-step method: read the question carefully, choose the correct algebraic technique, perform the calculation with careful attention to signs, and verify your result by a different method. This cycle works for algebraic fractions, inequalities, and even calculus problems.
例题1.6.1不仅关乎某一个二次函数。它展示了一个通用的四步方法：仔细读题，选择正确的代数技巧，谨慎地关注符号进行运算，并用不同方法验证结果。这一循环适用于代数分式、不等式乃至微积分问题。
AQA A-Level mathematics rewards method and clarity. Even if the final answer is wrong, a clear method of substitution and factorisation may earn you method marks. Write each step neatly, keep equations aligned, and never hide important algebra in your head. In this way, Example 1.6.1 becomes a springboard for tackling harder questions with confidence.
AQA A-Level数学重视方法与清晰度。即使最终答案错误，清晰的代入法和因式分解过程也可能会为你赢得方法分。整齐地写出每一步，保持方程对齐，永远不要把重要的代数步骤藏在脑子里。这样，例题1.6.1就能成为你自信攻克更难题目的跳板。
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