IGCSE OCR Maths: Quadratic Functions – Exam Focus | IGCSE OCR 数学：二次函数考点精讲

📚 IGCSE OCR Maths: Quadratic Functions – Exam Focus | IGCSE OCR 数学：二次函数考点精讲

Quadratic functions form one of the largest and most important topics in the IGCSE OCR Mathematics syllabus. They appear in almost every exam paper, from straightforward factorising tasks to challenging graph-sketching and problem-solving questions. This article breaks down every key skill you need, linking concepts to typical OCR-style questions and giving you a clear pathway from basic definitions to high-mark applications. Whether you are aiming for a solid pass or a top grade, mastering quadratics is essential.

二次函数是 IGCSE OCR 数学大纲中篇幅最长、最重要的主题之一。它几乎出现在每份考卷中，从简单的因式分解到复杂的图像绘制和实际应用题。本文将拆解你需要掌握的每一个关键技能，将概念与典型的 OCR 风格题目联系起来，为你提供一条从基础定义到高分应用的清晰路径。无论你的目标是稳过还是高分，掌握二次函数都至关重要。

1. What is a Quadratic Function? | 什么是二次函数？

A quadratic function is any function of the form f(x) = ax² + bx + c, where a, b and c are constants and a ≠ 0. The term ‘quadratic’ comes from the Latin word for square, because the highest power of x is 2. In IGCSE OCR, you will see these functions written as y = ax² + bx + c, and you will be asked to work with them in many ways.

二次函数是形如 f(x) = ax² + bx + c 的函数，其中 a、b 和 c 为常数且 a ≠ 0。’二次’一词源于拉丁语中的’平方’，因为 x 的最高次数是 2。在 IGCSE OCR 考试中，你会看到这类函数写成 y = ax² + bx + c 的形式，并且需要以多种方式处理它们。

The value of a determines the shape and direction of the curve called a parabola. If a > 0, the parabola opens upwards (a ‘smile’ shape); if a < 0, it opens downwards (a 'frown' shape). The constant c gives the y-intercept, because when x = 0, y = c. The term bx affects the position of the line of symmetry and the vertex.

系数 a 决定了这条曲线（抛物线）的形状和开口方向。若 a > 0，抛物线开口向上（呈 ‘微笑’ 形状）；若 a < 0，则开口向下（呈 '皱眉' 形状）。常数 c 代表了 y 轴截距，因为当 x = 0 时 y = c。bx 这一项影响对称轴和顶点的位置。

You must recognise the general form, factorised form y = a(x – p)(x – q) and completed square form y = a(x + h)² + k. Each form reveals different information and is used for different question types.

你必须能够识别一般式、因式分解式 y = a(x – p)(x – q) 和配方式 y = a(x + h)² + k。每种形式揭示不同的信息，用于不同类型的题目。

2. Expanding and Simplifying Quadratics | 展开与化简二次式

Expanding brackets is a fundamental skill. You will often need to multiply two linear expressions, such as (x + 3)(x – 5). In OCR exams, you must show full working, usually by applying the distributive property (FOIL: First, Outer, Inner, Last) or a grid method.

展开括号是一项基本技能。你经常需要将两个线性表达式相乘，例如 (x + 3)(x – 5)。在 OCR 考试中，你必须展示完整的步骤，通常使用分配律（首外内尾法）或网格法。

Example: Expand (x + 2)(x – 7). Multiply First terms: x × x = x². Outer: x × (-7) = -7x. Inner: 2 × x = 2x. Last: 2 × (-7) = -14. Then simplify: x² – 7x + 2x – 14 = x² – 5x – 14. Be careful with signs, especially when subtracting negative numbers.

示例：展开 (x + 2)(x – 7)。计算首项：x × x = x²。外项：x × (-7) = -7x。内项：2 × x = 2x。尾项：2 × (-7) = -14。然后化简：x² – 7x + 2x – 14 = x² – 5x – 14。注意正负号，特别是在减去负数时要格外小心。

OCR also expects you to expand squared brackets, like (3x – 4)². Write it as (3x – 4)(3x – 4) and expand: 9x² – 12x – 12x + 16 = 9x² – 24x + 16. A common mistake is to simply square each term individually, which is incorrect. Always use the full expansion technique.

OCR 还要求展开平方括号，例如 (3x – 4)²。将其写成 (3x – 4)(3x – 4) 并展开：9x² – 12x – 12x + 16 = 9x² – 24x + 16。一个常见的错误是简单地对每一项分别平方，这是不正确的。一定要使用完全的展开方法。

3. Factorising Quadratic Expressions | 因式分解二次式

Factorising is the reverse of expanding. You need to rewrite ax² + bx + c as a product of two binomials. For simple quadratics where a = 1, find two numbers that multiply to c and add to b. For example, factorise x² + 5x + 6. The numbers 2 and 3 multiply to 6 and add to 5, so the factors are (x + 2)(x + 3).

因式分解是展开的逆运算。你需要将 ax² + bx + c 重写为两个二项式的乘积。对于 a = 1 的简单二次式，找到两个数，使它们的乘积为 c、和为 b。例如，对 x² + 5x + 6 进行因式分解。数字 2 和 3 的乘积是 6、和是 5，因此因式为 (x + 2)(x + 3)。

When a > 1, use the ‘ac method’ or grouping. For 2x² + 7x + 3, multiply a and c: 2 × 3 = 6. Find two numbers that multiply to 6 and add to 7: 6 and 1. Rewrite: 2x² + 6x + x + 3. Group: 2x(x + 3) + 1(x + 3) = (x + 3)(2x + 1). Practice this method until it feels automatic; OCR marks schemes require clear evidence of method.

当 a > 1 时，使用 ‘ac 法’ 或分组法。对于 2x² + 7x + 3，将 a 与 c 相乘：2 × 3 = 6。找到两个数，其乘积为 6、和为 7：6 和 1。重写：2x² + 6x + x + 3。分组：2x(x + 3) + 1(x + 3) = (x + 3)(2x + 1)。通过练习使该方法变得熟练；OCR 评分方案要求展示清晰的方法步骤。

Always look for a common factor first. If all terms share a factor, take it out before factorising the quadratic part. For example, 3x² + 9x + 6 = 3(x² + 3x + 2) = 3(x + 1)(x + 2). Skipping this step can make factorising much harder and may lose method marks.

务必首先寻找公因式。如果所有项都有共同的因子，先将其提出再分解二次部分。例如，3x² + 9x + 6 = 3(x² + 3x + 2) = 3(x + 1)(x + 2)。跳过这一步会使因式分解困难得多，并可能丢失方法分。

4. Solving Quadratics by Factorising | 通过因式分解解二次方程

Once a quadratic expression is factorised, you can solve the equation by setting each bracket equal to zero. This relies on the fact that if a product is zero, at least one factor must be zero. For example, solve (x – 3)(x + 5) = 0. Either x – 3 = 0 ⇒ x = 3, or x + 5 = 0 ⇒ x = –5.

一旦将二次表达式分解因式，就可以通过令每个括号等于零来解方程。这依据的原理是：若乘积为零，则至少有一个因式为零。例如，解 (x – 3)(x + 5) = 0。要么 x – 3 = 0 ⇒ x = 3，要么 x + 5 = 0 ⇒ x = –5。

OCR questions often require rearranging the equation first. Do not forget to set the equation to zero before factorising. For instance, x² + 4x = 5 should be rearranged to x² + 4x – 5 = 0, then factorised to (x + 5)(x – 1) = 0, giving x = –5 and x = 1. A very common mistake is to factorise without moving all terms to one side – always check the equation equals zero.

OCR 的题目通常要求先对方程进行重新排列。不要忘记在因式分解前将方程设为零。例如，x² + 4x = 5 应重新排列为 x² + 4x – 5 = 0，然后因式分解为 (x + 5)(x – 1) = 0，得到 x = –5 和 x = 1。一个非常常见的错误是不把各项移到同一边就直接因式分解——务必检查方程是否等于零。

Some quadratic equations may have only one solution (a repeated root) when the factor is squared, e.g., (x – 2)² = 0 gives x = 2 only. Remember to state ‘x = 2 (repeated)’ if asked for all solutions.

有些二次方程可能只有一个解（重根），当因式为平方形式时，例如 (x – 2)² = 0 仅给出 x = 2。如果要求写出所有解，记得标明 ‘x = 2（重根）’。

5. The Quadratic Formula | 二次求根公式

When a quadratic cannot be factorised easily, or when the question explicitly asks for exact solutions or answers to a given number of decimal places, you must use the quadratic formula. The formula is derived from completing the square and is given on the OCR formula sheet: if ax² + bx + c = 0, then

当二次式不容易因式分解，或题目明确要求精确解或保留指定位数的小数时，你必须使用二次求根公式。该公式由配方法推导而来，并在 OCR 公式表中给出：若 ax² + bx + c = 0，则

x = [-b ± √(b² – 4ac)] / (2a)

You must first identify a, b and c correctly, paying special attention to negative signs. Substitute them into the formula with brackets to avoid calculator errors. For example, 2x² – 3x – 4 = 0 has a = 2, b = –3, c = –4.

你必须首先正确地确定 a、b 和 c，尤其注意负号。用括号将它们代入公式，以避免计算器错误。例如，2x² – 3x – 4 = 0 中 a = 2, b = –3, c = –4。

x = [-(-3) ± √((-3)² – 4 × 2 × (-4))] / (2 × 2) = [3 ± √(9 + 32)] / 4 = [3 ± √41] / 4. This gives two irrational solutions. If required, round to e.g. 3 significant figures or 2 decimal places as instructed. Always show the step before rounding to gain method marks.

x = [-(-3) ± √((-3)² – 4 × 2 × (-4))] / (2 × 2) = [3 ± √(9 + 32)] / 4 = [3 ± √41] / 4。这给出两个无理数解。如有要求，按指示四舍五入到例如 3 位有效数字或 2 位小数。务必展示四舍五入前的步骤以获取方法分。

If the question states ‘giving your answers in exact form’, you must leave the surd √(b² – 4ac) in the answer. Do not approximate unless asked. OCR frequently mixes requests for exact answers with decimal approximations in the same question.

如果题目要求 ‘以精确形式给出答案’，你必须将根号 √(b² – 4ac) 保留在答案中。除非有要求，否则不要取近似值。OCR 经常在同一题中混合要求精确解和小数近似解。

6. Completing the Square | 配方法

Completing the square is a powerful technique that rewrites a quadratic into the form a(x + h)² + k. This form instantly gives the coordinates of the vertex and is used to solve equations, find maximum or minimum values, or sketch graphs. The OCR specification places strong emphasis on this method.

配方法是一种强大的技巧，它将二次式重写为 a(x + h)² + k 的形式。该形式能直接给出顶点坐标，并用于解方程、求最大值或最小值以及绘制图像。OCR 大纲非常重视这种方法。

For a simple quadratic x² + bx + c, take half of b, square it, and adjust the constant term. For x² + 6x + 5, half of 6 is 3, square is 9. Write (x + 3)² – 9 + 5 = (x + 3)² – 4. The vertex is (–3, –4). Remember the sign change inside the bracket: x² + 6x becomes (x + 3)².

对于简单的二次式 x² + bx + c，取 b 的一半、将其平方，并调整常数项。对于 x² + 6x + 5，6 的一半是 3，平方为 9。写成 (x + 3)² – 9 + 5 = (x + 3)² – 4。顶点为 (–3, –4)。记住括号内的符号变化：x² + 6x 变成 (x + 3)²。

When the coefficient of x² is not 1, factor it out first from the x terms. For 2x² – 8x + 3, take out the 2: 2[x² – 4x] + 3. Then complete the square on x² – 4x: half of –4 is –2, square 4, so (x – 2)² – 4. Put back: 2[(x – 2)² – 4] + 3 = 2(x – 2)² – 8 + 3 = 2(x – 2)² – 5. The vertex is (2, –5).

当 x² 的系数不为 1 时，首先将它从含 x 的项中提出来。对于 2x² – 8x + 3，提出 2：2[x² – 4x] + 3。然后对 x² – 4x 进行配方：–4 的一半是 –2，平方为 4，因此 (x – 2)² – 4。代入：2[(x – 2)² – 4] + 3 = 2(x – 2)² – 8 + 3 = 2(x – 2)² – 5。顶点为 (2, –5)。

OCR often asks ‘Write x² + 10x + 9 in the form (x + p)² + q’ or ‘Hence solve the equation’. The second part uses the completed square to find exact solutions without the quadratic formula: (x + 5)² – 25 + 9 = 0 → (x + 5)² = 16 → x + 5 = ±4 → x = –1 or x = –9. Practise linking these steps logically.

OCR 经常要求 ‘将 x² + 10x + 9 写成 (x + p)² + q 的形式’ 或 ‘由此解方程’。第二部分利用配方式求出精确解，无需使用求根公式：(x + 5)² – 25 + 9 = 0 → (x + 5)² = 16 → x + 5 = ±4 → x = –1 或 x = –9。练习将这些步骤逻辑地串联起来。

7. Sketching Quadratic Graphs | 绘制二次函数图像

Sketching means drawing the main features without plotting individual points. OCR asks you to identify the shape (U or ∩), the y-intercept, the roots (x-intercepts), and the vertex. Use the completed square form to find the vertex, and the factorised form to find the roots when possible.

绘制图像意味着画出主要特征而无需逐点描点。OCR 要求你确定形状（U 形或 ∩ 形）、y 轴截距、根（x 轴截距）以及顶点。使用配方式求顶点，可能的话使用因式分解式求根。

For y = x² – 4x + 3, factorised gives y = (x – 1)(x – 3), so roots are x = 1 and x = 3. Complete the square: y = (x – 2)² – 1, vertex (2, –1). y-intercept is 3 (set x = 0). The graph is a U-shape passing through (1,0), (3,0), (0,3) with minimum point at (2, –1). Label these clearly on the sketch and mark the axes.

对于 y = x² – 4x + 3，因式分解得 y = (x – 1)(x – 3)，因此根为 x = 1 和 x = 3。配方：y = (x – 2)² – 1，顶点为 (2, –1)。y 轴截距为 3（令 x = 0）。图像为开口向上的 U 形，经过 (1,0)、(3,0)、(0,3)，最低点为 (2, –1)。在草图上清楚地标出这些点并标记坐标轴。

If the quadratic is given as y = –x² + 2x + 3, the negative sign flips the parabola to a maximum shape (∩). Find roots: –(x² – 2x – 3) = –(x – 3)(x + 1) gives x = 3 or x = –1. Vertex via completing square: –[(x – 1)² – 4] = –(x – 1)² + 4, vertex (1, 4). Always state whether the vertex is a maximum or minimum.

如果二次函数给出形式为 y = –x² + 2x + 3，负号将抛物线翻转为开口向下的最大值形状 (∩)。求根：–(x² – 2x – 3) = –(x – 3)(x + 1) 得 x = 3 或 x = –1。通过配方求顶点：–[(x – 1)² – 4] = –(x – 1)² + 4，顶点为 (1, 4)。务必说明顶点是最大值还是最小值。

8. The Discriminant and Nature of Roots | 判别式与根的性质

The discriminant is the part under the square root in the quadratic formula: D = b² – 4ac. It tells you about the nature of the roots without solving the equation entirely. This is a common OCR assessment topic, often appearing in multiple-choice or short-answer questions.

判别式是求根公式中平方根号下的部分：D = b² – 4ac。它能在不完全解方程的情况下告诉你根的性质。这是 OCR 常见考点，经常出现在选择题或简答题中。

If D > 0, there are two distinct real roots. If D = 0, there is exactly one real root (a repeated root, the graph touches the x-axis). If D < 0, there are no real roots; the equation cannot be solved for real numbers and the graph does not cross the x-axis. These results are linked to the graph’s behaviour.

如果 D > 0，有两个相异的实数根。如果 D = 0，恰好有一个实数根（重根，图像与 x 轴相切）。如果 D < 0，没有实数根；方程在实数范围内无解，图像不与 x 轴相交。这些结论与图像的行为相关。

OCR might ask: ‘Find the value of k for which the equation kx² + 4x + 1 = 0 has equal roots.’ For equal roots, set the discriminant to zero: b² – 4ac = 4² – 4 × k × 1 = 16 – 4k = 0 ⇒ k = 4. Always show the substitution clearly and state the condition you are using.

OCR 可能问：’求使方程 kx² + 4x + 1 = 0 有等根时 k 的值。’ 对于等根，令判别式为零：b² – 4ac = 4² – 4 × k × 1 = 16 – 4k = 0 ⇒ k = 4。务必清晰地展示代入过程并说明你使用的条件。

Sometimes you must interpret statements like ‘the graph does not cut the x-axis’ as D < 0 and solve the resulting inequality for k. Make sure you can move between the geometric interpretation and the algebraic condition.

有时你需要将诸如 ‘图像不与 x 轴相交’ 的表述解释为 D < 0，并解出关于 k 的不等式。确保你能在几何含义和代数条件之间进行转换。

9. Solving Quadratic Inequalities | 解二次不等式

Quadratic inequalities are solved by first rearranging to the form ax² + bx + c < 0 (or ≤, >, ≥), then finding the critical values where the expression equals zero. These values split the number line into intervals. You then test each interval or use a sketch of the graph to decide which regions satisfy the inequality.

解二次不等式的方法是首先将其整理为 ax² + bx + c < 0（或 ≤, >, ≥）的形式，然后找到使表达式等于零的临界值。这些值将数轴分成若干区间。然后你可以测试每个区间，或利用图像草图来判断哪些区域满足不等式。

Example: Solve x² – 5x + 6 ≥ 0. Factorise: (x – 2)(x – 3) = 0 gives roots 2 and 3. The parabola is U-shaped, so the expression is ≥ 0 outside the roots: x ≤ 2 or x ≥ 3. Write the final answer using set notation or inequalities as specified.

示例：解 x² – 5x + 6 ≥ 0。因式分解：(x – 2)(x – 3) = 0 得根 2 和 3。抛物线开口向上，因此在根的外部表达式 ≥ 0：x ≤ 2 或 x ≥ 3。按要求使用集合符号或不等式书写最终答案。

If the inequality is reversed, say x² – 5x + 6 < 0, the solution is the region between the roots: 2 < x < 3. Always draw a quick sketch to confirm. OCR penalises answers that include the roots when they should be excluded or vice versa.

如果不等式方向相反，例如 x² – 5x + 6 < 0，解为两根之间的区域：2 < x < 3。务必快速画出草图进行确认。OCR 会因错误地包含或排除根点而扣分。

When the coefficient of x² is negative, multiply through by –1 and reverse the inequality sign. For –x² + 4x – 3 ≤ 0, multiply by –1: x² – 4x + 3 ≥ 0, then solve. Many students lose marks by forgetting to flip the sign.

当 x² 系数为负时，对各项乘以 –1 并反转不等号方向。对于 –x² + 4x – 3 ≤ 0，乘以 –1：x² – 4x + 3 ≥ 0，然后求解。许多学生因忘记翻转不等号而失分。

10. Quadratic Word Problems | 二次方程应用题

OCR frequently embeds quadratics in real-life contexts: area of rectangles, product of consecutive numbers, projectile motion, or profit models. Typically, you must formulate an equation from the description, simplify to ax² + bx + c = 0, then solve and interpret the solution in context.

OCR 经常将二次方程嵌入现实生活情境：矩形面积、连续数的乘积、抛体运动或利润模型。通常，你需要根据描述列出方程，化简为 ax² + bx + c = 0，然后求解并在情境中解读答案。

Example: ‘A rectangle has length 3 cm longer than its width. The area is 54 cm². Find the dimensions.’ Let width = x, length = x + 3. Area = x(x + 3) = 54 → x² + 3x – 54 = 0 → (x + 9)(x – 6) = 0 → x = 6 (ignore negative). Width 6 cm, length 9 cm. Always reject solutions that make no physical sense.

示例：’一个矩形的长度比宽度长 3 cm。面积为 54 cm²。求尺寸。’ 设宽度 = x，则长度 = x + 3。面积 = x(x + 3) = 54 → x² + 3x – 54 = 0 → (x + 9)(x – 6) = 0 → x = 6（舍去负值）。宽 6 cm，长 9 cm。务必舍去在物理意义上无意义的解。

In projectile problems, height h(t) may be given as –5t² + vt + h₀. You might be asked when the object hits the ground (h = 0) or when it reaches a maximum height. Use the completed square to find the maximum or solve the quadratic for t. Remember to round time to a sensible precision and check units.

在抛体运动问题中，高度 h(t) 可能以 –5t² + vt + h₀ 的形式给出。你可能需要求物体何时落地（h = 0）或何时达到最大高度。使用配方法求最大值或解二次方程求 t。记得将时间四舍五入到合理的精度并检查单位。

11. Exam Tips and Common Mistakes | 考试技巧与常见错误

One of the most valuable habits is to always check whether an expression can be factorised before using the quadratic formula. If the coefficients are small integers, factorising saves time. If not, the formula is reliable. But always take out a common factor first.

最有价值的习惯之一是，在使用求根公式之前总是先检查表达式是否可以因式分解。如果系数是较小的整数，因式分解更省时。如果不可以，公式法是可靠的。但一定要先提取公因式。

A classic mistake is mishandling negative numbers when substituting into the formula or completing the square. Use brackets liberally: if b = –5, then –b = +5, and b² = 25. Write every step carefully. Another error is forgetting to set the equation to zero before factorising; OCR exam design often includes a non-zero constant on the right-hand side to trap the unwary.

一个经典错误是在代入公式或配方时错误处理负数。多使用括号：如果 b = –5，那么 –b = +5，而 b² = 25。仔细写下每一步。另一个错误是忘记在因式分解前将方程设为零；OCR 的试卷设计经常在右侧保留一个非零常数来考验不够仔细的学生。

When sketching, ensure you label the axes and key coordinates. A sketch without coordinates for intercepts and vertex loses marks. In inequality questions, be explicit about whether endpoints are included. Use solid dots for ≤ or ≥ and open circles for < or > on number lines.

在画草图时，确保你标注了坐标轴和关键坐标。没有标出截距和顶点坐标的草图会丢分。在不等式问题中，明确端点是否包含。在数轴上，用实心圆点表示 ≤ 或 ≥，用空心圆点表示 < 或 >。

Finally, manage time wisely. Quadratic questions range from 3 marks to 7 marks. If stuck on factorising, switch to the formula rather than spending too long searching for factors. Show all working – OCR awards method marks even if the final answer is wrong.

最后，合理管理时间。二次方程题的分值从 3 分到 7 分不等。如果因式分解卡住了，改用公式，而不是花太长时间寻找因式。展示所有步骤——即使最终答案错误，OCR 也会给方法分。

12. Summary | 总结

Quadratic functions are a cornerstone of IGCSE OCR maths. You must be fluent in expanding, factorising, using the quadratic formula, completing the square, interpreting the discriminant, solving inequalities, and applying these skills to graphs and word problems. Regular practice of past-paper questions under timed conditions will build confidence and expose you to the range of ways OCR examines this topic. Aim to understand not just the ‘how’ but also the ‘why’ behind each method – this deeper understanding makes it easier to adapt to unfamiliar question formats.

二次函数是 IGCSE OCR 数学的基石。你必须熟练掌握展开、因式分解、使用求根公式、配方法、解读判别式、解不等式，并将这些技能应用于图像和应用题。在限时条件下定期练习历年真题可以建立信心，并让你熟悉 OCR 考查该主题的各种方式。不仅要掌握 ‘怎么做’，还要理解每种方法背后的 ‘为什么’——这种更深入的理解有助于你更轻松地适应不熟悉的题型。

Published by TutorHao | IGCSE OCR Mathematics Revision Series | aleveler.com

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IGCSE OCR Maths: Quadratic Functions – Exam Focus | IGCSE OCR 数学：二次函数 考点精讲