IGCSE CIE Mathematics: Algebra and Functions – Key Points | IGCSE CIE 数学：代数和函数考点精讲

📚 IGCSE CIE Mathematics: Algebra and Functions – Key Points | IGCSE CIE 数学：代数和函数考点精讲

Algebra and functions form the backbone of the IGCSE CIE Mathematics syllabus, linking abstract reasoning with problem-solving techniques that appear in Papers 1 and 2. This revision guide breaks down essential topics – from simplifying expressions and solving quadratics to working with composite and inverse functions, as well as interpreting graph transformations. Mastering these areas will build your confidence and accuracy for the exam.

代数和函数是IGCSE CIE数学课程的核心骨架，将抽象推理与解题技巧紧密结合，出现在试卷1和试卷2中。本复习指南详细梳理了必考知识点——从表达式化简、二次方程求解到复合函数与反函数的运用，以及图像变换的解读。掌握这些内容将极大提升你的考试信心与正确率。

1. Algebraic Expressions and Simplification | 代数表达式与化简

Algebraic expressions contain numbers, letters and operation symbols. Simplifying means collecting like terms – terms that have exactly the same variable raised to the same power. For example, 3x + 5x simplifies to 8x, but 3x + 5y cannot be combined.

代数表达式包含数字、字母和运算符号。化简即合并同类项——同类项是指字母相同且指数也相同的项。例如，3x + 5x 可化为 8x，但 3x + 5y 无法合并。

Always follow the order of operations (BODMAS/BIDMAS) when simplifying: Brackets, Orders (powers), Division/Multiplication, Addition/Subtraction. For instance, 2(a + 3b) − a + 4b first expands to 2a + 6b − a + 4b, then like terms combine to a + 10b.

化简时务必遵循运算顺序（BODMAS/BIDMAS）：括号、乘方、乘除、加减。例如，2(a + 3b) − a + 4b 先展开为 2a + 6b − a + 4b，再合并同类项得到 a + 10b。

Be careful with negative signs. −(2x − 5) means −1 × (2x − 5) = −2x + 5. A common mistake is to write −2x − 5, so always distribute the minus sign across all terms inside the bracket.

注意负号的处理。−(2x − 5) 表示 −1 × (2x − 5) = −2x + 5。常见错误是写成 −2x − 5，因此一定要将负号分配给括号内的每一项。

2. Expanding Brackets | 去括号展开

Expanding brackets means multiplying each term inside the bracket by the term outside. For a single bracket: a(b + c) = ab + ac. With two brackets, use the FOIL method (First, Outer, Inner, Last) or a systematic grid. For (x + 3)(x − 2), you get x² − 2x + 3x − 6 = x² + x − 6.

去括号展开是指将括号外的项与括号内的每一项相乘。单个括号：a(b + c) = ab + ac。遇到两个括号，可使用FOIL法则（首、外、内、尾）或网格法。计算 (x + 3)(x − 2) 得 x² − 2x + 3x − 6 = x² + x − 6。

Special products often appear in exams: (a + b)² = a² + 2ab + b², (a − b)² = a² − 2ab + b², and (a + b)(a − b) = a² − b². Recognising these can save time.

考试中常出现特殊乘积：(a + b)² = a² + 2ab + b², (a − b)² = a² − 2ab + b², 以及 (a + b)(a − b) = a² − b²。能够识别这些形式可以节省大量时间。

When expanding with more than two terms, apply the distributive law repeatedly. For example, (x + 2)(x² − x + 1) gives x³ − x² + x + 2x² − 2x + 2, which simplifies to x³ + x² − x + 2.

当括号内不止两项时，反复应用分配律即可。例如 (x + 2)(x² − x + 1) 得到 x³ − x² + x + 2x² − 2x + 2，化简后为 x³ + x² − x + 2。

3. Factorising Expressions | 因式分解

Factorising is the reverse of expanding. Start by looking for a common factor: 4x + 8 = 4(x + 2). For quadratic expressions like x² + 5x + 6, find two numbers that multiply to the constant term (6) and add to the coefficient of x (5). Here, 2 and 3 work, so (x + 2)(x + 3).

因式分解是展开的逆运算。首先寻找公因数：4x + 8 = 4(x + 2)。对于二次式如 x² + 5x + 6，找出两个数，它们的乘积等于常数项(6)，和等于x的系数(5)。此处2和3符合，因此分解为 (x + 2)(x + 3)。

When the coefficient of x² is not 1, use a methodical approach. For 2x² + 7x + 3, multiply 2 and 3 to get 6. Find two numbers that multiply to 6 and add to 7: 6 and 1. Then split the middle term: 2x² + 6x + x + 3, and factor in pairs: 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3).

若x²的系数不为1，需采用系统方法。例如 2x² + 7x + 3，将2和3相乘得6。找出乘积为6且和为7的两个数：6和1。然后拆项：2x² + 6x + x + 3，再分组分解：2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)。

Difference of two squares: a² − b² = (a + b)(a − b). For example, x² − 9 = (x + 3)(x − 3), and 4x² − 25 = (2x + 5)(2x − 5). Always check if a common factor can be taken out first.

平方差公式：a² − b² = (a + b)(a − b)。例如，x² − 9 = (x + 3)(x − 3)，4x² − 25 = (2x + 5)(2x − 5)。始终先检查是否可以提取公因数。

4. Solving Linear Equations | 解线性方程

To solve a linear equation, isolate the variable by performing inverse operations on both sides. For 3x − 7 = 11, add 7 to both sides (3x = 18), then divide by 3 to get x = 6.

解线性方程时，通过等式两边进行逆运算来隔离变量。如 3x − 7 = 11，两边加7得 3x = 18，再除以3得 x = 6。

If the equation contains brackets, expand them first. For 2(x + 3) = 4x − 8, expand to 2x + 6 = 4x − 8, then bring variable terms to one side: 6 + 8 = 4x − 2x → 14 = 2x, so x = 7.

若方程含有括号，先展开。如 2(x + 3) = 4x − 8，展开得 2x + 6 = 4x − 8，再将含变量项移到一边：6 + 8 = 4x − 2x → 14 = 2x，故 x = 7。

Equations with fractions can be cleared by multiplying every term by the lowest common denominator. For x/3 + (x − 2)/4 = 2, multiply by 12 to get 4x + 3(x − 2) = 24, which solves to x = 30/7.

含有分数的方程可乘以最小公分母消去分母。如 x/3 + (x − 2)/4 = 2，乘以12得 4x + 3(x − 2) = 24，解得 x = 30/7。

Always check your solution by substituting it back into the original equation. This verifies that both sides are equal.

务必通过将解代入原方程进行检验，确保等式两边相等。

5. Solving Quadratic Equations by Factorising | 因式分解法解二次方程

A quadratic equation takes the form ax² + bx + c = 0. If the quadratic expression factorises, you can set each factor equal to zero. For x² − 5x + 6 = 0, factorise as (x − 2)(x − 3) = 0, giving x = 2 or x = 3.

二次方程的标准形式为 ax² + bx + c = 0。若二次式可分解，则可令每个因式等于零。如 x² − 5x + 6 = 0，分解得 (x − 2)(x − 3) = 0，因此 x = 2 或 x = 3。

When the product of two expressions is zero, at least one of them must be zero. This is the key principle behind the method. Apply it even when one factor is a simple term, e.g., x(x + 4) = 0 gives x = 0 or x = −4.

当两个表达式的乘积为零时，至少有一个为零。这是该方法的根本原理。即使其中一个因式是单项也同样适用，如 x(x + 4) = 0 给出 x = 0 或 x = −4。

Always bring all terms to one side so the equation equals zero before factorising. For 2x² = 8x, rearrange to 2x² − 8x = 0, then factorise 2x(x − 4) = 0 to get x = 0 or x = 4.

在分解前，务必将所有项移到一边使方程等于零。对于 2x² = 8x，先整理为 2x² − 8x = 0，再提取公因数 2x(x − 4) = 0，得 x = 0 或 x = 4。

6. The Quadratic Formula and Completing the Square | 二次公式与配方法

When factorising is difficult or impossible, use the quadratic formula. For ax² + bx + c = 0, the solutions are given by:

x = (−b ± √(b² − 4ac)) / (2a)

当因式分解困难或不可行时，使用二次公式。对于 ax² + bx + c = 0，解为上述公式。

The discriminant D = b² − 4ac tells you the nature of the roots. If D > 0, there are two distinct real roots. If D = 0, there is one repeated root. If D < 0, there are no real roots. This is frequently tested.

判别式 D = b² − 4ac 揭示了根的性质。若 D > 0，有两个不等实根；若 D = 0，有一个重根；若 D < 0，无实数根。这是常考知识点。

Completing the square transforms a quadratic into the form a(x + p)² + q. For x² + 6x + 5, take half of 6 (which is 3) and square it (9) to write (x + 3)² − 9 + 5 = (x + 3)² − 4. This form is useful for finding the vertex of a parabola and solving equations.

配方法将二次式转化为 a(x + p)² + q 的形式。例如 x² + 6x + 5，取6的一半（3）并平方（9），写成 (x + 3)² − 9 + 5 = (x + 3)² − 4。这种形式便于求抛物线的顶点和解方程。

To solve by completing the square, set the expression equal to zero and isolate the squared bracket. For (x + 3)² − 4 = 0, we get (x + 3)² = 4, so x + 3 = ±2, giving x = −1 or x = −5.

使用配方法解方程时，令表达式等于零并隔离完全平方项。如 (x + 3)² − 4 = 0，得 (x + 3)² = 4，因此 x + 3 = ±2，解为 x = −1 或 x = −5。

7. Functions and Function Notation | 函数与函数记号

A function is a rule that takes an input and produces exactly one output. Notation f(x) is read as ‘f of x’. For f(x) = 2x + 1, f(3) means substitute x = 3, giving 2(3) + 1 = 7. The input is the x-value, the output is the y-value or f(x).

函数是将一个输入按照规则转换成唯一输出的关系。记号 f(x) 读作“f of x”。对于 f(x) = 2x + 1，f(3) 表示代入 x = 3，得到 2(3) + 1 = 7。输入是x值，输出是y值或 f(x)。

Functions can be represented by equations, graphs, mapping diagrams or tables. In IGCSE, you will often be given f(x) = … and asked to evaluate f(a) or find x such that f(x) = k. To solve f(x) = 5 for f(x) = 3x − 4, set 3x − 4 = 5, giving x = 3.

函数可用方程式、图像、映射图或表格表示。IGCSE 考试中常给出 f(x) = …，要求计算 f(a) 或求解 f(x) = k 中的 x。若 f(x) = 3x − 4，解 f(x) = 5 即 3x − 4 = 5，得 x = 3。

Domain is the set of all possible input values (x-values). Range is the set of all possible output values (y-values). For f(x) = x², the domain is all real numbers, but the range is y ≥ 0. Be mindful of restrictions like division by zero or square roots of negative numbers.

定义域是所有可能输入值（x值）的集合，值域是所有可能输出值（y值）的集合。对于 f(x) = x²，定义域为全体实数，值域为 y ≥ 0。需注意分母不为零或负数开平方等限制。

8. Composite Functions | 复合函数

A composite function combines two functions, applying one after the other. fg(x) means first apply g, then apply f to the result. It is defined as f(g(x)). For example, if f(x) = x² and g(x) = 2x + 1, then fg(x) = f(2x + 1) = (2x + 1)².

复合函数是将两个函数按顺序组合。fg(x) 表示先作用 g，再对结果作用 f，即 f(g(x))。例如 f(x) = x²，g(x) = 2x + 1，则 fg(x) = f(2x + 1) = (2x + 1)²。

gf(x) may be different. Using the same functions, gf(x) = g(x²) = 2x² + 1. Always work from the innermost function outward. Write out the inner function, then substitute it into the outer function.

gf(x) 可能不同。用同样函数，gf(x) = g(x²) = 2x² + 1。始终由内层函数向外运算。写出内层函数表达式，再代入外层函数。

When asked to solve fg(x) = k, first form the composite function, then set it equal to k and solve. For f(x) = 3x, g(x) = x − 2, fg(x) = 3(x − 2) = 3x − 6. Setting 3x − 6 = 12 gives x = 6.

若题目要求解 fg(x) = k，先构造复合函数，再令其等于 k 并求解。设 f(x) = 3x，g(x) = x − 2，则 fg(x) = 3(x − 2) = 3x − 6。令 3x − 6 = 12 得 x = 6。

You can also evaluate composite functions at a specific value. For the above, fg(5) = 3(5) − 6 = 9. Remember that order matters: fg(x) is not generally the same as gf(x).

也可在特定值上计算复合函数。如上，fg(5) = 3(5) − 6 = 9。记住顺序很重要：fg(x) 通常不等于 gf(x)。

9. Inverse Functions | 反函数

The inverse function f⁻¹(x) reverses the effect of f(x). If f(x) = 2x + 3, to find the inverse, write y = 2x + 3, swap x and y to get x = 2y + 3, then solve for y: y = (x − 3)/2, so f⁻¹(x) = (x − 3)/2.

反函数 f⁻¹(x) 逆转 f(x) 的作用。若 f(x) = 2x + 3，求反函数时，写出 y = 2x + 3，交换 x 和 y 得 x = 2y + 3，解出 y = (x − 3)/2，因此 f⁻¹(x) = (x − 3)/2。

Not all functions have an inverse unless they are one-to-one (each output comes from a unique input). The domain may need to be restricted, e.g., f(x) = x² for x ≥ 0 has inverse f⁻¹(x) = √x.

并非所有函数都有反函数，除非是一一对应的（每个输出对应唯一输入）。可能需限制定义域，例如 f(x) = x² （x ≥ 0）的反函数为 f⁻¹(x) = √x。

A useful check: f(f⁻¹(x)) = x and f⁻¹(f(x)) = x. For the example above, f(f⁻¹(x)) = 2((x − 3)/2) + 3 = x − 3 + 3 = x, confirming the inverse is correct.

一个有用的检验：f(f⁻¹(x)) = x 且 f⁻¹(f(x)) = x。在上例中，f(f⁻¹(x)) = 2((x − 3)/2) + 3 = x − 3 + 3 = x，验证反函数正确。

Graphically, the inverse is a reflection of the original function in the line y = x. If (a, b) lies on f(x), then (b, a) lies on f⁻¹(x). This geometric link is often tested.

图像上，反函数是原函数关于直线 y = x 的反射。若 (a, b) 在 f(x) 上，则 (b, a) 在 f⁻¹(x) 上。这种几何联系常被考查。

10. Graphs and Transformations of Functions | 函数图像与变换

Understanding graphs of linear (y = mx + c) and quadratic functions (y = ax² + bx + c) is essential. The graph of y = mx + c is a straight line with gradient m and y-intercept c. The quadratic graph is a parabola; if a > 0 it opens upwards (∪), if a < 0 it opens downwards (∩).

理解一次函数图像（y = mx + c）和二次函数图像（y = ax² + bx + c）至关重要。y = mx + c 的图像是一条斜率为 m、y轴截距为 c 的直线。二次图像是抛物线；若 a > 0 开口向上（∪），若 a < 0 开口向下（∩）。

Key points on a quadratic graph include the y-intercept (x = 0), the x-intercepts (roots, by solving ax² + bx + c = 0), and the vertex (turning point). Completing the square gives the vertex in the form ( −p, q ) from a(x + p)² + q.

二次图像的关键点包括y轴截距（x = 0）、x轴截距（根，通过解 ax² + bx + c = 0 得到）和顶点（转折点）。配方法可从 a(x + p)² + q 得出顶点坐标为 ( −p, q )。

Transformations of the graph of y = f(x):

Vertical translation: y = f(x) + a moves the graph up by a units if a > 0, down if a < 0.
水平平移 (中文): y = f(x + a) 将图像向左平移 a 个单位（若 a > 0），向右平移 |a| 个单位（若 a < 0）。注意符号方向。

Transformations: y = f(x) + a is a vertical shift; y = f(x + a) is a horizontal shift in the opposite direction to the sign. A common mistake is to think x + 2 shifts right, but it actually shifts left by 2.

图像变换：y = f(x) + a 是垂直平移；y = f(x + a) 是水平平移，方向与符号相反。常见错误是以为 x + 2 向右移，实际是向左移动2个单位。

Reflection in x-axis: y = −f(x) reflects the graph vertically. Points above the x-axis go below, and vice versa.
关于x轴反射 (中文): y = −f(x) 将图像沿x轴上下翻转，原来在x轴上方的点映射到下方，反之亦然。
Reflection in y-axis: y = f(−x) reflects the graph horizontally.
关于y轴反射: y = f(−x) 将图像水平翻转。
Vertical stretch: y = a f(x) with a > 1 stretches the graph vertically by factor a; 0 < a < 1 compresses it.
垂直伸缩: y = a f(x)，a > 1 为垂直拉伸 a 倍，0 < a < 1 为压缩。

When multiple transformations are applied, follow the order: horizontal shifts, stretches/reflections, then vertical shifts. However, always refer to the specific function form. For example, y = 2f(x − 1) + 3 involves a horizontal shift right by 1, a vertical stretch by factor 2, and a vertical shift up by 3.

当进行多个变换时，通常顺序为：水平平移、伸缩/反射、最后垂直平移。但需结合具体函数形式。例如，y = 2f(x − 1) + 3 包括向右平移1单位、纵向拉伸2倍、向上平移3单位。

Sketching transformed graphs accurately requires identifying the new positions of key points such as intercepts and turning points. Apply transformations to the coordinates of these points methodically.

准确描绘变换后的图像需要确定关键点（截距、顶点）的新位置，按规则逐步对其坐标施加变换。

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

IGCSE CIE Mathematics: Algebra and Functions – Key Points | IGCSE CIE 数学：代数和函数 考点精讲