GCSE Edexcel Maths: Algebra and Functions Key Points | GCSE Edexcel 数学：代数和函数考点精讲

📚 GCSE Edexcel Maths: Algebra and Functions Key Points | GCSE Edexcel 数学：代数和函数考点精讲

Algebra and functions are at the heart of the GCSE Edexcel Maths syllabus. Mastering manipulation of expressions, solving equations, understanding sequences and working with function notation will secure a large proportion of the marks on both the Foundation and Higher tier papers. This revision guide breaks down every key topic with clear explanations, examples and exam tips.

代数与函数是 GCSE Edexcel 数学大纲的核心。掌握表达式变形、方程求解、数列规律以及函数符号的应用，将为你锁定 Foundation 与 Higher 卷中大量分数。本复习指南通过清晰的解释、例题和考试技巧，逐一梳理每一个重要考点。

1. Simplifying Algebraic Expressions | 代数表达式化简

You must be able to collect like terms, expand brackets and factorise simple expressions. Always apply the rules of indices when multiplying or dividing terms with the same base, e.g. x² × x³ = x⁵.

你必须能够合并同类项、展开括号以及分解简单表达式。当底数相同时进行乘除运算，务必使用指数法则，例如 x² × x³ = x⁵。

When expanding two brackets, use the FOIL method (First, Outer, Inner, Last) or the grid method. Remember (a + b)² = a² + 2ab + b² and (a – b)² = a² – 2ab + b², as well as the difference of two squares: a² – b² = (a + b)(a – b).

展开两个括号时，使用 FOIL 方法（首项、外项、内项、尾项）或网格法。牢记完全平方公式 (a + b)² = a² + 2ab + b²，(a – b)² = a² – 2ab + b²，以及平方差公式 a² – b² = (a + b)(a – b)。

Factorising is the reverse of expanding. Look for a common factor first, then check for quadratic forms. For x² + bx + c, find two numbers that multiply to c and add to b.

因式分解是展开的逆运算。首先提取公因式，再检查是否可分解为二次三项式。对于 x² + bx + c 型，寻找两个数其积为 c，和为 b。

2. Solving Linear Equations | 解线性方程

A linear equation has the variable to the power of 1. Solve by performing the same operation on both sides to isolate the unknown. Always aim to undo addition/subtraction before multiplication/division.

线性方程中变量的次数为 1。通过等式两边进行相同运算来隔离未知数。通常先处理加减，再处理乘除。

For equations with unknowns on both sides, collect the variable terms on one side and the constants on the other. Example: 3x + 5 = 2x – 1 → 3x – 2x = -1 – 5 → x = -6.

当未知数出现在等式两边时，将所有含变量的项移到一边，常数项移到另一边。例：3x + 5 = 2x – 1 → 3x – 2x = -1 – 5 → x = -6。

Always check your solution by substituting it back into the original equation. If you obtain a statement like 0 = 5 during solving, the equation has no solution.

务必将解代回原方程检验。若在求解过程中出现 0 = 5 这样的矛盾等式，则该方程无解。

3. Solving Quadratic Equations | 解二次方程

Quadratics are of the form ax² + bx + c = 0. Three main methods: factorising, using the quadratic formula, and completing the square. Factorising is fastest when the quadratic breaks down into integer factors.

二次方程的一般形式为 ax² + bx + c = 0。主要有三种方法：因式分解法、公式法和配方法。若二次式能分解为整系数因式，因式分解法最为快捷。

The quadratic formula:

x = (-b ± √(b² – 4ac)) / 2a

works for any quadratic, even when it does not factorise. The discriminant b² – 4ac tells you the number of real solutions: >0 gives two distinct roots, =0 gives one repeated root, <0 gives no real roots.

求根公式

x = (-b ± √(b² – 4ac)) / 2a

适用于所有二次方程，即使不能因式分解也可用。判别式 b² – 4ac 揭示了实根个数：>0 两个不等实根，=0 两个相等实根，<0 无实根。

Completing the square rewrites the quadratic in the form (x + p)² + q, which is useful for finding the vertex of a parabola and solving when factorising is not straightforward. Example: x² + 6x + 5 = (x + 3)² – 4.

配方法将二次式写成 (x + p)² + q 的形式，便于求抛物线顶点，也用于不易因式分解的情形。例如 x² + 6x + 5 = (x + 3)² – 4。

4. Simultaneous Equations | 联立方程

Simultaneous equations involve finding values that satisfy two or more equations at the same time. The elimination method is efficient when coefficients of one variable can be made equal. Multiply one or both equations so that adding or subtracting eliminates that variable.

联立方程要求找到同时满足两个或多个方程的数值。当某个变量的系数可以配成相等时，加减消元法十分高效。将方程乘以适当倍数，使加减后该变量抵消。

For linear and quadratic pairs, substitution is the standard approach. Rearrange the linear equation for one variable and substitute into the quadratic. This yields a quadratic in one variable which you then solve, and substitute back to find the other variable.

处理一个线性方程与一个二次方程的组合时，通常采用代入法。将线性方程变形为用含一个变量的式子表示另一个变量，然后代入二次方程。得到一个一元二次方程，求解后再回代求另一变量。

Always give your final answer as pairs of (x, y) coordinates. Check each pair in both original equations.

最终答案应以 (x, y) 数对形式给出。将每一组解代回两个原方程中检验。

5. Inequalities | 不等式

Inequalities use the symbols <, >, ≤, ≥. Solving linear inequalities mimics solving equations, but multiplying or dividing by a negative number reverses the inequality sign. Example: -2x > 6 becomes x < -3.

不等式使用 <, >, ≤, ≥ 符号。解线性不等式与解方程类似，但乘或除以负数时，不等号方向要改变。例：-2x > 6 变为 x < -3。

Represent inequalities on a number line using open circles for strict inequalities (<, >) and closed circles for inclusive inequalities (≤, ≥). For quadratic inequalities, sketch the graph of the quadratic to identify intervals where the inequality holds.

在数轴上表示不等式时，严格不等号（<, >）用空心圆，包含等号（≤, ≥）用实心圆。解二次不等式时，画出二次函数草图，确定不等式成立的区间。

Solving x² – 4x – 5 > 0: factorise to (x – 5)(x + 1) > 0, roots at x = 5 and x = -1. The parabola opens upwards, so the solution is x < -1 or x > 5. Use set notation where required.

解 x² – 4x – 5 > 0：因式分解得 (x – 5)(x + 1) > 0，根为 x = 5 和 x = -1。抛物线开口向上，故解为 x < -1 或 x > 5。必要时使用集合记法。

6. Sequences and the nth Term | 数列与第 n 项

An arithmetic sequence has a constant difference between consecutive terms. The nth term is given by a + (n – 1)d, where a is the first term and d is the common difference. You can also write it as dn + c, finding c using the first term.

等差数列相邻两项的差为常数。第 n 项公式为 a + (n – 1)d，其中 a 为首项，d 为公差。也可写成 dn + c 的形式，利用首项求出 c。

For quadratic sequences, the second difference is constant. To find the nth term of the form an² + bn + c, halve the second difference to get a, then use simultaneous equations to find b and c. Example: 2, 6, 12, 20… second difference = 2, so a = 1, giving n² + n.

二次数列的二阶差分为常数。形如 an² + bn + c 的第 n 项，先将二阶差分除以 2 得到 a，再通过联立方程求 b 和 c。例：2, 6, 12, 20… 二阶差分为 2，故 a = 1，得 n² + n。

Other sequences include geometric sequences (ratio between terms constant) and Fibonacci-style sequences. Read the question carefully to identify the rule. Use term-to-term rules when given, e.g., uₙ₊₁ = 2uₙ + 3.

其他数列包括等比数列（相邻项比值为常数）和斐波那契形式的数列。仔细读题辨别规律。若给出递推关系，如 uₙ₊₁ = 2uₙ + 3，按要求计算。

7. Functions and Function Notation | 函数与函数符号

A function is a rule that maps each input to exactly one output. f(x) = 3x + 2 means for any input x, the output is 3x + 2. f(4) = 3(4) + 2 = 14. Always replace x with the given value.

函数是一种将每个输入映射到唯一输出的规则。f(x) = 3x + 2 表示对于任意输入 x，输出为 3x + 2。f(4) = 3(4) + 2 = 14。始终用给定值替换表达式中的 x。

The domain is the set of all possible input values. The range is the set of all possible output values. For GCSE, common restrictions come from square roots (inside must be ≥ 0) and fractions (denominator cannot be zero).

定义域是所有可能输入值的集合，值域是所有可能输出值的集合。在 GCSE 中，常见的限制来自平方根（根号内须 ≥ 0）和分式（分母不能为零）。

Use graphs to understand functions: the x-axis shows the domain, the y-axis the range. A relation is not a function if a vertical line crosses the graph more than once.

借助图像理解函数：x 轴体现定义域，y 轴体现值域。若一条竖直线与图像有多个交点，则该关系不是函数。

8. Composite Functions | 复合函数

A composite function combines two functions, applying one after the other. fg(x) means apply g first, then f. So fg(x) = f(g(x)). Always work from the inside out: calculate g(x), then substitute that result into f.

复合函数将两个函数组合，依次执行。fg(x) 表示先执行 g，再执行 f。因此 fg(x) = f(g(x))。始终由内而外计算：先算出 g(x)，再把结果代入 f。

Example: f(x) = 2x + 1, g(x) = x². Then fg(x) = f(x²) = 2(x²) + 1 = 2x² + 1. gf(x) = g(2x + 1) = (2x + 1)² = 4x² + 4x + 1. Note that fg(x) and gf(x) are usually different.

例：f(x) = 2x + 1，g(x) = x²。则 fg(x) = f(x²) = 2(x²) + 1 = 2x² + 1。gf(x) = g(2x + 1) = (2x + 1)² = 4x² + 4x + 1。注意 fg(x) 与 gf(x) 通常不同。

When evaluating composite functions for a specific value, e.g., fg(3), you can either find fg(x) first then substitute x = 3, or calculate g(3) and then apply f to the result. Both methods are valid.

当需要计算复合函数的具体值时，如 fg(3)，可以先求出 fg(x) 表达式再代入 x = 3，也可先算 g(3) 再把结果代入 f。两种方法都正确。

9. Inverse Functions | 反函数

The inverse function f⁻¹(x) reverses the effect of f(x). To find the inverse, write y = f(x), swap x and y, and then make y the subject. The input of the inverse is the output of the original function, so the domain and range swap.

反函数 f⁻¹(x) 逆转 f(x) 的作用。求反函数时，写出 y = f(x)，交换 x 与 y，再将 y 解出。反函数的输入是原函数的输出，因此定义域与值域互换。

Example: f(x) = (x – 4)/3. Let y = (x – 4)/3, swap to x = (y – 4)/3, multiply by 3: 3x = y – 4, so y = 3x + 4. Hence f⁻¹(x) = 3x + 4. Always check that f(f⁻¹(x)) = x.

例：f(x) = (x – 4)/3。设 y = (x – 4)/3，交换得 x = (y – 4)/3，两边乘 3：3x = y – 4，所以 y = 3x + 4。即 f⁻¹(x) = 3x + 4。务必验证 f(f⁻¹(x)) = x。

The graph of y = f⁻¹(x) is a reflection of y = f(x) in the line y = x. Only one-to-one functions have an inverse that is also a function. If the original function fails the horizontal line test, restrict its domain to make it invertible.

y = f⁻¹(x) 的图像是 y = f(x) 关于直线 y = x 的反射。只有一一对应函数才存在反函数。若原函数不满足水平线检测，可限制其定义域使之可逆。

10. Transformations of Functions | 函数图像变换

Graph transformations allow you to sketch related functions without plotting points. Learn these key transformations of y = f(x):

图像变换让你无需逐点描画就能绘制相关函数图像。牢记 y = f(x) 的下列关键变换：

f(x) + a: translation upwards by a units.

f(x) + a：向上平移 a 个单位。
f(x + a): translation to the left by a units. (Opposite direction to the sign)

f(x + a)：向左平移 a 个单位。（与符号方向相反）
-f(x): reflection in the x-axis (flips vertically).

-f(x)：关于 x 轴反射（纵向翻转）。
f(-x): reflection in the y-axis (flips horizontally).

f(-x)：关于 y 轴反射（横向翻转）。
af(x): vertical stretch by scale factor a. Points on the x-axis remain fixed.

af(x)：纵向拉伸 a 倍。位于 x 轴上的点不动。
f(ax): horizontal stretch by scale factor 1/a (compresses if a > 1). Points on y-axis stay fixed.

f(ax)：横向拉伸 1/a 倍（若 a > 1 则为压缩）。位于 y 轴上的点不动。

When multiple transformations are combined, apply stretches and reflections before translations to avoid mistakes. Always pay close attention to the order and bracket placement, e.g., f(2x – 6) should be treated as f(2(x – 3)), a horizontal compression by factor 1/2 and a translation 3 units right.

当多种变换结合时，先进行拉伸与反射，再进行平移，避免出错。务必注意变换顺序与括号的位置，例如 f(2x – 6) 应写作 f(2(x – 3))，表示先横向压缩至 1/2，再向右平移 3 个单位。

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GCSE Edexcel Maths: Algebra and Functions Key Points | GCSE Edexcel 数学：代数和函数 考点精讲