📚 GCSE CCEA Maths: Algebra and Functions Key Points | 代数和函数 考点精讲
Algebra and functions form the core of GCSE Mathematics, building the skills to manipulate expressions, solve equations and inequalities, and understand the language of functions. This revision guide covers the key topics in the CCEA specification, with clear examples and bilingual explanations to boost your confidence.
代数和函数是 GCSE 数学的核心，培养你化简表达式、解方程和不等式、理解函数语言的能力。这篇考点精讲涵盖了 CCEA 考试大纲中的重点专题，并配有清晰示例和双语解释，助你增强信心。
1. Algebraic Notation and Simplifying | 代数符号与化简
Algebraic terms are called 'like terms' if they have exactly the same variable parts. Like terms can be collected by adding or subtracting their coefficients, while unlike terms cannot be combined.
如果代数项具有完全相同的变量部分，则称为“同类项”。同类项可以通过加减系数来合并，而不同类项则不能合并。
ke terms, so 3x + 5x = 8x. However, 4x and 2y are not like terms and must remain separate.
例如，3x 和 5x 是同类项，所以 3x + 5x = 8x。然而 4x 和 2y 不是同类项，必须保持分开。
letters are simply written together, e.g., p × q = pq. The order is usually alphabetical.
进行乘法运算时，要运用指数规则：a × a = a²，不同字母直接写在一起，如 p × q = pq。字母通常按字母表顺序书写。
| Expression | Simplified form |
|---|---|
| 7m + 3m – m | 9m |
| 3a × 2b × a | 6a²b |
| 2x + 5y – x + 3y | x + 8y |
ow the distributive law: a(b + c) = ab + ac. Any sign outside multiplies each term inside.
代数中的括号遵循分配律：a(b + c) = ab + ac。括号外的符号要与括号内每一项相乘。
2. Expanding Brackets | 展开括号
iply the term outside by every term inside, taking care with negative signs. Example: 5(2x – 3) = 10x – 15.
展开单个括号时，将括号外的项与括号内每一项相乘，注意负号。示例：5(2x – 3) = 10x – 15。
thod (First, Outer, Inner, Last): (x + a)(x + b) = x² + bx + ax + ab = x² + (a + b)x + ab.
对于两个括号相乘，使用 FOIL 方法（首、外、内、尾）：(x + a)(x + b) = x² + bx + ax + ab = x² + (a + b)x + ab。
Squaring a binomial requires applying (a + b)² = a² + 2ab + b². Don't forget the middle term. Example: (2x + 3)² = 4x² + 12x + 9.
二项式的平方必须运用 (a + b)² = a² + 2ab + b²。千万不要遗漏中间项。示例：(2x + 3)² = 4x² + 12x + 9。
Expanding expressions like (x + 2)(x – 5) gives x² – 3x – 10. Always combine like terms after expansion.
展开类似 (x + 2)(x – 5) 的表达式得到 x² – 3x – 10。展开后一定要合并同类项。
3. Factorising Expressions | 因式分解
fying the highest common factor (HCF) of all terms and factor it out: 8x + 12 = 4(2x + 3).
因式分解是展开的逆运算。首先找出所有项的最高公因式（HCF）并将其提取出来：8x + 12 = 4(2x + 3)。
o c and add to b. Example: x² + 7x + 10 = (x + 2)(x + 5) because 2 × 5 = 10 and 2 + 5 = 7.
对于形如 x² + bx + c 的二次式，寻找两个数，它们的乘积等于 c，和等于 b。示例：x² + 7x + 10 = (x + 2)(x + 5)，因为 2 × 5 = 10 且 2 + 5 = 7。
trial. Split the middle term: 2x² + x + 6x + 3 = x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3).
当 x² 的系数不是 1 时，如 2x² + 7x + 3，可采用分组分解法或系统尝试。将中间项拆分：2x² + x + 6x + 3 = x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3)。
The difference of two squares factorises as a² – b² = (a + b)(a – b). Recognise it quickly: x² – 9 = (x + 3)(x – 3); 25y² – 4 = (5y + 2)(5y – 2).
平方差公式为 a² – b² = (a + b)(a – b)。要能快速识别：x² – 9 = (x + 3)(x – 3)；25y² – 4 = (5y + 2)(5y – 2)。
4. Solving Linear Equations | 解线性方程
The goal is to isolate the unknown variable using inverse operations. Always perform the same operation on both sides to keep the equation balanced.
目标是通过逆运算将未知数孤立出来。始终在方程两边执行同样的运算以保持平衡。
a two-step equation like 2x + 5 = 13, subtract 5 first, then divide by 2: 2x = 8 → x = 4.
对于如 2x + 5 = 13 的两步方程，先减去 5，再除以 2：2x = 8 → x = 4。
n appears on both sides, collect variable terms on one side and constants on the other: 3x – 4 = x + 8 → 2x = 12 → x = 6.
当未知数出现在等号两边时，将含变量的项移到一边，常数项移到另一边：3x – 4 = x + 8 → 2x = 12 → x = 6。
With brackets, expand first: 4(2x – 1) = 20 → 8x – 4 = 20 → 8x = 24 → x = 3.
有括号时先展开：4(2x – 1) = 20 → 8x – 4 = 20 → 8x = 24 → x = 3。
5. Solving Inequalities | 解不等式
s, but multiplying or dividing by a negative number reverses the inequality sign. Example: –2x < 10 → x > –5.
解不等式与解方程类似，但当乘以或除以一个负数时，不等号的方向会反转。示例：–2x < 10 → x > –5。
or < or >, and a closed circle for ≤ or ≥. Shade the region that satisfies the inequality.
在数轴上表示解集：对于 < 或 > 使用空心圆，对于 ≤ 或 ≥ 使用实心圆。将满足不等式的区域涂上阴影。
Double inequalities like 1 < 3x + 4 ≤ 10 can be solved by subtracting 4 throughout: –3 < 3x ≤ 6, then dividing by 3: –1 < x ≤ 2.
像 1 < 3x + 4 ≤ 10 这样的双向不等式可以整体减去 4：–3 < 3x ≤ 6，再除以 3：–1 < x ≤ 2。
6. Solving Quadratic Equations by Factorising | 用因式分解解二次方程
e left-hand side factorises, apply the null factor law: if A × B = 0, then A = 0 or B = 0.
二次方程的形式为 ax² + bx + c = 0。如果左边可以因式分解，则运用零因子律：若 A × B = 0，则 A = 0 或 B = 0。
Example: x² – 5x + 6 = 0 factorises to (x – 2)(x – 3) = 0, giving solutions x = 2 and x = 3.
示例：x² – 5x + 6 = 0 分解为 (x – 2)(x – 3) = 0，解得 x = 2 或 x = 3。
Always rearrange the equation so that one side equals zero. If given x² = 4x – 3, bring all terms to one side: x² – 4x + 3 = 0, then factorise to (x – 1)(x – 3) = 0, so x = 1 or 3.
始终将方程移项，使一边为零。如给出 x² = 4x – 3，将所有项移到一边：x² – 4x + 3 = 0，然后分解为 (x – 1)(x – 3) = 0，所以 x = 1 或 3。
7. Quadratic Equations: The Quadratic Formula | 二次方程：求根公式
ising is not possible, use the quadratic formula. For ax² + bx + c = 0, the solutions are:
当无法因式分解时，使用求根公式。对于 ax² + bx + c = 0，解为：
x = (−b ± √b² − 4ac) / 2a
The expression under the square root, b² – 4ac, is called the discriminant. It tells us about the nature of the roots.
根号下的表达式 b² – 4ac 称为判别式。它告诉我们根的性质。
If b² – 4ac > 0, the equation has two distinct real roots. If it equals 0, there is one repeated real root. If negative, there are no real roots.
如果 b² – 4ac > 0，方程有两个不同实根。若等于 0，有一个重根。若为负数，则没有实数根。
Example: Solve 2x² + 3x – 5 = 0. Here a = 2, b = 3, c = –5. Substitute: x = (−3 ± √9 + 40) / 4 = (−3 ± √49) / 4 = (−3 ± 7) / 4, giving x = 1 or x = –2.5.
示例：解 2x² + 3x – 5 = 0。这里 a = 2, b = 3, c = –5。代入公式：x = (−3 ± √9 + 40) / 4 = (−3 ± √49) / 4 = (−3 ± 7) / 4，解得 x = 1 或 x = –2.5。
8. Completing the Square | 配方法
This method rewrites a quadratic in the form (x + p)² + q, making it easier to find the vertex of the graph or solve equations. To complete the square for x² + bx + c, use (x + b/2)² – (b/2)² + c.
这种方法将二次式改写为 (x + p)² + q 的形式，便于求图像顶点或解方程。对于 x² + bx + c，配方为 (x + b/2)² – (b/2)² + c。
For example, x² + 6x + 1 becomes (x + 3)² – 9 + 1 = (x + 3)² – 8.
例如，x² + 6x + 1 可化为 (x + 3)² – 9 + 1 = (x + 3)² – 8。
oefficient of x² is not 1, factor it out first: 2x² + 8x + 3 = 2(x² + 4x) + 3 = 2[(x + 2)² – 4] + 3 = 2(x + 2)² – 5.
当 x² 的系数不是 1 时，先将其提取出来：2x² + 8x + 3 = 2(x² + 4x) + 3 = 2[(x + 2)² – 4] + 3 = 2(x + 2)² – 5。
Solving by completing the square yields x = –p ± √−q, which connects to the quadratic formula.
用配方法解方程可得 x = –p ± √−q，这与求根公式相通。
9. Sequences and nth Term | 数列与第 n 项
A linear sequence has a constant difference between consecutive terms. The nth term is given by an + b, where a is the common difference and a + b is the first term.
线性数列相邻两项的差为常数。第 n 项表达式为 an + b，其中 a 是公差，a + b 是首项。
8, 11, 14, ..., the difference a = 3. Then 3(1) + b = 5 gives b = 2, so nth term = 3n + 2.
对于数列 5, 8, 11, 14, ...，公差 a = 3。由 3(1) + b = 5 得 b = 2，因此第 n 项 = 3n + 2。
rm rule, e.g., u₁ = 2, uₙ₊₁ = 2uₙ + 1. You generate terms by applying the rule repeatedly.
某些数列通过递推关系定义，例如 u₁ = 2，uₙ₊₁ = 2uₙ + 1。重复应用该递推式即可生成各项。
Quadratic sequences have a second difference that is constant. The nth term will be of the form an² + bn + c, where 2a equals the second difference.
二次数列的二级差为常数。其第 n 项表达式为 an² + bn + c，其中 2a 等于二级差。
10. Introduction to Functions | 函数简介
nction is a rule that assigns each input exactly one output. It is often written as f(x) = …, read as 'f of x'. f(x) is not multiplication; it means the value of the function at x.
函数是一种规则，它为每个输入值指定唯一的输出值。通常写作 f(x) = …，读作 “f of x”。f(x) 不是乘法，它表示函数在 x 处的值。
To evaluate a function, substitute the input value: if f(x) = 3x² – 1, then f(2) = 3(2)² – 1 = 11; f(–4) = 3(16) – 1 = 47.
计算函数值时，代入输入值：若 f(x) = 3x² – 1，则 f(2) = 3(2)² – 1 = 11；f(–4) = 3(16) – 1 = 47。
Functions can be expressed as mapping diagrams, algebraic rules, or graphs. You must be able to move between these forms.
函数可以用映射图、代数规则或图像来表示。你需要能够在这几种形式之间进行转换。
11. Inverse and Composite Functions | 反函数与复合函数
and y, then solve for y. Always check that the original function is a one-to-one mapping.
反函数 f⁻¹(x) 能逆转 f(x) 的作用。求反函数时，令 y = f(x)，交换 x 和 y，再解出 y。需确保原函数是一一映射。
Example: f(x) = 2x + 3. Write y = 2x + 3 → swap: x = 2y + 3 → solve: 2y = x – 3 → y = (x – 3)/2, so f⁻¹(x) = (x – 3)/2.
例如：f(x) = 2x + 3。令 y = 2x + 3 → 交换：x = 2y + 3 → 求解：2y = x – 3 → y = (x – 3)/2，所以 f⁻¹(x) = (x – 3)/2。
Composite functions combine two functions. (f ∘ g)(x) or fg(x) means apply g first, then f: fg(x) = f(g(x)).
复合函数将两个函数组合在一起。(f ∘ g)(x) 或 fg(x) 表示先施加 g，再施加 f：fg(x) = f(g(x))。
If f(x) = x² and g(x) = 3x – 1, then fg(x) = f(3x – 1) = (3x – 1)². And gf(x) = g(x²) = 3x² – 1. Note that order matters.
若 f(x) = x²，g(x) = 3x – 1，则 fg(x) = f(3x – 1) = (3x – 1)²。而 gf(x) = g(x²) = 3x² – 1。注意顺序不同结果也会不同。
12. Graphs of Functions | 函数图像
roduce straight lines with gradient m and y-intercept c. Parallel lines have equal gradients.
形式为 y = mx + c 的线性函数产生直线，斜率为 m，y 轴截距为 c。平行直线斜率相等。
Quadratic functions y = x² create a U-shaped parabola; y = –x² gives an inverted U. The turning point can be found by completing the square.
二次函数 y = x² 产生一条 U 形抛物线；y = –x² 则是倒 U 形。通过配方可求得其转折点。
reciprocal functions y = 1/x have two separate branches approaching the axes (asymptotes).
三次函数（如 y = x³）呈现 S 形曲线，而反比例函数 y = 1/x 有两个分别趋近坐标轴的分支（渐近线）。
Exponential functions of the type y = aˣ (a > 0) rise or decay rapidly. Recognise their shapes; they never touch the x-axis.
形如 y = aˣ (a > 0) 的指数函数迅速增长或衰减。要能识别其形状，它们永远不触及 x 轴。
Transformations: f(x) + a shifts the graph up by a; f(x + a) shifts left by a; –f(x) reflects in the x-axis; f(–x) reflects in the y-axis.
图像变换：f(x) + a 使图像上移 a 个单位；f(x + a) 向左移 a 个单位；–f(x) 关于 x 轴反射；f(–x) 关于 y 轴反射。
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