📚 Algebra and Functions: Key Concepts for IGCSE OCR | 代数和函数考点精讲
In the IGCSE OCR Mathematics specification, Algebra and Functions form the backbone of many problem-solving scenarios. This article revisits the essential topics, from simplifying expressions and solving equations to understanding function notation, composite and inverse functions, and interpreting graphs. By mastering these concepts, you will strengthen your ability to handle both routine and application-style questions with confidence.
在 IGCSE OCR 数学考试中，代数和函数是解决许多实际问题的基础。本文回顾了核心考点，包括表达式化简与展开、各类方程的求解、函数符号与运算、复合函数与反函数以及函数图像的分析。掌握这些内容能帮助你从容应对常规题和应用题，提升解题信心。
1. Algebraic Expressions and Simplification | 代数表达式与化简
An algebraic expression consists of terms combined by addition, subtraction, multiplication or division. Simplifying an expression means collecting like terms – those with identical variable parts. For example, 3x + 5y – 2x + y simplifies to x + 6y. Always pay attention to the signs and ensure you only combine terms that share exactly the same combination of variables and exponents.
代数表达式由加、减、乘、除连接的项组成。化简表达式就是合并同类项，即变量部分完全相同的项。例如，3x + 5y – 2x + y 可化简为 x + 6y。化简时要注意每一项的符号，并且只合并变量和指数完全一致的项。
Another key skill is applying index laws when variables are multiplied or divided. For instance, a² × a³ = a⁵ and (b²)³ = b⁶. When dividing, a⁵ ÷ a² = a³. Recognising these patterns allows swift simplification of longer expressions, reducing the risk of arithmetic errors.
另一项关键技能是运用指数法则处理变量的乘除。例如，a² × a³ = a⁵，(b²)³ = b⁶；做除法时，a⁵ ÷ a² = a³。掌握这些模式有助于快速化简更长的表达式，减少计算错误。
2. Expanding and Factorising | 展开与因式分解
Expanding brackets involves multiplying each term inside the bracket by the term outside. For example, 4(2x – 3) = 8x – 12. With double brackets, (x + 5)(x – 2) expands to x² – 2x + 5x – 10, which simplifies to x² + 3x – 10. You can use the FOIL (First, Outer, Inner, Last) method or inspection to keep track of all products.
展开括号要将括号外的项与括号内的每一项相乘。例如 4(2x – 3) = 8x – 12。对于两个括号相乘，(x + 5)(x – 2) 展开得到 x² – 2x + 5x – 10，合并后为 x² + 3x – 10。你可以使用 FOIL（首、外、内、末）法则或逐项观察来确保不漏乘。
Factorising is the reverse process. Start by looking for the highest common factor: in 6x² + 9x, both terms share 3x, giving 3x(2x + 3). For quadratics like x² + 7x + 12, find two numbers that multiply to 12 and add to 7, namely 3 and 4, so it factorises to (x + 3)(x + 4). Factorising helps solve equations and simplify algebraic fractions.
因式分解是展开的逆运算。先寻找最大公因式：在 6x² + 9x 中，两项都含有 3x，因此可分解为 3x(2x + 3)。对于二次式如 x² + 7x + 12，寻找两个数相乘得 12、相加得 7，即 3 和 4，于是分解为 (x + 3)(x + 4)。因式分解是解方程和化简代数分式的基础。
3. Solving Linear Equations | 解线性方程
A linear equation involves the variable raised to the power of 1. To solve, perform the same operation on both sides to isolate the unknown. For example, 2x – 5 = 9 → add 5: 2x = 14 → divide by 2: x = 7. Always check your solution by substituting it back into the original equation.
线性方程中的变量次数为 1。解法是对等式两边施行相同的运算，以求出未知数。例如 2x – 5 = 9 → 两边加 5：2x = 14 → 两边除以 2：x = 7。求出解后务必代回原方程检验。
When equations contain brackets or fractions, clear them first. For (x/3) + 2 = 5, subtract 2: x/3 = 3, then multiply by 3: x = 9. Equations with variables on both sides require moving terms: 5x + 2 = 3x + 10 → 5x – 3x = 10 – 2 → 2x = 8 → x = 4.
若方程含有括号或分式，应先去括号、去分母。例如 (x/3) + 2 = 5，先减 2 得 x/3 = 3，再乘 3 得 x = 9。两边都有变量的方程需要移项：5x + 2 = 3x + 10 → 5x – 3x = 10 – 2 → 2x = 8 → x = 4。
4. Formulae and Subject Change | 公式与换主项
A formula shows how two or more quantities are related. Changing the subject means rearranging the formula so that a different variable is isolated on one side. For example, in v = u + at, making a the subject: subtract u: v – u = at, then divide by t: a = (v – u) / t.
公式表达两个或多个量之间的关系。换主项就是重新整理公式，使另一个变量单独位于等号一侧。例如 v = u + at，将 a 变为主项：先减 u 得 v – u = at，再除以 t 得 a = (v – u) / t。
When the required variable appears in a denominator or under a root, undo the operation step by step. For E = ½mv², to make v the subject: multiply both sides by 2: 2E = mv², divide by m: 2E/m = v², then take the square root: v = √(2E/m). Remember to consider both ± roots when context requires.
若所求变量出现在分母或根号内，要逐步反向运算。例如 E = ½mv²，使 v 为主项：两边乘 2 得 2E = mv²，除以 m 得 2E/m = v²，再开平方得 v = √(2E/m)。在适当时要注意正负根号的取舍。
5. Linear Inequalities | 线性不等式
Inequalities are solved similarly to equations, but with one critical difference: if you multiply or divide by a negative number, you must reverse the inequality sign. Solve 3 – 2x ≤ 7: subtract 3: –2x ≤ 4, divide by –2 and flip the sign: x ≥ –2.
不等式的解法与方程类似，但有一个关键区别：如果两边乘以或除以一个负数，必须翻转不等号。解 3 – 2x ≤ 7：减 3 得 –2x ≤ 4，除以 –2 并翻转不等号得 x ≥ –2。
Solutions are often represented on a number line. A filled circle indicates ≥ or ≤, while an open circle shows > or
不等式的解常用数轴表示。实心圆点代表 ≥ 或 ≤，空心圆点代表 > 或
6. Quadratic Expressions and Equations | 二次表达式与方程
Quadratic expressions have the form ax² + bx + c. Solving quadratic equations can be done by factorising, using the quadratic formula, or completing the square. When factorising x² – 5x + 6 = 0, find numbers –2 and –3 because (–2) × (–3) = 6 and (–2) + (–3) = –5, giving (x – 2)(x – 3) = 0, so x = 2 or x = 3.
二次表达式的一般形式为 ax² + bx + c。解二次方程可以通过因式分解、公式法或配方法进行。例如解 x² – 5x + 6 = 0，因式分解时需要找到两个数相乘得 6、相加得 –5，即 –2 和 –3，所以 (x – 2)(x – 3) = 0，得 x = 2 或 x = 3。
The quadratic formula x = [–b ± √(b² – 4ac)] / 2a is essential when factorising is not straightforward. The discriminant Δ = b² – 4ac tells you the nature of roots: Δ > 0 gives two distinct real roots, Δ = 0 gives one repeated root, and Δ < 0 gives no real roots.
当无法直接因式分解时，可使用求根公式 x = [–b ± √(b² – 4ac)] / 2a。判别式 Δ = b² – 4ac 决定了根的性质：Δ > 0 时有两个不等实根，Δ = 0 时有一个重根，Δ < 0 时无实数根。
7. Simultaneous Equations | 联立方程
Simultaneous equations involve two (or more) unknowns and must be solved together. The elimination method works well when coefficients can be matched. For 2x + y = 7 and x – y = 2, add the equations: 3x = 9 → x = 3, then substitute back: 3 – y = 2 → y = 1.
联立方程包含两个（或多个）未知数，需同时求解。当系数容易匹配时，消元法十分有效。例如 2x + y = 7 和 x – y = 2，两式相加得 3x = 9 → x = 3，再代入得 3 – y = 2 → y = 1。
When neither variable has matching coefficients, multiply one or both equations before subtracting. For 3x + 2y = 12 and 4x – y = 5, multiply the second equation by 2: 8x – 2y = 10, then add to the first: 11x = 22 → x = 2, and y = 4 × 2 – 5 = 3. Always check both original equations.
若两个方程中变量系数都不匹配，需要先将一个或两个方程乘以适当的数，再进行加减。例如 3x + 2y = 12 和 4x – y = 5，将第二个方程乘以 2 得 8x – 2y = 10，与第一个相加得 11x = 22 → x = 2，再求得 y = 3。解完后一定要代回原方程检验。
8. Functions and Notation | 函数与记号
A function is a rule that assigns exactly one output for each input. The notation f(x) is read as ‘f of x’. For example, if f(x) = 2x + 3, then f(4) = 2(4) + 3 = 11. The letter inside the bracket is the input variable, and the expression defines how the output is calculated.
函数是一个规则，它为每一个输入值分配唯一的输出值。记号 f(x) 读作 ‘f of x’。例如，若 f(x) = 2x + 3，则 f(4) = 2(4) + 3 = 11。括号内的字母是输入变量，表达式则定义了如何计算输出值。
You can also use other letters, like g(t) = t² – 1. The domain is the set of all possible inputs, and the range is the set of all possible outputs. For f(x) = √(x – 2), the domain is x ≥ 2 because you cannot take the square root of a negative number.
也可以使用其他字母，如 g(t) = t² – 1。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。例如 f(x) = √(x – 2) 的定义域为 x ≥ 2，因为不能对负数开平方根。
9. Composite Functions | 复合函数
A composite function applies one function to the output of another. Given f(x) = 3x and g(x) = x + 4, the composite fg(x) means first apply g, then f: fg(x) = f(g(x)) = f(x + 4) = 3(x + 4) = 3x + 12. The order matters – gf(x) would be g(3x) = 3x + 4, which is different.
复合函数是将一个函数应用于另一个函数的输出。设 f(x) = 3x，g(x) = x + 4，则 fg(x) 表示先作用 g 再作用 f：fg(x) = f(g(x)) = f(x + 4) = 3(x + 4) = 3x + 12。顺序很重要，gf(x) = g(3x) = 3x + 4，结果不同。
When dealing with squares or fractions, carefully substitute. If f(x) = x² and g(x) = 2x – 1, then gf(x) = g(x²) = 2x² – 1, while fg(x) = f(2x – 1) = (2x – 1)² = 4x² – 4x + 1. Always write out the substitution completely to avoid mistakes.
处理平方或分式时，代入要格外仔细。如果 f(x) = x²，g(x) = 2x – 1，则 gf(x) = g(x²) = 2x² – 1，而 fg(x) = f(2x – 1) = (2x – 1)² = 4x² – 4x + 1。完整写出代入过程有助于避免错误。
10. Inverse Functions | 反函数
The inverse function reverses the effect of the original function. If f(x) = 4x – 5, to find f⁻¹(x), write y = 4x – 5, swap x and y: x = 4y – 5, then solve for y: x + 5 = 4y → y = (x + 5)/4. So f⁻¹(x) = (x + 5)/4. The composite ff⁻¹(x) should return x.
反函数的作用是逆转原函数的运算。若 f(x) = 4x – 5，求 f⁻¹(x) 的步骤是：设 y = 4x – 5，交换 x 与 y 得 x = 4y – 5，解出 y：x + 5 = 4y → y = (x + 5)/4，因此 f⁻¹(x) = (x + 5)/4。验证 ff⁻¹(x) 应得到 x。
Not all functions have inverses over their entire domain. A function must be one-to-one (each output from only one input). You can restrict the domain to make it invertible – for instance, f(x) = x² with domain x ≥ 0 has inverse f⁻¹(x) = √x.
并非所有函数在其整个定义域内都有反函数。函数必须是一一对应的（每个输出只对应一个输入）。可以通过限制定义域来获得反函数，例如 f(x) = x² 限制定义域为 x ≥ 0，其反函数为 f⁻¹(x) = √x。
11. Graphs of Functions | 函数图像
The graph of a function provides a visual representation of the relationship between x and f(x). For linear functions f(x) = mx + c, m is the gradient and c the y-intercept. To sketch, identify the y-intercept and use the gradient to find another point. Quadratic graphs y = ax² + bx + c are parabolas; the sign of a determines whether they open upwards (a > 0) or downwards (a < 0).
函数的图像直观地表示了 x 与 f(x) 之间的关系。对于线性函数 f(x) = mx + c，m 是斜率，c 是纵轴截距。画图时，先标出 y 轴截距，再根据斜率找到另一点。二次函数 y = ax² + bx + c 的图像为抛物线；a 的符号决定开口朝上 (a > 0) 还是朝下 (a < 0)。
Important features include x-intercepts (roots), the vertex (turning point), and the line of symmetry. The roots can be found by solving f(x) = 0. For y = x² – 4x + 3, factorising gives roots at x = 1 and x = 3; the vertex is halfway at x = 2, and substituting gives y = –1. Plotting these helps shape the parabola accurately.
重要的特征包括与 x 轴的交点（方程的根）、顶点（转折点）以及对称轴。通过解 f(x) = 0 可以找到根。对于 y = x² – 4x + 3，因式分解得根为 x = 1 和 x = 3；顶点位于中点 x = 2，代入得 y = –1。标出这些点有助于准确描绘抛物线。
12. Algebraic Fractions | 代数分式
Algebraic fractions are fractions with polynomials in the numerator and/or denominator. Simplify them by factorising and cancelling common factors. For (x² – 4)/(x – 2), the numerator factorises as (x – 2)(x + 2), then cancel (x – 2) to get x + 2, provided x ≠ 2.
代数分式是指分子和（或）分母中含有多项式的分数。通过因式分解并约去公因式进行化简。例如 (x² – 4)/(x – 2)，分子因式分解为 (x – 2)(x + 2)，约去 (x – 2) 得到 x + 2，但要注意 x ≠ 2。
When adding or subtracting, find a common denominator. For 1/(x + 1) + 2/(x – 1), the common denominator is (x + 1)(x – 1) = x² – 1. Write as (x – 1)/(x² – 1) + 2(x + 1)/(x² – 1) = (x – 1 + 2x + 2)/(x² – 1) = (3x + 1)/(x² – 1). Always state restrictions like x ≠ 1, –1.
进行加减运算时，要先通分。例如 1/(x + 1) + 2/(x – 1)，公分母为 (x + 1)(x – 1) = x² – 1。通分后为 (x – 1)/(x² – 1) + 2(x + 1)/(x² – 1) = (x – 1 + 2x + 2)/(x² – 1) = (3x + 1)/(x² – 1)。必须注明分母不为零的限制，如 x ≠ 1, –1。
Solving equations with algebraic fractions often involves multiplying both sides by the common denominator to eliminate fractions. For 2/(x) + 3/(x + 1) = 1, multiply by x(x + 1): 2(x + 1) + 3x = x(x + 1), then solve the resulting quadratic. Check for extraneous solutions.
解含有代数分式的方程时，通常两边同乘公分母以消去分母。例如 2/(x) + 3/(x + 1) = 1，两边乘 x(x + 1) 得 2(x + 1) + 3x = x(x + 1)，然后解所得的二次方程，并检验是否有增根。
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