ESAT Mathematics: Algebra and Functions Targeted Breakthrough | ESAT 数学：代数与函数专题突破

📚 ESAT Mathematics: Algebra and Functions Targeted Breakthrough | ESAT 数学：代数与函数专题突破

This article provides a focused revision of algebra and functions for the ESAT mathematics section. It follows the style of the exam, with concise techniques and worked examples so that you can apply them under timed conditions.

本文为 ESAT 数学部分的代数与函数内容提供针对性复习。我们按照考试风格，通过简明技巧和例题帮助你快速突破，并在限时条件下准确运用。

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1. Core Algebraic Manipulation | 代数核心运算

Algebraic fluency underpins every ESAT question. You must be able to expand, factorise and simplify expressions quickly and without a calculator.

代数运算熟练度是所有 ESAT 题目的基础。你必须能够快速且准确地展开、因式分解和化简表达式，并且不依赖计算器。

You should know the three most useful identities by heart:

你应该牢记以下三个最常用的公式：

(a + b)² = a² + 2ab + b²

(a − b)² = a² − 2ab + b²

(a + b)(a − b) = a² − b²

Example: Simplify (x + 3)² − (x − 2)(x + 2). Expanding gives (x² + 6x + 9) − (x² − 4) = 6x + 13. Always arrange terms carefully to avoid sign errors.

例如：化简 (x + 3)² − (x − 2)(x + 2)。展开得 (x² + 6x + 9) − (x² − 4) = 6x + 13。排列各项时务必仔细，避免符号错误。

When factorising, always take out the greatest common factor first, then look for a difference of squares or a quadratic trinomial that can be written as a product of two binomials.

因式分解时，永远先提公因式，再观察是否为平方差或可以写成两个二项式乘积的二次三项式。

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2. Quadratic Equations and Discriminant | 二次方程与判别式

Quadratic equations appear constantly in ESAT. The solution to ax² + bx + c = 0 is found by the quadratic formula:

二次方程在 ESAT 中反复出现。方程 ax² + bx + c = 0 的解由求根公式给出：

x = (−b ± √(b² − 4ac)) / 2a

The discriminant is Δ = b² − 4ac. If Δ > 0, there are two distinct real roots. If Δ = 0, there is one repeated root. If Δ < 0, there are no real roots.

判别式为 Δ = b² − 4ac。若 Δ > 0，则有两个不同实根；若 Δ = 0，则有一个重根；若 Δ < 0，则没有实根。

For a quadratic with roots α and β, the sum of roots is α + β = −b/a and the product is αβ = c/a. These relationships can solve problems that ask for expressions involving the roots.

对于根为 α 和 β 的二次方程，两根之和满足 α + β = −b/a，两根之积满足 αβ = c/a。这些关系可用于求解涉及根的表达式。

Example: Find the values of k for which x² + kx + 9 = 0 has equal roots. The condition is Δ = k² − 36 = 0, so k = ±6.

例如：求使 x² + kx + 9 = 0 有等根的 k 值。条件为 Δ = k² − 36 = 0，所以 k = ±6。

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3. Inequalities and Sign Diagrams | 不等式与符号图

Linear inequalities are solved like equations, but you must reverse the inequality sign when multiplying or dividing by a negative number.

线性不等式的解法与方程类似，但在乘以或除以负数时，必须改变不等号方向。

Quadratic inequalities are best solved by factorising and drawing a sign diagram. For example, solve x² − 5x + 6 < 0. Factorise to (x − 2)(x − 3) < 0.

二次不等式通常通过因式分解和画符号图来求解。例如，解 x² − 5x + 6 < 0。因式分解得 (x − 2)(x − 3) < 0。

Interval	x < 2	2 < x < 3	x > 3
(x − 2)	−	+	+
(x − 3)	−	−	+
Product	+	−	+

Because the product must be negative, the solution is 2 < x < 3.

因为乘积必须为负，所以解为 2 < x < 3。

Remember to write the final answer in set notation or interval notation if the question asks for it.

如果题目要求，请用集合记号或区间记号写出最终答案。

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4. Simultaneous Equations | 联立方程

ESAT often combines a linear equation with a quadratic equation. The usual method is substitution: solve the linear equation for one variable and substitute into the quadratic.

ESAT 常将一次方程与二次方程联立。常用方法是代入消元：从一次方程中解出一个变量，再代入二次方程。

Example: Solve y = 2x + 1 and y = x² + x − 1 together. Substitute the linear expression into the quadratic:

例如：联立 y = 2x + 1 与 y = x² + x − 1。将一次表达式代入二次方程：

2x + 1 = x² + x − 1

This simplifies to x² − x − 2 = 0, so (x − 2)(x + 1) = 0. Hence x = 2 gives y = 5, and x = −1 gives y = −1.

化简得 x² − x − 2 = 0，所以 (x − 2)(x + 1) = 0。因此 x = 2 时 y = 5，x = −1 时 y = −1。

Graphically, the discriminant of the substituted quadratic tells you how many intersection points exist: two, one or zero.

从图形角度看，代入后所得二次方程的判别式决定了交点数量：两个、一个或零个。

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5. Functions: Notation and Evaluation | 函数：记号与求值

A function is a rule that assigns exactly one output to each valid input. The notation f(x) is read as “f of x”.

函数是将每个有效输入唯一对应到一个输出的规则。记号 f(x) 读作“x 的 f”。

For example, if f(x) = 2x² − 3, then f(2) = 2(2)² − 3 = 5, and f(−1) = 2(−1)² − 3 = −1.

例如，若 f(x) = 2x² − 3，则 f(2) = 2(2)² − 3 = 5，f(−1) = 2(−1)² − 3 = −1。

Given a graph, you can use the vertical line test: a graph represents a function if every vertical line cuts it at no more than one point.

对于图像，可以使用竖线检验：若任何竖直线与图像相交不超过一个点，则该图像表示一个函数。

Be careful when evaluating piecewise functions: choose the correct branch based on the value of x.

分段函数求值时务必小心：根据 x 的值选择正确的分支。

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6. Domain and Range | 定义域与值域

The domain is the set of all possible

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