📚 IGCSE Maths Teacher's Book 700: Quadratic Equations and Graphs | IGCSE 数学教师用书 700：二次方程与图像
Welcome to this IGCSE Maths revision article based on the Teacher's Book 700 topic. Quadratic expressions, equations and graphs appear in both the Core and Extended syllabuses, and they are a frequent source of marks because they connect algebra, coordinate geometry and problem solving. This article explains the most important ideas in a structured way, with examples, formulas and exam tips.
欢迎阅读这篇基于教师用书 700 主题的 IGCSE 数学复习文章。二次表达式、二次方程和二次图像同时出现在核心课程和扩展课程中，并且因为连接代数、坐标几何与问题解决，常常是得分来源。本文以结构化方式讲解最重要的概念，并提供例题、公式和考试技巧。
1. Recognizing Quadratic Functions | 识别二次函数
A quadratic expression in x has the standard form ax² + bx + c, where a, b and c are constants and a ≠ 0. If a were zero, the x² term would disappear and the expression would become linear, so the condition a ≠ 0 is essential. In IGCSE questions, you should be able to identify quadratic expressions such as 2x² + 3x - 5, 4 - x - x² and x² - 9.
关于 x 的二次表达式标准形式为 ax² + bx + c，其中 a、b、c 为常数且 a ≠ 0。如果 a 为零，x² 项会消失，表达式就变成一次式，因此 a ≠ 0 的条件非常重要。在 IGCSE 题目中，你应能识别诸如 2x² + 3x - 5、4 - x - x² 和 x² - 9 这样的二次表达式。
atic function can also be written as y = ax² + bx + c, and its graph is always a parabola.
当二次表达式等于零时，就形成二次方程：ax² + bx + c = 0。方程的解称为根。二次函数也可以写作 y = ax² + bx + c，它的图像总是一条抛物线。
2. The Shape of a Parabola | 抛物线的形状
a maximum turning point. The value of c gives the y-intercept, because when x = 0, y = c.
y = ax² + bx + c 的图像是一条称为抛物线的曲线。如果 a > 0，抛物线开口向上，具有最低点；如果 a < 0，开口向下，具有最高点。c 的值给出 y 轴截距，因为当 x = 0 时，y = c。
value makes it wider. For example, y = 3x² is narrower than y = x², and y = ½x² is wider.
系数 a 还控制抛物线的宽度。a 的绝对值越大，抛物线越窄；绝对值越小，抛物线越宽。例如，y = 3x² 比 y = x² 窄，而 y = ½x² 比 y = x² 宽。
3. Solving by Factorising | 因式分解法求解
= 0 becomes (x - 2)(x - 3) = 0, so x = 2 or x = 3. Always check by expanding the brackets.
对于可因式分解的二次式，将表达式写成 (px + q)(rx + s) 的形式，并令每个括号等于零。例如，x² - 5x + 6 = 0 变为 (x - 2)(x - 3) = 0，因此 x = 2 或 x = 3。务必通过展开括号进行检查。
- Step 1: Write the quadratic in standard form. 第一步：将二次式写成标准形式。
- Step 2: Find two numbers that multiply to ac and add to b. 第二步：找到两个数，使它们的积为 ac，和为 b。
- Step 3: Split the middle term and factorise by grouping. 第三步：拆分中间项并分组因式分解。
- Step 4: Set each factor equal to zero and solve. 第四步：令每个因式等于零并求解。
4. The Quadratic Formula | 二次公式
the quadratic formula. For ax² + bx + c = 0, the solutions are given by the formula below.
当无法因式分解时，使用二次公式。对于 ax² + bx + c = 0，解由以下公式给出。
x = (-b ± √b² - 4ac) ÷ 2a
3x - 2 = 0, we have a = 2, b = 3 and c = -2. Substituting into the formula gives x = (-3 ± √9 + 16) ÷ 4 = (-3 ± √25) ÷ 4 = (-3 ± 5) ÷ 4. Therefore x = 2 ÷ 4 = 1/2 or x = -8 ÷ 4 = -2.
以 2x² + 3x - 2 = 0 为例，a = 2，b = 3，c = -2。代入公式得到 x = (-3 ± √9 + 16) ÷ 4 = (-3 ± √25) ÷ 4 = (-3 ± 5) ÷ 4。因此 x = 2 ÷ 4 = 1/2 或 x = -8 ÷ 4 = -2。
5. Completing the Square | 配方法
is 3, so x² + 6x + 5 = (x + 3)² - 9 + 5 = (x + 3)² - 4. The vertex is therefore (-3, -4).
配方法将 ax² + bx + c 改写为 a(x + p)² + q 的形式。这种形式对于求顶点和解方程特别有用。以 x² + 6x + 5 为例，x 的系数的一半为 3，所以 x² + 6x + 5 = (x + 3)² - 9 + 5 = (x + 3)² - 4。因此顶点为 (-3, -4)。
The general form for completing the square when a = 1 is shown below.
当 a = 1 时，配方法的一般形式如下所示。
x² + bx + c = (x + b/2)² - (b/2)² + c
t. For example, 2x² + 8x + 3 becomes 2(x² + 4x) + 3 = 2[(x + 2)² - 4] + 3 = 2(x + 2)² - 5.
如果 a 不等于 1，先提取 a。例如，2x² + 8x + 3 变为 2(x² + 4x) + 3 = 2[(x + 2)² - 4] + 3 = 2(x + 2)² - 5。
6. The Discriminant | 判别式
e of the roots without solving the equation. This is useful when a question asks about the number of solutions.
判别式是二次公式中根号下的部分：D = b² - 4ac。它可以在不求解方程的情况下判断根的性质。当题目要求判断解的数量时，这非常有用。
| Discriminant D = b² - 4ac | Nature of roots | 根的性质 |
|---|---|
| D > 0 | Two distinct real roots | 两个不同的实根 |
| D = 0 | One repeated real root | 一个重根 |
| D < 0 | No real roots | 没有实根 |
- 4 × 1 × 5 = 16 - 20 = -4, so it has no real roots. Its graph does not cross the x-axis.
例如，x² - 4x + 5 = 0 的判别式 D = (-4)² - 4 × 1 × 5 = 16 - 20 = -4，因此它没有实根。它的图像不与 x 轴相交。
7. Sketching Quadratic Graphs | 绘制二次函数图像
To sketch y = ax² + bx + c, find the y
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