📚 Mastering Quadratic Equations and Graphs | 掌握二次方程与图像
Quadratic equations and their graphs are central to the IGCSE Mathematics syllabus. They appear in algebra, coordinate geometry, and real-life modelling questions. This revision article explains the key methods, common exam techniques, and graphical interpretation needed to answer questions confidently.
二次方程及其图像是 IGCSE 数学大纲中的核心内容。它们出现在代数、坐标几何和实际建模题中。本复习文章将讲解关键方法、常见考试技巧和图像解读，帮助你自信地解答相关题目。
1. What Is a Quadratic Expression? | 什么是二次表达式？
A quadratic expression is an algebraic expression of degree 2. Its standard form is ax² + bx + c, where a, b and c are constants and a ≠ 0.
二次表达式是一个次数为 2 的代数式。它的标准形式为 ax² + bx + c，其中 a、b、c 是常数，且 a ≠ 0。
y = ax² + bx + c
constant term. The condition a ≠ 0 is essential: if a = 0, the expression becomes linear.
ax² 称为二次项，bx 称为一次项，c 称为常数项。a ≠ 0 这个条件非常重要：如果 a = 0，表达式就变成了一次式。
For example, x² + 5x + 6, 2x² - 3x + 1 and -x² + 4 are all quadratic expressions.
例如，x² + 5x + 6、2x² - 3x + 1 和 -x² + 4 都是二次表达式。
2. Expanding Double Brackets | 展开双重括号
Expanding is often the first step when simplifying a quadratic expression. To expand (x + p)(x + q), multiply each term in the first bracket by each term in the second bracket.
展开通常是化简二次表达式的第一步。要展开 (x + p)(x + q)，需要把第一个括号中的每一项分别乘以第二个括号中的每一项。
(x + p)(x + q) = x² + px + qx + pq = x² + (p + q)x + pq
For example, expand (x + 2)(x - 3). Multiply x by x and -3, then multiply 2 by x and -3.
例如，展开 (x + 2)(x - 3)。先用 x 乘以 x 和 -3，再用 2 乘以 x 和 -3。
(x + 2)(x - 3) = x² - 3x + 2x - 6 = x² - x - 6
The same process works when the coefficient of x is not 1, such as (2x + 1)(x - 4).
当 x 的系数不是 1 时，同样的方法也适用，例如 (2x + 1)(x - 4)。
(2x + 1)(x - 4) = 2x² - 8x + x - 4 = 2x² - 7x - 4
3. Factorising Quadratic Expressions | 二次式的因式分解
ses expansion. When a = 1, look for two numbers that multiply to give c and add to give b.
因式分解是展开的逆运算。当 a = 1 时，需要找到两个数，它们的乘积等于 c，和等于 b。
To factorise x² + 7x + 10, find two numbers whose product is 10 and sum is 7. The numbers are 2 and 5.
对 x² + 7x + 10 进行因式分解，需要找到乘积为 10、和为 7 的两个数。这两个数是 2 和 5。
x² + 7x + 10 = (x + 2)(x + 5)
For x² - 2x - 8, the two numbers must multiply to -8 and add to -2, so they are -4 and 2.
对于 x² - 2x - 8，这两个数必须乘积为 -8，和为 -2，因此它们是 -4 和 2。
x² - 2x - 8 = (x - 4)(x + 2)
When a ≠ 1, factorising requires more care. For 2x² + 7x + 3, split the middle term or use trial and error to obtain (2x + 1)(x + 3).
当 a ≠ 1 时，因式分解需要更加仔细。对于 2x² + 7x + 3，可以拆分中间项或通过试错法得到 (2x + 1)(x + 3)。
4. Solving Quadratic Equations by Factorisation | 因式分解法解二次方程
If the left side can be factorised, the zero product property gives the solutions easily.
二次方程的形式为 ax² + bx + c = 0。如果左边可以因式分解，零乘积性质就能快速求出解。
If AB = 0, then A = 0 or B = 0.
For example, solve x² - 5x + 6 = 0. Factorise to get (x - 2)(x - 3) = 0.
例如，解方程 x² - 5x + 6 = 0。因式分解得到 (x - 2)(x - 3) = 0。
(x - 2)(x - 3) = 0 → x - 2 = 0 or x - 3 = 0 → x = 2 or x = 3
For example, x² + 2x = 8 becomes x² + 2x - 8 = 0, which factorises to (x + 4)(x - 2) = 0.
在因式分解之前，一定要把方程整理为等于 0 的形式。例如，x² + 2x = 8 应先化为 x² + 2x - 8 = 0，再因式分解为 (x + 4)(x - 2) = 0。
5. Completing the Square | 配方法
Completing the square transforms a quadratic expression into the form a(x - h)² + k. This method is useful for solving equations and finding graph features.
配方法可以把二次表达式转化为 a(x - h)² + k 的形式。这种方法在解方程和寻找图像特征时非常有用。
To complete the square for x² + 6x + 1, halve the coefficient of x to get 3, square it to get 9, and adjust the constant term.
对 x² + 6x + 1 配方：将 x 的系数除以 2 得到 3，平方得到 9，然后调整常数项。
x² + 6x + 1 = (x + 3)² - 9 + 1 = (x + 3)² - 8
To solve x² + 6x + 1 = 0, set (x + 3)² - 8 = 0. Then (x + 3)² = 8, so x + 3 = ±√8.
解方程 x² + 6x + 1 = 0 时，令 (x + 3)² - 8 = 0。于是 (x + 3)² = 8，所以 x + 3 = ±√8。
x = -3 ± √8 = -3 ± 2√2
The completed square form also reveals the turning point immediately: for (x + 3)² - 8, the vertex is (-3, -8).
配方法还可以直接显示顶点：对于 (x + 3)² - 8，顶点坐标为 (-3, -8)。
6. The Quadratic Formula | 二次公式
mula solves any quadratic equation ax² + bx + c = 0, even when factorisation is difficult.
二次公式可以解任意一个二次方程 ax² + bx + c = 0，即使因式分解很困难。
x = (-b ± √b² - 4ac) / 2a
For example, solve 2x² + 3x - 2 = 0. Here a = 2, b = 3, c = -2.
例如，解方程 2x² + 3x - 2 = 0。这里 a = 2，b = 3，c = -2。
x = (-3 ± √(3² - 4 × 2 × (-2))) / (2 × 2) = (-3 ± √25) / 4 = (-3 ± 5) / 4
This gives x = 2/4 = 1/2 or x = -8/4 = -2.
因此 x = 2/4 = 1/2 或 x = -8/4 = -2。
Remember to write the formula clearly and substitute the values of a, b and c with their signs.
记住要清晰地写出公式，并代入带有正负号的 a、b、c 值。
7. The Discriminant | 判别式
The discriminant is the part of the quadratic formula under the square root: b² - 4ac. It tells us how many real solutions an equation has.
判别式是二次公式中平方根下的部分：b² - 4ac。它告诉我们方程有多少个实数解。
Δ = b² - 4ac
| Discriminant | Number of real roots | Graph behaviour |
|---|---|---|
| Δ > 0 | 2 distinct real roots | Parabola crosses the x-axis twice |
| Δ = 0 | 1 repeated real root | Parabola touches the x-axis at one point |
| Δ < 0 | No real roots | Parabola does not meet the x-axis |
For example, x² - 4x + 4 = 0 has Δ = (-4)² - 4 × 1 × 4 = 0, so it has one repeated root x = 2.
例如，x² - 4x + 4 = 0 的判别式 Δ = (-4)² - 4 × 1 × 4 = 0，因此它有一个重根 x = 2。
8. Graphs of Quadratic Functions | 二次函数图像
. If a > 0, the parabola opens upwards and has a minimum point. If a < 0, it opens downwards and has a maximum point.
y = ax² + bx + c 的图像是一条抛物线。如果 a > 0，抛物线开口向上，有最小值点；如果 a < 0，抛物线开口向下，有最大值点。
is c, because when x = 0, y = c. The x-intercepts are the real roots of ax² + bx + c = 0.
y 轴截距为 c，因为当 x = 0 时，y = c。x 轴截距就是方程 ax² + bx + c = 0 的实根。
(x - 1)(x - 3), so the curve crosses the x-axis at x = 1 and x = 3. The y-intercept is 3.
画 y = x² - 4x + 3 的图像时，可以先因式分解求出 x 轴截距：y = (x - 1)(x - 3)，所以曲线在 x = 1 和 x = 3 处穿过 x 轴。y 轴截距为 3。
line halfway between the roots, x = 2. Substituting x = 2 gives the minimum point (2, -1).
对称轴是两根中间位置的竖直线 x = 2。代入 x = 2 可得到最小值点 (2, -1)。
9. Turning Points and Symmetry | 顶点与对称性
nt, also called the vertex. The x-coordinate of the turning point is given by x = -b / 2a.
每个二次函数图像都有一个顶点，也称为转折点。顶点的 x 坐标由 x = -b / 2a 给出。
x = -b / 2a, y = f(-b / 2a)
x = -8 / (2 × 2) = -2. Substituting x = -2 gives y = 2(-2)² + 8(-2) + 5 = 8 - 16 + 5 = -3.
对于 y = 2x² + 8x + 5，a = 2，b = 8，所以 x = -8 / (2 × 2) = -2。代入 x = -2 得到 y = 2(-2)² + 8(-2) + 5 = 8 - 16 + 5 = -3。
The turning point is (-2, -3). The axis of symmetry is the vertical line x = -2.
顶点坐标为 (-2, -3)。对称轴是竖直线 x = -2。
eting the square is another fast route to the vertex: y = a(x - h)² + k has vertex (h, k).
配方法也是求顶点的另一种快捷方法：y = a(x - h)² + k 的顶点为 (h, k)。
10. Real-Life Applications | 实际应用
Quadratic models appear in area problems, projectile motion, and business profit questions. A typical IGCSE question asks you to form a quadratic equation from a word problem and then solve it.
二次模型出现在面积问题、抛体运动和商业利润问题中。典型的 IGCSE 题目会要求你根据文字题建立二次方程并求解。
For example, a rectangle has length 2 m more than its width, and its area is 48 m². Let the width be x m, so the length is x + 2 m.
例如
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