📚 IGCSE Maths Teacher's Guide 600: Quadratics in Depth | IGCSE 数学教师指南 600：深度解析二次函数
A strong command of quadratic expressions, equations and graphs is essential for success in IGCSE Mathematics. This teacher-focused guide breaks down the key teaching points, common errors and effective classroom strategies for the topic of quadratics.
扎实掌握二次表达式、二次方程及其图像是 IGCSE 数学取得高分的关键。本教师指南详细梳理了二次函数部分的核心教学要点、常见错误以及有效的课堂教学策略。
1. Understanding Quadratic Expressions | 理解二次表达式
s and a ≠ 0. If a = 0, the expression becomes linear, so the condition a ≠ 0 is essential.
二次表达式是形如 ax² + bx + c 的式子，其中 a、b、c 是常数且 a ≠ 0。如果 a = 0，式子就变成了一次表达式，因此 a ≠ 0 这一条件至关重要。
When expanding two linear brackets such as (x + p)(x + q), the result is x² + (p + q)x + pq. This relationship is the reverse of factorising and helps students see why the middle coefficient is a sum and the constant term is a product.
将两个一次括号相乘，例如 (x + p)(x + q)，得到 x² + (p + q)x + pq。这一关系正是因式分解的逆过程，能帮助学生理解为什么中间系数是和，而常数项是积。
2. Factorising Quadratics with a = 1 | 首项系数为 1 的二次式因式分解
When the coefficient of x² is 1, factorising x² + bx + c means finding two numbers whose product is c and whose sum is b. For example, to factorise x² + 7x + 12, we need two numbers that multiply to 12 and add to 7, namely 3 and 4.
当 x² 的系数为 1 时，对 x² + bx + c 进行因式分解，就是找到两个数，使它们的乘积为 c，和为 b。例如，分解 x² + 7x + 12 时，需要找到两个积为 12、和为 7 的数，即 3 和 4。
eachers should emphasise checking by expanding the brackets, as this builds confidence and reduces sign errors.
因此，x² + 7x + 12 = (x + 3)(x + 4)。教师应强调通过展开括号来进行检验，这能建立学生信心并减少符号错误。
3. Factorising Quadratics with a ≠ 1 | 首项系数不为 1 的二次式因式分解
When a ≠ 1, the factorisation process is more demanding. One reliable method is the ac method: multiply a and c, find two numbers with that product and sum b, then split the middle term and factor by grouping.
当 a ≠ 1 时，因式分解过程要求更高。一种可靠的方法是 ac 法：先将 a 与 c 相乘，找到两个数使它们的积为 ac、和为 b，然后拆分中间项并进行分组分解。
and add to 7, so 2x² + 7x + 3 = 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3).
例如，分解 2x² + 7x + 3 时，将 2 和 3 相乘得 6。数字 6 和 1 的积为 6、和为 7，所以 2x² + 7x + 3 = 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)。
4. Solving Quadratic Equations by Factorising | 利用因式分解解二次方程
zero. If (x + 3)(x + 4) = 0, then either x + 3 = 0 or x + 4 = 0, giving x = −3 or x = −4.
二次表达式完成因式分解后，就可以令每个因式等于零来解方程。若 (x + 3)(x + 4) = 0，则 x + 3 = 0 或 x + 4 = 0，得到 x = −3 或 x = −4。
This method works because if the product of two numbers is zero, at least one of them must be zero. Teachers should remind students never to divide both sides by an unknown factor, as this can lose a valid solution.
这一方法成立的原因是：如果两个数的乘积为零，那么其中至少有一个数必须为零。教师应提醒学生切勿在方程两边同时除以未知因式，因为这样可能会丢失有效解。
5. Completing the Square | 配方法
Completing the square rewrites a quadratic expression in the form (x + p)² + q. This is particularly useful for finding the vertex of a graph and for deriving the quadratic formula.
配方法可将二次表达式改写为 (x + p)² + q 的形式。这种方法在求图像顶点以及推导求根公式时特别有用。
For x² + 6x + 5, we take half of 6, which is 3, and write (x + 3)² − 9 + 5, simplifying to (x + 3)² − 4. The vertex is therefore at (−3, −4).
对于 x² + 6x + 5，取 6 的一半，即 3，写成 (x + 3)² − 9 + 5，化简为 (x + 3)² − 4。因此顶点坐标为 (−3, −4)。
x² + bx + c = (x + b/2)² − (b/2)² + c
The centred formula above summarises the general completing-the-square process. It is a useful reference point for learners who struggle with the algebraic steps.
上面居中的公式概括了一般的配方法过程。对于掌握代数步骤有困难的学生，这是一个很有用的参考公式。
6. The Quadratic Formula | 求根公式
. It is particularly helpful when factorisation is difficult or impossible using integers.
求根公式可以求解任何形如 ax² + bx + c = 0 的方程。当因式分解难以进行或无法用整数分解时，求根公式尤其有用。
x = (−b ± √b² − 4ac) ÷ 2a
c before substituting. For example, in 2x² + 3x − 2 = 0, we have a = 2, b = 3 and c = −2.
学生应先写下 a、b、c 的值再代入。例如在 2x² + 3x − 2 = 0 中，a = 2，b = 3，c = −2。
Substitution gives x = (−3 ± √(3² − 4 × 2 × (−2))) ÷ (2 × 2) = (−3 ± √9 + 16) ÷ 4 = (−3 ± 5) ÷ 4, so x = 1/2 or x = −2.
代入后得到 x = (−3 ± √(3² − 4 × 2 × (−2))) ÷ (2 × 2) = (−3 ± √9 + 16) ÷ 4 = (−3 ± 5) ÷ 4，所以 x = 1/2 或 x = −2。
7. The Discriminant and the Nature of Roots | 判别式与根的性质
. It tells us how many real roots a quadratic equation has without requiring full solving.
判别式是求根公式中平方根号下的部分：D = b² − 4ac。它可以在不完整求解的情况下判断二次方程有多少个实根。
- If D > 0, the equation has two distinct real roots.
- If D = 0, the equation has one repeated real root.
- If D < 0, the equation has no real roots.
- 如果 D > 0，方程有两个不同的实根。
- 如果 D = 0，方程有一个重复实根。
- 如果 D < 0，方程没有实根。
Teachers can use the discriminant to help students check their answers quickly. If a student obtains two real roots but D < 0, they know an error has occurred.
教师可以利用判别式帮助学生快速检查答案。如果学生得出两个实根，但 D < 0，就说明计算过程中出现了错误。
8. Graphs of Quadratic Functions | 二次函数的图像
. When a > 0, the parabola opens upwards and has a minimum point. When a < 0, it opens downwards and has a maximum point.
y = ax² + bx + c 的图像是一条抛物线。当 a > 0 时，抛物线开口向上，图像有最低点。当 a < 0 时，抛物线开口向下，图像有最高点。
= c. The x-intercepts are the roots of the equation ax² + bx + c = 0, if real roots exist.
图像的 y 截距为 c，因为当 x = 0 时，y = c。x 截距就是方程 ax² + bx + c = 0 的根，前提是存在实根。
Encourage students to sketch parabolas by first identifying the y-intercept, then the roots, and finally the turning point. This gives a reliable framework for accurate graphs.
鼓励学生画抛物线草图时，先确定 y 截距，再确定根，最后确定转折点。这样能提供可靠的分析框架，有助于准确作图。
9. Turning Points and Symmetry | 转折点与对称性
a quadratic graph lies on the axis of symmetry. Its x-coordinate is given by x = −b ÷ 2a.
二次函数图像的转折点，即顶点，位于对称轴上。其 x 坐标由 x = −b ÷ 2a 给出。
e is −(−4) ÷ 2 = 2, and the y-coordinate is 2² − 4 × 2 + 1 = −3, so the vertex is (2, −3).
知道了 x 坐标后，将其代入原方程即可求出 y 坐标。对于 y = x² − 4x + 1，x 坐标为 −(−4) ÷ 2 = 2，y 坐标为 2² − 4 × 2 + 1 = −3，因此顶点为 (2, −3)。
This symmetry means that if two points on the graph have the same y-value, their x-coordinates average to the x-coordinate of the vertex. This is a useful check when plotting.
这种对称性意味着，如果图像上有两个点的 y 值相同，那么它们 x 坐标的平均值就是顶点的 x 坐标。这在作图时是非常有用的检验方法。
10. Quadratic Inequalities | 二次不等式
0, first factorise to get (x − 2)(x − 3) > 0. The critical values are x = 2 and x = 3.
解二次不等式，例如 x² − 5x + 6 > 0，首先因式分解得到 (x − 2)(x − 3) > 0。临界值为 x = 2 和 x = 3。
Test values in the intervals x < 2, 2 < x < 3 and x > 3. For x < 2, both factors are negative, so their product is positive. For 2 < x < 3, one factor is positive and one is negative. For x > 3, both are positive.
在区间 x < 2、2 < x < 3 和 x > 3 中代入检验值。当 x < 2 时，两个因式都为负，乘积为正。当 2 < x < 3 时，一个因式为正，一个为负。当 x > 3 时，两个因式都为正。
Therefore, the solution is x < 2 or x > 3. Teachers should encourage sketching the parabola to visualise where the curve lies above or below the x-axis.
因此，解集为 x < 2 或 x > 3。教师应鼓励学生画出抛物线草图，直观观察曲线在 x 轴上方或下方所在的位置。
11. Applications of Quadratics | 二次函数的实际应用
Quadratic equations commonly model projectile motion, area problems, product pricing and many optimisation situations. A typical area problem might ask for the dimensions of a rectangle given its area and perimeter.
二次方程常用于描述抛体运动、面积问题、产品定价以及许多最优化情形。一个典型的面积问题可能会给出矩形的面积和周长，要求其边长。
ten be modelled by h = −5t² + vt + h₀. Solving h = 0 gives the time when the object lands.
在抛体运动中，物体在 t 秒后的高度 h 通常可以用 h = −5t² + vt + h₀ 来建模。解方程 h = 0 即可得到物体落地的时间。
When solving real-world problems, students must interpret answers in context. A negative time or length is usually not valid, even if it is a mathematically correct root of the equation.
在解决实际问题时，学生必须结合情境解读答案。负的时间或长度通常是不合理的，即使它是方程在数学意义上的正确根。
12. Common Misconceptions and Exam Tips | 常见误区与考试技巧
y x, which loses the root x = 0. Always factorise first and set each factor equal to zero.
许多学生忘记二次方程可以有两个解。他们有时会在方程两边同时除以 x，从而丢失 x = 0 这个根。务必先进行因式分解，再令每个因式等于零。
Another common error is mishandling signs when using the quadratic formula, especially when b or c is negative. Writing down the values of a, b and c clearly reduces this risk.
另一个常见错误是在使用求根公式时处理符号不当，特别是在 b 或 c 为负数时。清晰地写出 a、b、c 的值可以降低出错风险。
In exams, marks are often awarded for method as well as final answers. Show clear substitution, factorisation or graph-sketching steps, and always check solutions by substituting them back into the original equation.
考试中，通常不仅对最终答案给分，也对解题过程给分。要清晰地展示代入、因式分解或作图步骤，并始终把解代回原方程进行检查。
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