📚 Quadratic Functions and Equations | 二次函数与方程
Quadratics appear in nearly every IGCSE Mathematics paper, whether as pure algebra, graph sketching, or a word problem at the end of Paper 4. A quadratic expression has the general form ax² + bx + c, where a, b and c are constants and a ≠ 0. In this revision guide we will connect every topic you need: expanding, factorising, solving, graphing, inequalities and applications.
二次函数几乎出现在每一份IGCSE数学试卷中，无论是纯粹的代数、图像绘制，还是Paper 4最后的应用题。二次表达式的标准形式为ax² + bx + c，其中a、b、c为常数，且a ≠ 0。本修订指南将把您需要掌握的所有内容串联起来：展开、因式分解、求解、作图、不等式和应用题。
1. Expanding Double Brackets | 展开双括号
The first skill is expanding (x + p)(x + q). Use the distributive law, often remembered as FOIL (First, Outer, Inner, Last). For example, (x + 3)(x + 5) = x² + 5x + 3x + 15 = x² + 8x + 15. Notice that the coefficient of x is 3 + 5 = 8, and the constant term is 3 × 5 = 15. This pattern is the key to reverse factorising later.
第一个技能是展开(x + p)(x + q)。使用乘法分配律，常记为FOIL（First, Outer, Inner, Last，即首项、外项、内项、末项）。例如，(x + 3)(x + 5) = x² + 5x + 3x + 15 = x² + 8x + 15。注意x的系数是3 + 5 = 8，常数项是3 × 5 = 15。这一规律是后续反向因式分解的关键。
When a coefficient is not 1, such as (2x + 1)(x − 4), multiply term by term: 2x × x = 2x², 2x × (−4) = −8x, 1 × x = x, and 1 × (−4) = −4. Combining like terms gives 2x² − 7x − 4. Special cases include the difference of two squares: (x + a)(x − a) = x² − a², and the perfect square: (x + a)² = x² + 2ax + a².
当系数不为1时，例如(2x + 1)(x − 4)，逐项相乘：2x × x = 2x²，2x × (−4) = −8x，1 × x = x，1 × (−4) = −4。合并同类项得2x² − 7x − 4。特殊情况包括平方差公式：(x + a)(x − a) = x² − a²，以及完全平方公式：(x + a)² = x² + 2ax + a²。
2. Factorising Quadratics | 因式分解二次式
Factorising is the reverse process. To factorise x² + 5x + 6, find two numbers that multiply to 6 and add to 5. The numbers are 2 and 3, so x² + 5x + 6 = (x + 2)(x + 3). Always check by expanding. If the constant term is negative, the two numbers have opposite signs. For x² − x − 12, we need two numbers that multiply to −12 and add to −1; those are 3 and −4, so x² − x − 12 = (x + 3)(x − 4).
因式分解是展开的逆过程。要分解x² + 5x + 6，找到两个乘积为6且和为5的数。这两个数是2和3，所以x² + 5x + 6 = (x + 2)(x + 3)。务必通过展开来验证。如果常数项为负，则两个数符号相反。对于x² − x − 12，我们需要两个乘积为−12且和为−1的数，即3和−4，所以x² − x − 12 = (x + 3)(x − 4)。
When the coefficient of x² is not 1, use the method of factoring by grouping or trial and error. For 2x² + 7x + 3, multiply a × c = 2 × 3 = 6. Find two numbers that multiply to 6 and add to 7: these are 1 and 6. Rewrite the middle term: 2x² + x + 6x + 3, then group: x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3). This method is reliable for every quadratic that factorises over the integers.
当x²的系数不为1时，可采用分组分解法或试错法。对于2x² + 7x + 3，计算a × c = 2 × 3 = 6。找到乘积为6且和为7的两个数：1和6。改写中间项：2x² + x + 6x + 3，然后分组：x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3)。这种方法对所有整数范围内可分解的二次式都可靠。
3. Solving by Factorisation | 用因式分解法求解
A quadratic equation in the form ax² + bx + c = 0 can often be solved by factorising. The rule is that if the product of two expressions is zero, then at least one must be zero. For example, solve x² − 3x − 10 = 0. Factorising gives (x − 5)(x + 2) = 0. Therefore x − 5 = 0 or x + 2 = 0, so x = 5 or x = −2. Always write both solutions clearly.
形如ax² + bx + c = 0的二次方程通常可通过因式分解来求解。规则是：若两个表达式的乘积为零，则至少有一个为零。例如，解x² − 3x − 10 = 0。因式分解得(x − 5)(x + 2) = 0。因此x − 5 = 0或x + 2 = 0，所以x = 5或x = −2。务必清楚写出两个解。
Before factorising, always rearrange the equation so that one side is zero. For example, x² = 5x can only be solved after bringing all terms to one side: x² − 5x = 0, then factor out x: x(x − 5) = 0, giving x = 0 or x = 5. A common mistake is to divide both sides by x, which loses the solution x = 0. Never divide by an expression that contains a variable unless you are certain it is non-zero.
在因式分解之前，务必先整理方程，使一边为零。例如，x² = 5x必须先移项：x² − 5x = 0，然后提出公因式x：x(x − 5) = 0，得x = 0或x = 5。常见错误是两边同时除以x，这会导致丢失解x = 0。切勿除以含有变量的表达式，除非你确定它不为零。
4. Solving by Completing the Square | 配方法
Completing the square rewrites a quadratic in the form a(x + p)² + q. This form reveals the vertex of the parabola and allows solving equations without factorising. For a monic quadratic x² + bx + c, take half of b, square it, and adjust the constant:
配方法将二次式改写为a(x + p)² + q的形式。这种形式揭示了抛物线的顶点，并可在不因式分解的情况下解方程。对于首项系数为1的二次式x² + bx + c，取b的一半并平方，然后调整常数项：
x² + bx + c = (x + b/2)² − (b/2)² + c
For example, complete the square for x² + 6x + 11. Half of 6 is 3, and 3² = 9, so x² + 6x + 11 = (x + 3)² − 9 + 11 = (x + 3)² + 2. To solve, set the expression equal to zero: (x + 3)² + 2 = 0, hence (x + 3)² = −2. Since a square cannot be negative, this equation has no real solutions. Notice that this quadratic never touches the x-axis.
例如，对x² + 6x + 11配方。6的一半是3，3² = 9，所以x² + 6x + 11 = (x + 3)² − 9 + 11 = (x + 3)² + 2。要求解，令表达式为零：(x + 3)² + 2 = 0，因此(x + 3)² = −2。由于平方不可能为负，该方程无实数解。注意这个二次函数永远不会接触x轴。
5. The Quadratic Formula | 二次公式
The quadratic formula solves any quadratic equation ax² + bx + c = 0, even when factorisation is difficult or impossible. The formula is:
二次公式可以求解任何形式为ax² + bx + c = 0的二次方程，即使因式分解困难或不可能。公式为：
x = (−b ± √(b² − 4ac)) / (2a)
Consider 2x² − 4x − 3 = 0. Here a = 2, b = −4, c = −3. Substitute carefully: x = (4 ± √((−4)² − 4 × 2 × (−3))) / (2 × 2) = (4 ± √(16 + 24)) / 4 = (4 ± √40) / 4. Since √40 = 2√10, x = (4 ± 2√10) / 4 = (2 ± √10) / 2. These two values are approximately 2.58 and −0.58. Leave values in surd form unless the question asks for decimals.
考虑方程2x² − 4x − 3 = 0。这里a = 2，b = −4，c = −3。仔细代入：x = (4 ± √((−4)² − 4 × 2 × (−3))) / (2 × 2) = (4 ± √(16 + 24)) / 4 = (4 ± √40) / 4。由于√40 = 2√10，x = (4 ± 2√10) / 4 = (2 ± √10) / 2。这两个值约为2.58和−0.58。除非题目要求小数，否则保留根式形式。
To use the formula efficiently, remember three steps: identify a, b and c; substitute into the formula; and simplify. Watch the sign of b: if b is negative, then −b becomes positive. Also remember the square applies to the whole of b, so (−4)² = 16, not −16. This tiny error is one of the most common marks lost in IGCSE exams.
高效使用该公式需记住三步：识别a、b和c；代入公式；化简。注意b的符号：若b为负，则−b变为正。同时注意平方作用于整个b，因此(−4)² = 16，而不是−16。这个细微错误是IGCSE考试中丢分最常见的原因之一。
6. The Discriminant | 判别式
The expression b² − 4ac inside the square root is called the discriminant, denoted by Δ. Its value determines how many real roots the equation has:
根号内的表达式b² − 4ac称为判别式，记作Δ。其值决定了方程有多少个实数根：
| 判别式 Δ | 实数根数量 | 图像与x轴的关系 |
| Δ > 0 | 两个不同实数根 | 与x轴有两个交点 |
| Δ = 0 | 一个重根（两个相等） | 与x轴相切于一点 |
| Δ < 0 | 没有实数根 | 与x轴无交点 |
For example, for x² + 2x + 5 = 0, the discriminant is 2² − 4 × 1 × 5 = 4 − 20 = −16, which is negative, so there are no real roots. For x² − 6x + 9 = 0, the discriminant is 36 − 36 = 0, giving one repeated root x = 3. Questions may ask you to find the range of k such that a quadratic has two real roots; set b² − 4ac > 0 and solve for k.
例如，对于x² + 2x + 5 = 0，判别式为2² − 4 × 1 × 5 = 4 − 20 = −16，为负，因此没有实数根。对于x² − 6x + 9 = 0，判别式为36 − 36 = 0，得到唯一重根x = 3。题目有时会要求找出k的取值范围，使得某个二次方程有两个实数根；只需令b² − 4ac > 0并解出k。
7. Sketching Quadratic Graphs | 绘制二次函数图像
The graph of y = ax² + bx + c is a parabola. Its shape is controlled by a: if a > 0, the parabola opens upwards like a smile; if a < 0, it opens downwards like a frown. To sketch a quadratic, you need four key features: the y-intercept, the roots, the turning point, and the general shape.
函数y = ax² + bx + c的图像是抛物线。其形状由a决定：若a > 0，抛物线开口向上，如微笑；若a < 0，开口向下，如皱眉。绘制二次函数草图需要四个关键特征：y轴截距、根、顶点坐标及整体形状。
For y = x² − 4x + 3, the y-intercept is at (0, 3), because substituting x = 0 gives c. To find the roots, solve x² − 4x + 3 = 0, which factorises to (x − 1)(x − 3) = 0, so the graph crosses the x-axis at x = 1 and x = 3. The x-coordinate of the turning point is the average of the roots: x = (1 + 3)/2 = 2. Substituting x = 2 gives y = 2² − 8 + 3 = −1, so the vertex is at (2, −1). Plot these points and label the curve y = x² − 4x + 3.
对于y = x² − 4x + 3，y轴截距为(0, 3)，因为代入x = 0得c。求根时解x² − 4x + 3 = 0，因式分解得(x − 1)(x − 3) = 0，因此图像在x = 1和x = 3处穿过x轴。顶点的x坐标是两个根的平均值：x = (1 + 3)/2 = 2。代入x = 2得y = 2² − 8 + 3 = −1，所以顶点为(2, −1)。描出这些点并标出曲线y = x² − 4x + 3。
8. Finding the Turning Point by Completing the Square | 用配方法求顶点
When a quadratic is written as y = a(x + p)² + q, the turning point is at (−p, q). This is because (x + p)² is always ≥ 0, so the minimum or maximum value occurs when x + p = 0, i.e. x = −p. For y = (x − 3)² + 4, the vertex is (3, 4) and the minimum value is 4. This method works for any quadratic and avoids relying on factorisation.
当二次函数写成y = a(x + p)² + q时，顶点为(−p, q)。这是因为(x + p)²始终 ≥ 0，所以最小值或最大值在x + p = 0时取得，即x = −p。对于y = (x − 3)² + 4，顶点为(3, 4)，最小值为4。这种方法适用于任何二次函数，无需依赖因式分解。
For a non-monic quadratic such as y = 2x² + 8x + 5, first factor out the 2 from the first two terms: y = 2(x² + 4x) + 5. Complete the square inside: x² + 4x = (x + 2)² − 4. Then y = 2[(x + 2)² − 4] + 5 = 2(x + 2)² − 8 + 5 = 2(x + 2)² − 3. Hence the vertex is (−2, −3), and since a = 2 > 0, this is a minimum point.
对于首项系数不为1的二次函数，如y = 2x² + 8x + 5，先从第一、二项中提出2：y = 2(x² + 4x) + 5。在括号内配方：x² + 4x = (x + 2)² − 4。然后y = 2[(x + 2)² − 4] + 5 = 2(x + 2)² − 8 + 5 = 2(x + 2)² − 3。因此顶点为(−2, −3)，由于a = 2 > 0，这是一个最小值点。
9. Solving Quadratic Inequalities | 解二次不等式
A quadratic inequality such as x² − x − 6 < 0 is best solved by sketching or by a sign diagram. First factorise: (x − 3)(x + 2) < 0. The critical values are x = −2 and x = 3. Since the coefficient of x² is positive, the parabola opens upwards and is below the x-axis between the roots. Therefore the solution is −2 < x < 3.
形如x² − x − 6 < 0的二次不等式最好通过画图或符号表来求解。首先因式分解：(x − 3)(x + 2) < 0。临界值为x = −2和x = 3。由于x²的系数为正，抛物线开口向上，在两个根之间位于x轴下方。因此解为−2 < x < 3。
For the inequality x² − 2x − 3 ≥ 0, factorise to (x − 3)(x + 1) ≥ 0. The graph is above or on the x-axis when x ≤ −1 or x ≥ 3. The critical difference between quadratic and linear inequalities is that the solution is often two separate regions, not one interval. Remember to test a value from each region to confirm the sign.
对于不等式x² − 2x − 3 ≥ 0，因式分解得(x − 3)(x + 1) ≥ 0。图像在x ≤ −1或x ≥ 3时位于x轴上方或与x轴相交。二次不等式与一次不等式的一个关键区别是：解往往是两个不相连的区域，而不是一个区间。记得从每个区域取一个数值来验证符号。
10. Applications of Quadratics | 二次函数的应用
Quadratic equations appear from geometry, physics and finance. For example, the product of two consecutive integers is 506, find the integers. Let the integers be n and n + 1, then n(n + 1) = 506, so n² + n − 506 = 0. Factorise to (n + 22)(n − 23) = 0, giving n = 23 or n = −22. Thus the pairs are (23, 24) or (−22, −21). Always interpret the answer in the context of the question.
二次方程出现在几何、物理和金融问题中。例如，两个连续整数的乘积是506，求这两个整数。设这两个整数为n和n + 1，则n(n + 1) = 506，即n² + n − 506 = 0。因式分解得(n + 22)(n − 23) = 0，故n = 23或n = −22。因此这两组整数分别为(23, 24)或(−22, −21)。务必结合题目情境解释答案。
Projectile motion is another classic application. If a ball is thrown upward with height h metres after t seconds given by h = 20t − 5t², find when it hits the ground. Set h = 0: 20t − 5t² = 0, factorise 5t(4 − t) = 0, so t = 0 or t = 4. The ball hits the ground after 4 seconds. The reader should also be able to find the maximum height by completing the square: h = −5(t − 2)² + 20, so the maximum height is 20 m at t = 2 s.
抛体运动是另一个经典应用。若一个球被向上抛出，t秒后高度h米满足h = 20t − 5t²，求球何时落地。令h = 0：20t − 5t² = 0，因式分解5t(4 − t) = 0，得t = 0或t = 4。球在4秒后落地。读者还应能通过配方法求最大高度：h = −5(t − 2)² + 20，因此在t = 2秒时最大高度为20米。
11. Common Pitfalls and Exam Tips | 常见错误与考试技巧
Five errors lose the most marks on quadratic questions in IGCSE papers. First, forgetting to rearrange the equation to zero before solving. Second, losing the root x = 0 when dividing by x. Third, substituting a negative b into the quadratic formula incorrectly. Fourth, drawing the parabola with the wrong opening direction. Fifth, omitting labels such as intercepts and turning points on a sketch.
IGCSE试卷中二次函数题有五类最容易丢分的错误。第一，求解前忘记将方程整理为零侧。第二，除以x时丢失根x = 0。第三，将负的b代入二次公式时出错。第四，抛物线开口方向画反。第五，草图上漏标截距和顶点等关键点。
Always check your solutions by substituting back into the original equation. When sketching, write the equation of the graph next to the curve. When using a calculator for decimals, give answers to three significant figures unless instructed otherwise. Consistent practice with both factorising and the formula will make these routines automatic under exam pressure.
始终通过回代原方程来检验解。绘图时，在曲线旁写出函数方程。使用计算器求小数时，除非另有要求，否则保留三位有效数字。坚持练习因式分解法和公式法，使这些过程在考试压力下自动完成。
12. Quick Reference: Solving Strategies | 快速参考：求解策略
Choose the fastest method for each equation. For x² + 5x + 6 = 0, factorisation is quick. For x² = 81, simply take square roots: x = ±9. Use completing the square when you need the turning point or when the equation has a surd root. Use the quadratic formula when asked for solutions to any quadratic or when factorisation fails. The table below summarises when each method is most useful.
为每个方程选择最快的方法。对于x² + 5x + 6 = 0，因式分解最快。对于x² = 81，直接开平方：x = ±9。当需要求顶点或方程含有根式解时使用配方法。当题目要求解任意二次方程或因式分解失败时使用二次公式。下表总结了各方法最适用的情境。
| 方法 | 最适用情境 | 优点 |
| 因式分解 | 系数简单、整数根 | 最快、直观 |
| 配方法 | 需要顶点坐标 | 同时给出最值 |
| 二次公式 | 任意二次方程 | 通用、可靠 |
Finally, remember the connection between the three representations: the factorised form shows the roots, the completed-square form shows the vertex, and the expanded form shows the y-intercept. Mastering the links between these representations is exactly what the IGCSE examiner expects. Keep practising, keep checking your answers, and the marks will follow.
最后，记住三种表示形式之间的联系：因式分解形式显示根，配方法形式显示顶点，展开形式显示y轴截距。掌握这些表示形式之间的联系正是IGCSE考官期望看到的能力。坚持练习，坚持检验答案，分数自然会随之而来。
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