📚 IGCSE Maths: Quadratic Equations, Functions and Graphs | IGCSE 数学：二次方程、函数与图像
Quadratic equations appear throughout the IGCSE Mathematics syllabus, from algebraic manipulation and solving equations to sketching graphs and interpreting real-life problems. A quadratic expression has the general form ax² + bx + c, where a, b and c are constants and a ≠ 0. This article brings together the methods you need to confidently handle quadratic equations, quadratic functions and their graphs, and it shows you how to choose the fastest method under exam pressure.
二次方程贯穿 IGCSE 数学课程，从代数变形、解方程到画图像和解释实际问题都会出现。二次表达式的一般形式为 ax² + bx + c，其中 a、b、c 为常数且 a ≠ 0。本文将汇总你处理二次方程、二次函数及其图像所需的各类方法，并说明在考试压力下如何选择最快的方法。
1. Recognising a Quadratic Expression | 识别二次表达式
A quadratic expression is any expression that can be written in the standard form ax² + bx + c, where the highest power of x is 2. The coefficient a is called the quadratic coefficient, b is the linear coefficient and c is the constant term. Typical examples include x² + 5x + 6, 2x² − 3x + 1 and 4 − x², all of which fit the same pattern even though their signs look different.
二次表达式是任何可以写成标准形式 ax² + bx + c 的表达式，其中 x 的最高次数为 2。系数 a 称为二次项系数，b 为一次项系数，c 为常数项。常见的例子包括 x² + 5x + 6、2x² − 3x + 1 和 4 − x²，它们的符号虽然不同，但都符合同一个模式。
Not every expression containing an x² is a quadratic. Expressions such as x³ − 1, 3x + 2, 1 ÷ x and √x are not quadratics because their highest power of x is not exactly 2. An expression written as (x + 3)(x − 5) is quadratic in disguise: expand it and you get x² − 2x − 15, which is clearly in standard form.
并不是所有含 x² 的表达式都是二次式。像 x³ − 1、3x + 2、1 ÷ x 和 √x 都不是二次式，因为它们 x 的最高次数不是恰好为 2。写成 (x + 3)(x − 5) 的表达式是"伪装"的二次式：展开后得到 x² − 2x − 15，很清楚就是标准形式。
Recognising the standard form is the first step in almost every quadratic question. Once you can identify a, b and c confidently, factorising, completing the square, using the formula and sketching the graph all become routine procedures rather than guesses.
识别标准形式几乎是每一道二次题目的第一步。一旦你能准确找出 a、b、c，因式分解、配方、套公式和画图像都会变成固定流程，而不再是靠猜。
2. Standard Form, Coefficients and the Condition a ≠ 0 | 标准形式、系数与 a ≠ 0 的条件
or example, 7 − 2x² + 3x becomes −2x² + 3x + 7, so a = −2, b = 3 and c = 7. Getting this ordering right prevents the most common sign errors in later work.
在看系数之前，必须先把表达式整理成 x² 项在前、x 项居中、常数项在后的顺序。例如 7 − 2x² + 3x 应写成 −2x² + 3x + 7，于是 a = −2、b = 3、c = 7。把顺序整理正确，可以避免后续运算中最常见的符号错误。
which is linear, not quadratic. So the condition a ≠ 0 is essential. It is the single feature that separates a straight line from a parabola, and examiners often test it by asking whether a given function is quadratic.
如果 a = 0，表达式就变成 bx + c，这是一次式而不是二次式，因此 a ≠ 0 这个条件很重要。它正是区分直线与抛物线的唯一特征，考官也常通过"判断某函数是否为二次函数"来考查这一点。
The coefficients can be positive, negative, fractions or decimals, and the terms may appear in any order. So ½x² − 0.4x + 3 and −x² + 8 are just as valid as x² + 5x + 6. What matters is that the highest power of x is 2 and that its coefficient is not zero.
系数可以为正数、负数、分数或小数，各项的顺序也可以不同。因此 ½x² − 0.4x + 3 与 −x² + 8 和 x² + 5x + 6 一样都是合法的二次式。关键是 x 的最高次数为 2，且其系数不为零。
3. Solving by Factorising | 用因式分解法解方程
wo factors is zero, then at least one factor must be zero. So once you write (x + p)(x + q) = 0, you can immediately conclude x = −p or x = −q.
当根为整数或简单分数时，因式分解通常是解二次方程最快的方法。核心思想是零积法则：若两个因式的乘积为零，则至少有一个因式为零。所以一旦写成 (x + p)(x + q) = 0，就可以立刻得出 x = −p 或 x = −q。
because 2 × 3 = 6 and 2 + 3 = 5, giving (x + 2)(x + 3) = 0 and therefore x = −2 or x = −3.
对于 a = 1 的 x² + bx + c，要找两个数：它们的乘积为 c，和为 b。对 x² + 5x + 6，2 和 3 满足条件，因为 2 × 3 = 6 且 2 + 3 = 5，于是 (x + 2)(x + 3) = 0，解得 x = −2 或 x = −3。
torisation is (2x − 1)(x − 1), giving x = ½ or x = 1. Always expand mentally to check your factors before solving.
当 a ≠ 1 时，先提取公因式，再拆分中间项，或者对 a 与 c 的因数对进行试验调整。对 2x² − 3x + 1，因式分解为 (2x − 1)(x − 1)，得 x = ½ 或 x = 1。解方程前一定要在脑中展开检验因式是否正确。
² − 49 = 0 gives (x + 7)(x − 7) = 0 and x = ±7. Spotting this pattern saves a lot of time.
一个值得牢记的特殊情况是平方差公式：a² − b² = (a + b)(a − b)。因此 x² − 49 = 0 可写成 (x + 7)(x − 7) = 0，得 x = ±7。识别出这一模式可以节省大量时间。
4. Completing the Square | 配方法
Completing the square rewrites a quadratic in the form a(x + p)² + q. This form is not mainly used for solving; its real value is that it reveals the vertex of the parabola immediately, since the minimum or maximum occurs when the bracket equals zero.
配方法把二次式改写成 a(x + p)² + q 的形式。这种形式主要不是用来解方程的；它真正的价值在于能立刻揭示抛物线的顶点，因为当括号内为零时取到最小值或最大值。
For a quadratic written as x² + bx + c, half the coefficient of x and square it:
对于写成 x² + bx + c 的二次式，取一次项系数的一半并平方：
x² + bx + c = (x + b ÷ 2)² − (b ÷ 2)² + c
−8) and the minimum value of the expression is −8. Solving (x + 3)² − 8 = 0 gives x = −3 ± √8, which simplifies to x = −3 ± 2√2.
例如 x² + 6x + 1 = (x + 3)² − 9 + 1 = (x + 3)² − 8。因此顶点为 (−3, −8)，该表达式的最小值为 −8。解 (x + 3)² − 8 = 0 得 x = −3 ± √8，化简为 x = −3 ± 2√2。
he vertex is (2, −3) and the minimum value is −3. This is the standard route to maximum and minimum problems.
当 a ≠ 1 时，先只从前两项提取 a：2x² − 8x + 5 = 2(x² − 4x) + 5 = 2[(x − 2)² − 4] + 5 = 2(x − 2)² − 3。顶点为 (2, −3)，最小值为 −3。这是解决最大值与最小值问题的标准路径。
5. The Quadratic Formula and the Discriminant | 求根公式与判别式
When a quadratic does not factorise nicely, use the quadratic formula. It works for every quadratic equation and is the most reliable tool in the topic.
当二次式不易因式分解时，就使用求根公式。它对所有二次方程都适用，是本主题中最可靠的工具。
x = (−b ± √b² − 4ac) ÷ 2a
always be placed in brackets. For 2x² − 4x − 3 = 0, a = 2, b = −4 and c = −3, so x = (4 ± √16 + 24) ÷ 4 = (4 ± √40) ÷ 4 = 1 ± √10 ÷ 2, roughly 2.58 or −0.58.
分三步进行：先把方程写成标准形式，读出 a、b、c，再小心代入。负数一定要加括号。对 2x² − 4x − 3 = 0，a = 2、b = −4、c = −3，于是 x = (4 ± √16 + 24) ÷ 4 = (4 ± √40) ÷ 4 = 1 ± √10 ÷ 2，约等于 2.58 或 −0.58。
e discriminant, and it tells you how many real roots the equation has before you solve it.
根号下的表达式 Δ = b² − 4ac 称为判别式，它能在求解之前就告诉你方程有几个实根。
| Value of Δ = b² − 4ac | Nature of the roots | Graph |
| Δ > 0 | Two distinct real roots | Curve crosses the x-axis twice |
| Δ = 0 | One repeated real root | Curve touches the x-axis |
| Δ < 0 | No real roots | Curve never meets the x-axis |
Leave answers in surd form when the question asks for exact values, and round only at the end when a decimal answer is required. Always give both roots unless the repeated-root case applies.
题目要求精确值时，答案应保留根式形式；需要小数时，只在最后一步取近似值。除重根情况外，一定要给出两个根。
6. Choosing the Best Method | 选择最合适的方法
Examiners reward efficiency. Spending two minutes on factorising when the formula takes thirty seconds costs marks elsewhere in the paper. Use the guide below to make a fast decision.
考官看重效率。如果本可用三十秒公式解决的问题却花两分钟因式分解，其他题目就会丢分。用下面的表格快速做决定。
| Situation | Best method |
| Small integer roots, a = 1 | Factorising |
| Difference of two squares | Factorising |
| Question asks for vertex or maximum | Completing the square |
| Awkward numbers or surd answers | Quadratic formula |
| Question asks about number of roots | Discriminant only |
A useful habit is to check the discriminant first. If Δ is not a perfect square, the roots will be irrational, so factorising into integers is impossible and you should go straight to the formula.
一个有用的习惯是先算判别式。如果 Δ 不是完全平方数，根就是无理数，无法分解成整数因式，那就直接使用公式。
7. Quadratic Functions and the Shape of a Parabola | 二次函数与抛物线的形状
. If a > 0 the curve opens upwards and has a minimum point; if a < 0 it opens downwards and has a maximum point. The larger the magnitude of a, the narrower the curve.
y = ax² + bx + c 的图像是一条抛物线。若 a > 0，抛物线开口向上，有最低点；若 a < 0，开口向下，有最高点。a 的绝对值越大，图像越窄。
The vertex, or turning point, lies on the axis of symmetry. Its x-coordinate can be found without completing the square:
顶点（即转折点）位于对称轴上。它的横坐标无需配方即可求出：
x = −b ÷ 2a
very parabola also has a y-intercept at (0, c) and roots where y = 0. These three features — vertex, intercepts and direction of opening — is the axis of symmetry: it is the vertical line x = −b ÷ 2a, not the vertex point itself.
每条抛物线还有 y 轴截距 (0, c) 以及 y = 0 时的根。顶点、截距和开口方向这三个特征足以画出一张合格的草图。一个常见的考试陷阱是对称轴：它是竖直线 x = −b ÷ 2a，而不是顶点这个点本身。
8. Sketching and Transforming Quadratic Graphs | 绘制与变换二次函数图像
ots by solving y = 0, find the vertex by completing the square or using x = −b ÷ 2a, then plot the points and draw a smooth curve. Do not use a ruler between the roots.
绘制二次函数图像时遵循固定流程：由 a 的符号判断开口方向，令 x = 0 求 y 轴截距，令 y = 0 解根，用配方或 x = −b ÷ 2a 求顶点，然后描点并画出光滑曲线。两根之间不要用直尺画直线。
+ p)² + q. A change to q moves the curve vertically, a change to p moves it horizontally in the opposite direction to the sign, and a change to a stretches or compresses it.
图像变换遵循以顶点式 y = a(x + p)² + q 为基础的固定规则。改变 q 使曲线上下平移，改变 p 使曲线左右平移（方向与符号相反），改变 a 则使图像拉伸或压缩。
| Transformation | Effect on y = x² |
| y = x² + k | Translation k units up |
| y = (x + k)² | Translation k units left |
| y = ax², |a| > 1 | Vertical stretch, narrower curve |
| y = −x² | Reflection in the x-axis |
This symmetry is useful for checking your sketch: if your points are not mirror images about the axis of symmetry, something has gone wrong.
由于 x² = (−x)²，基本抛物线 y = x² 关于 y 轴对称。利用这一对称性可以检验草图：如果各点关于对称轴不成镜像，说明出现了错误。
9. Finding a Quadratic Equation from Its Graph | 由图像求二次函数表达式
)(x − r₂) and then substitute any other known point to find the value of k. This method is fast and avoids solving simultaneous equations.
如果已知根，就把方程写成因式形式 y = k(x − r₁)(x − r₂)，再代入任意一个已知点求出 k 的值。这种方法快捷，可以避免解联立方程。
3) passing through (0, 5) gives 5 = a(0 − 2)² − 3, so 4a = 8 and a = 2, meaning y = 2(x − 2)² − 3.
如果已知顶点 (h, k)，就使用顶点式 y = a(x − h)² + k，再代入另一个已知点。例如顶点为 (2, −3) 且过点 (0, 5) 的抛物线满足 5 = a(0 − 2)² − 3，所以 4a = 8，a = 2，即 y = 2(x − 2)² − 3。
ree simultaneous equations and solve for a, b and c. Always finish by expanding your answer into the standard form if the question asks for it.
如果已知三个任意点，就把每对坐标代入 y = ax² + bx + c，得到三个联立方程，解出 a、b、c。如果题目要求标准形式，最后一定要把答案展开整理。
10. Intersections, Simultaneous Equations and Inequalities | 交点、联立方程与不等式
To find where a line meets a parabola, substitute the linear equation into the quadratic one and solve the resulting quadratic equation. Each real solution gives one intersection point, so two solutions mean two crossing points, one repeated solution means the line is a tangent, and no real solution means the line misses the curve entirely.
求直线与抛物线的交点时，把一次方程代入二次方程，解所得的二次方程。每个实根对应一个交点：两个解表示有两个交点，一个重根表示直线与曲线相切，无实根表示直线与曲线不相交。
lue of the constant for which a line just touches a curve, a very common higher-tier question.
判别式可以快速判断相切。令 Δ = 0，即可求出直线恰好与曲线相切时的常数取值，这是高阶试卷中非常常见的题型。
erval. For x² − x − 6 < 0 the roots are x = −2 and x = 3, and a quick test shows the expression is negative between them, so the solution is −2 < x < 3.
解二次不等式时，先解对应方程求出临界值，再判断各区间内表达式的符号。对 x² − x − 6 < 0，根为 x = −2 和 x = 3，简单代入可知两根之间表达式为负，因此解为 −2 < x < 3。
ical values themselves are included, so use solid dots on a number line. If a < 0, it is usually safest to multiply by −1 and reverse the inequality sign before testing intervals.
如果不等式使用 ≥ 或 ≤，临界值本身包含在解集内，在数轴上要用实心点表示。若
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