📚 Quadratic Equations and Graphs: Factorising, Formula and Sketching | 二次方程与图像：因式分解、公式法与草图
This revision guide explains how to expand, factorise, solve and sketch quadratic functions, with a focus on the methods most often tested in IGCSE Mathematics. Every technique here is linked back to the shape of the graph, so that algebra and sketching support each other rather than being learned as two separate topics.
本复习指南讲解如何展开、因式分解、求解并绘制二次函数图像，重点介绍 IGCSE 数学考试中最常考查的方法。文中每个技巧都与图像的形状相互联系，让代数运算与画草图彼此支撑，而不是当作两个独立的知识点来死记。
1. Recognising Quadratic Expressions | 识别二次表达式
A quadratic expression is any expression of the form ax² + bx + c, where a, b and c are constants and a is not zero. If a were zero the x² term would disappear and the expression would become linear. The highest power of the variable is therefore always 2.
二次表达式是形如 ax² + bx + c 的表达式，其中 a、b、c 为常数且 a 不等于零。如果 a 为零，x² 项就会消失，表达式就变成一次式。因此变量的最高次数始终为 2。
tion true are called roots or solutions. A quadratic equation can have two distinct real roots, one repeated root, or no real roots at all.
将二次表达式设为零，就得到二次方程：ax² + bx + c = 0。使方程成立的 x 值称为根或解。二次方程可以有两个不同的实根、一个重根，或者没有实根。
point). The constant c is always the y-intercept, because substituting x = 0 leaves y = c.
二次函数图像的标准形状是抛物线。当 a > 0 时开口向上（存在最低点），当 a < 0 时开口向下（存在最高点）。常数 c 始终是 y 轴截距，因为代入 x = 0 后得到 y = c。
- Quadratic expression: ax² + bx + c with a ≠ 0 — 二次表达式：ax² + bx + c，且 a ≠ 0。
- Roots: the x-values that satisfy the equation — 根：使方程成立的 x 值。
- Parabola: the curved graph of a quadratic — 抛物线：二次函数的曲线图像。
2. Expanding Brackets into Standard Form | 展开括号化为一般式
Before you can solve or sketch anything, you often need the expression in standard form. Multiplying two linear brackets is done term by term; many students remember this as FOIL (First, Outer, Inner, Last).
在求解或画图之前，通常需要先把表达式化为一般式。两个一次括号相乘要逐项展开，很多同学用 FOIL 口诀记忆（首项、外项、内项、末项）。
(x + p)(x + q) = x² + (p + q)x + pq
Two special cases appear constantly in exam papers. When the two brackets are identical you get a perfect square, and when they differ only in sign you get the difference of two squares.
考试中反复出现两种特殊情形。当两个括号完全相同时得到完全平方，当两个括号只有符号不同时得到平方差。
(x + p)² = x² + 2px + p² (x + p)(x − p) = x² − p²
Worked example: expand (2x − 3)(x + 5). Multiply out to get 2x² + 10x − 3x − 15, then collect like terms: 2x² + 7x − 15. Always check that the number of terms collected matches the pattern, and keep track of negative signs.
例题：展开 (2x − 3)(x + 5)。逐项相乘得到 2x² + 10x − 3x − 15，再合并同类项：2x² + 7x − 15。做完后要检查同类项是否合并完整，并特别留意负号。
3. Factorising Simple Quadratics | 简单二次式的因式分解
Always look for a common factor first. For example 6x² + 9x factorises as 3x(2x + 3), and 4x² − 16 factorises as 4(x + 2)(x − 2). Removing the common factor makes everything that follows much easier.
第一步永远先看有没有公因式。例如 6x² + 9x 可以分解为 3x(2x + 3)，4x² − 16 可以分解为 4(x + 2)(x − 2)。先提出公因式，后面的运算会简单很多。
need two numbers that multiply to give c and add to give b. Then the factorised form is (x + p)(x + q), where p and q are those two numbers.
对于 a = 1 的形式 x² + bx + c，需要找到两个数：它们的乘积等于 c，和等于 b。那么因式分解结果就是 (x + p)(x + q)，其中 p、q 就是这两个数。
4. Product −24 and sum −5 gives −8 and 3, so the answer is (x − 8)(x + 3). When c is negative the two numbers have opposite signs; when c is positive and b is negative both numbers are negative.
例：x² + 7x + 12。乘积为 12、和为 7 的两个数是 3 和 4，所以 x² + 7x + 12 = (x + 3)(x + 4)。例：x² − 5x − 24。乘积 −24、和 −5 的两个数是 −8 和 3，答案为 (x − 8)(x + 3)。当 c 为负时两数异号；当 c 为正而 b 为负时两数同为负。
4. Factorising When a ≠ 1 | 当 a ≠ 1 时的因式分解
When the coefficient of x² is not 1, a reliable method is to multiply a and c, then find two numbers that multiply to ac and add to b. Split the middle term using these two numbers and factorise in pairs.
当 x² 的系数不是 1 时，可靠的方法是先算 a × c，再找两个数使其乘积为 ac、和为 b。用这两个数拆开中间项，然后两两分组分解。
nd b = 11, so the two numbers are 15 and −4. Rewrite the middle term: 6x² + 15x − 4x − 10.
例题：分解 6x² + 11x − 10。此处 ac = 6 × (−10) = −60，b = 11，所以两个数为 15 和 −4。改写中间项：6x² + 15x − 4x − 10。
Now group: 3x(2x + 5) − 2(2x + 5). Both groups share the bracket (2x + 5), so the answer is (2x + 5)(3x − 2). A quick expansion check should return the original expression; never skip this step in an exam.
接着分组：3x(2x + 5) − 2(2x + 5)。两组都含有括号 (2x + 5)，所以答案是 (2x + 5)(3x − 2)。快速展开验算应还原原式，考试中不要省略这一步。
6x² + 11x − 10 = (2x + 5)(3x − 2)
5. Completing the Square | 配方法
Completing the square rewrites a quadratic in the form a(x + p)² + q. This form immediately reveals the turning point of the graph, which is why it is so useful for sketching questions as well as for solving equations.
配方法把二次式改写为 a(x + p)² + q 的形式。这种形式能直接看出图像的转折点，因此既适合解方程，也非常适合做画图题。
x² + bx + c = (x + b/2)² − (b/2)² + c
² − 8. The turning point is (−3, −8), a minimum because the coefficient of x² is positive.
例：x² + 6x + 1。6 的一半是 3，所以写成 (x + 3)² − 9 + 1 = (x + 3)² − 8。转折点为 (−3, −8)，因为 x² 系数为正，所以是最低点。
+ 7, write 2(x² − 6x) + 7 = 2[(x − 3)² − 9] + 7 = 2(x − 3)² − 11. The vertex is (3, −11).
当 a ≠ 1 时，先把前两项的系数 a 提到括号外再配方。对于 2x² − 12x + 7，写成 2(x² − 6x) + 7 = 2[(x − 3)² − 9] + 7 = 2(x − 3)² − 11。顶点为 (3, −11)。
6. The Quadratic Formula | 求根公式
la always works, provided the equation is first rearranged into the form ax² + bx + c = 0.
并非所有二次式都能用整数因式分解。只要先把方程整理成 ax² + bx + c = 0 的形式，求根公式总是有效的。
x = (−b ± √b² − 4ac) / (2a)
ve 2x² − 4x − 3 = 0. Here a = 2, b = −4, c = −3, so b² − 4ac = 16 + 24 = 40. Then x = (4 ± √40) / 4, which simplifies to x = (2 ± √10) / 2, giving x ≈ 2.58 or x ≈ −0.58 to three significant figures.
例题：解 2x² − 4x − 3 = 0。此处 a = 2，b = −4，c = −3，所以 b² − 4ac = 16 + 24 = 40。于是 x = (4 ± √40) / 4，化简为 x = (2 ± √10) / 2，得 x ≈ 2.58 或 x ≈ −0.58（保留三位有效数字）。
step, and never drop the ± symbol. Round only when the question asks for a decimal answer.
有两个习惯能保住分数：在最后一步之前保留精确的根式形式，以及绝不要丢掉 ± 符号。只有当题目要求小数答案时才进行四舍五入。
7. The Discriminant and the Nature of the Roots | 判别式与根的性质
The expression b² − 4ac is called the discriminant and is usually written as Δ. It tells you how many real roots the equation has without solving it, and therefore how many times the parabola meets the x-axis.
表达式 b² − 4ac 称为判别式，通常记作 Δ。它无需解方程就能告诉你方程有几个实根，从而说明抛物线与 x 轴有几个交点。
Δ = b² − 4ac
| Value of Δ | Δ 的取值 | Nature of roots | 根的性质 | Graph | 图像 |
|---|---|---|
| Δ > 0 | Two distinct real roots | 两个不同实根 | Crosses the x-axis twice | 与 x 轴有两个交点 |
| Δ = 0 | One repeated root | 一个重根 | Touches the x-axis once | 与 x 轴相切于一点 |
| Δ < 0 | No real roots | 无实根 | Does not meet the x-axis | 与 x 轴不相交 |
ntact, the discriminant is also the standard tool for questions about tangency, such as finding the value of k for which a line touches a curve.
由于 Δ = 0 只给出一个接触点，判别式也是处理相切问题的标准工具，例如求直线与曲线相切时 k 的取值。
8. Roots, x-Intercepts and Factorised Form | 根、x 轴截距与因式分解形式
the x-intercepts of the graph. The y-intercept is found by putting x = 0, giving y = apq.
如果二次函数能分解为 y = a(x − p)(x − q)，那么根就是 x = p 和 x = q，它们正好是图像的 x 轴截距。令 x = 0 可求出 y 轴截距，即 y = apq。
m is (2x − 3)(x + 5), so either 2x − 3 = 0 or x + 5 = 0. The roots are x = 3/2 and x = −5.
例：用因式分解解 2x² + 7x − 15 = 0。由第 2 节可知分解式为 (2x − 3)(x + 5)，因此 2x − 3 = 0 或 x + 5 = 0。根为 x = 3/2 与 x = −5。
/2. For this example the axis is x = (3/2 − 5)/2 = −7/4, which also equals −b/(2a) = −7/4.
由对称性可知，抛物线的对称轴正好位于两根的正中间，即 x = (p + q)/2。本例中对称轴为 x = (3/2 − 5)/2 = −7/4，这也等于 −b/(2a) = −7/4。
9. The Vertex and Axis of Symmetry | 顶点与对称轴
s x-value back into the equation gives the y-coordinate of the turning point, so the vertex is at (−b/(2a), y).
y = ax² + bx + c 的对称轴始终为 x = −b/(2a)。把这个 x 值代回方程即可得到转折点的纵坐标，因此顶点为 (−b/(2a), y)。
x = 3 gives y = 9 − 18 + 5 = −4. The vertex is (3, −4) and it is a minimum because a > 0.
例：y = x² − 6x + 5。此处 −b/(2a) = 6/2 = 3，代入 x = 3 得 y = 9 − 18 + 5 = −4。顶点为 (3, −4)，因 a > 0 故为最低点。
If the same expression is written in completed-square form, (x − 3)² − 4, the vertex can be read off directly. Both routes should agree; using one to check the other is an excellent exam habit.
若把同一表达式写成配方形式 (x − 3)² − 4，则可直接读出顶点。两种方法的结果应当一致，用一个去检验另一个是很好的考试习惯。
10. Sketching a Parabola Step by Step | 抛物线草图的分步画法
A quadratic sketch is not a plot from a table of values; it is a labelled diagram showing the key features. Examiners award marks for the correct shape, the intercepts, the turning point and the symmetry.
二次函数草图不是用数值表逐点描出来的图像，而是一张标出关键特征的示意图。评分点包括正确的形状、截距、转折点以及对称性。
- Step 1: check the sign of a to decide whether the parabola opens up or down — 第一步：看 a 的正负，决定抛物线开口向上还是向下。
- Step 2: find the y-intercept by putting x = 0 — 第二步：令 x = 0 求出 y 轴截距。
- Step 3: find the roots by factorising, the formula or completing the square — 第三步：用因式分解、公式或配方法求出根。
- Step 4: find the vertex using x = −b/(2a) or the completed-square form — 第四步：用 x = −b/(2a) 或配方形式求顶点。
- Step 5: draw a smooth curve through the points, label the coordinates and mark the axis of symmetry — 第五步：用平滑曲线连接各点，标出坐标并画出对称轴。
x − 5)(x + 1), giving roots x = 5 and x = −1; the axis is x = 2 and the vertex is (2, −9).
例：画出 y = x² − 4x − 5 的草图。开口向上；y 轴截距为 (0, −5)；分解为 (x − 5)(x + 1)，根为 x = 5 与 x = −1；对称轴为 x = 2，顶点为 (2, −9)。
11. Solving Equations and Inequalities Graphically | 用图像解方程与不等式
Graphs solve equations by intersection. The number of times a line meets the parabola equals the number of solutions of the corresponding equation, which is precisely the discriminant condition seen earlier.
图像通过交点来解方程。直线与抛物线的交点个数等于对应方程的解的个数，这正是前面判别式所描述的条件。
x = −1, and since the parabola opens upwards the expression is positive outside the roots.
解二次不等式时，先求根，再判断哪个区间满足不等式。解 x² − 4x − 5 > 0 得根 x = 5 与 x = −1，由于抛物线开口向上，表达式在两根之外为正。
x² − 4x − 5 > 0 ⟺ x < −1 or x > 5
