📚 A-Level Pure Maths Unit 4: Graphs and Transformations – Higher Homework | A-Level 纯数学第4单元：函数图像与变换——进阶作业
This revision article covers the core skills in Edexcel A-Level Pure Mathematics Unit 4, pages 64-99. The focus is graph sketching and the algebra of transformations for polynomial functions. Use it as a higher homework guide to build fluency with translations, stretches, reflections, and graphical intersections.
本篇复习文章覆盖 Edexcel A-Level 纯数学第4单元（64-99页）的核心技能。重点在于多项式函数的图像绘制与变换代数。请将本文作为进阶作业指南，帮助你熟练掌握平移、伸缩、对称以及图像交点等技能，并把它当作一次完整的图像变换专题复盘。
1. What Do We Mean by Transformation? | 什么是图像变换？
ormation is a rule that maps every point (x, y) on the graph of y = f(x) to a new point (x', y' osition, size or orientation of a curve while preserving its overall shape - a parabola stays a parabola, and a cubic stays a cubic.
变换是一条规则，它把图像 y = f(x) 上的每一点 (x, y) 映射到一个新点 (x', y')。在第4单元中，我们研究三大类变换：平移、伸缩与对称，以及它们的组合。它们都会改变曲线的位置、大小或方向，但保持整体形状不变——抛物线仍然是抛物线，三次曲线仍然是三次曲线。
lel to the y-axis with scale factor a gives (x, y) → (x, a y). A stretch parallel to the x-axis with scale factor 1 ÷ a gives (x, y) → (x ÷ a, y).
映射写法是描述"任意一点发生了什么"最清晰的方式。平移给出 (x, y) → (x + a, y + b)。平行于 y 轴、伸缩系数为 a 的单向伸缩给出 (x, y) → (x, a y)。平行于 x 轴、伸缩系数为 1 ÷ a 的伸缩给出 (x, y) → (x ÷ a, y)。
ges affect the y-coordinate while horizontal changes affect the x-coordinate. Remember that horizontal transformations often look 'backwards' ph a units to the left, not to the right. This single idea causes more lost marks than any other topic in Unit 4.
对于函数 y = f(x)，垂直方向的变化影响 y 坐标，而水平方向的变化影响 x 坐标。请记住，水平变换常常看起来是"反向"的：我们把 x 替换成 x + a，图像是向左平移 a 个单位，而不是向右。就是这一个观念，让第4单元中失分的学生比其他任何知识点都多。
Translation: (x, y) → (x + a, y + b) | Vertical stretch: (x, y) → (x, a y) | Horizontal stretch: (x, y) → (x ÷ a, y)
2. Translations: y = f(x) + a and y = f(x + a) | 平移变换：y = f(x) + a 与 y = f(x + a)
ola y = x² shifted up by 3 units, so its minimum point moves from (0, 0) to (0, 3), and every y-coordinate increases by exactly 3.
垂直平移由 y = f(x) + a 给出。若 a > 0，图像向上移动 a 个单位；若 a < 0，则向下移动 a 个单位。例如 y = x² + 3 就是抛物线 y = x² 向上平移 3 个单位，最低点从 (0, 0) 移到 (0, 3)，每个 y 坐标都恰好增加 3。
he right. So y = (x − 2)² moves the base parabola 2 units to the right, because the bracket is zero when x = 2 rather than when x = 0.
水平平移写作 y = f(x + a)。当 a > 0 时，y = f(x + a) 将图像向左平移 a 个单位；当 a < 0 时，则向右平移 a 个单位。因此 y = (x − 2)² 把基础抛物线向右移动 2 个单位，因为括号内为零的位置从 x = 0 变成了 x = 2。
In exam answers you should describe a translation with a column vector, not with a sentence such as 'moves right 2'. The standard wording is 'translation by the vector with top component 2 and bottom component 0', or 'translation by (2, 0)ᵀ'. Marks are awarded for the direction words and the numbers, so write both.
在考试作答中，你应该用列向量来描述平移，而不是只用"向右移动2"这类句子。标准表述是"按上分量为 2、下分量为 0 的向量平移"，或简写为"按 (2, 0)ᵀ 平移"。分数是给方向和数字的，所以两者都要写清楚。
- y = f(x) + a → vertical translation by (0, a)ᵀ | 垂直平移，向量为 (0, a)ᵀ
- y = f(x + a) → horizontal translation by (−a, 0)ᵀ | 水平平移，向量为 (−a, 0)ᵀ
- y = f(x − a) → horizontal translation by (a, 0)ᵀ | 水平平移，向量为 (a, 0)ᵀ
Vertical translation: y = f(x) + a | Horizontal translation: y = f(x + a)
3. Stretches: y = a f(x) and y = f(a x) | 伸缩变换：y = a f(x) 与 y = f(a x)
lt; 1, the graph is compressed vertically towards the x-axis. Points that already lie on the x-axis, where f(x) = 0, do not move at all under a vertical stretch.
垂直伸缩写作 y = a f(x)。若 a > 1，图像在垂直方向按系数 a 拉伸，各点远离 x 轴；若 0 < a < 1，图像则向 x 轴压扁。原本就在 x 轴上的点（即 f(x) = 0 的点）在垂直伸缩下完全不动，这一点是解题的关键检查点。
nd a = ½ produces a doubling. Under a horizontal stretch the y-intercept stays fixed, because (0, y) → (0 ÷ a, y) = (0, y).
水平伸缩写作 y = f(a x)。此时伸缩平行于 x 轴，伸缩系数为 1 ÷ a：a = 2 产生"缩小一半"的效果，a = ½ 则产生"拉长一倍"的效果。在水平伸缩下，y 轴截距保持不变，因为 (0, y) → (0 ÷ a, y) = (0, y)。
rst doubles every height but keeps the period at 360°, while the second keeps the amplitude at 1 but halves the period to 180°. Reading the coefficient' ction or inside the bracket - tells you immediately which axis the stretch is parallel to.
比较 y = 2 sin x 与 y = sin 2x：前者把每个高度加倍，但周期仍为 360°；后者振幅仍为 1，但周期减半为 180°。看清系数在函数外面还是在括号里面，就能立刻判断伸缩平行于哪条轴——这是本节最实用的读图技巧。
y = a f(x): stretch parallel to the y-axis, scale factor a | y = f(a x): stretch parallel to the x-axis, scale factor 1 ÷ a
4. Reflections: y = −f(x) and y = f(−x) | 对称变换：y = −f(x) 与 y = f(−x)
intercepts are unchanged. The roots of the equation f(x) = 0 are exactly the roots of −f(x) = 0.
关于 x 轴作对称写作 y = −f(x)。每一点 (x, y) 变为 (x, −y)，轴上方的峰变成轴下方的谷，而 x 轴截距保持不变。方程 f(x) = 0 的根与 −f(x) = 0 的根完全相同。
r an even function such as y = x² or y = cos x, the curve maps onto itself and the transformation looks like no change at all.
关于 y 轴作对称写作 y = f(−x)。此时 (x, y) 变为 (−x, y)，y 轴截距不变，但图像左右镜像翻转。对于偶函数，例如 y = x² 或 y = cos x，曲线与自身重合，变换后看起来"没有任何变化"——这正是判别偶函数的图形定义。
isely the algebraic test for odd symmetry about the origin, and examiners like to see it quoted.
对于奇函数，例如 y = x³，两种对称给出相同结果：−f(x) = −x³，而 f(−x) = (−x)³ = −x³。这个等式 −f(x) = f(−x) 正是关于原点奇对称的代数判据，考官很乐意看到考生引用它。
5. Combining Transformations and the Order of Operations | 复合变换与运算顺序
2 f(x + 1) − 3. The safe method is to track a single general point (x, y) through the recipe, or to follow the operations in the order in which they act on the variable x and then on the whole function.
大多数考题会组合两到三个变换，例如 y = 2 f(x + 1) − 3。最稳妥的方法有两种：一是追踪一个一般点 (x, y) 依次经历各个步骤；二是按照"先处理 x、再处理整个函数"的顺序逐步执行。
For y = a f(x + b) + c, the accepted order is: translate horizontally by (−b, 0)ᵀ, stretch vertically with scale factor a, then translate vertically by (0, c)ᵀ must be done before the vertical translation, otherwise the value of c gets stretched too.
对于 y = a f(x + b) + c，公认的顺序是：先按向量 (−b, 0)ᵀ 水平平移，再按系数 a 垂直伸缩，最后按向量 (0, c)ᵀ 垂直平移。注意水平步骤在括号内部，因此最先发生；而垂直伸缩必须排在垂直平移之前，否则 c 也会被一起拉伸，结果就错了。
his shows a translation of 3 ÷ 2 to the left followed by a stretch parallel to the x-axis with scale factor 1 ÷ 2. Forgetting to factorise is the single most common mistake with this type of question.
当系数在括号内部时，先把它提出来。对于 y = f(2x + 3)，改写成 y = f(2(x + 3 ÷ 2))。这样就能看出：先向左平移 3 ÷ 2，再平行于 x 轴按系数 1 ÷ 2 伸缩。忘记提取系数，是这类题目中最常见的错误。
+ 3 ÷ 2)) → translate left 3 ÷ 2, then stretch parallel to the x-axis, scale factor 1 ÷ 2
6. Key Features: Intercepts, Turning Points and Asymptotes | 图像关键特征：截距、极值点与渐近线
intercept comes from substituting x = 0; the x-intercepts come from solving f(x) = 0. Turning points are found by completing the square, by differentiation, or by symmetry about the axis of the parabola.
一幅完整的草图必须标出所有截距的坐标、所有极值点的坐标，以及任何渐近线的方程。y 轴截距由代入 x = 0 得到；x 轴截距由解 f(x) = 0 得到。极值点则可以用配方法、求导法，或利用抛物线关于对称轴对称来求得。
x₀ − b, y₀). For y = f(a x) it becomes (x₀ ÷ a, y₀), and for y = −f(x) it becomes (x₀, −y₀).
变换之后，极值点按可预测的方式移动。对于 y = a f(x) + c，点 (x₀, y₀) 变为 (x₀, a y₀ + c)；对于 y = f(x + b)，变为 (x₀ − b, y₀)；对于 y = f(a x)，变为 (x₀ ÷ a, y₀)；对于 y = −f(x)，变为 (x₀, −y₀)。
| Transformation | New turning point | Fixed feature |
| y = f(x) + c | (x₀, y₀ + c) | x-intercepts shift too |
| y = f(x + b) | (x₀ − b, y₀) | y-intercept shifts |
| y = a f(x) | (x₀, a y₀) | x-intercepts fixed |
| y = f(a x) | (x₀ ÷ a, y₀) | y-intercept fixed |
ion both asymptotes move with the curve, so y = 1 ÷ (x − 3) + 2 has asymptotes x = 3 and y = 2. Always write asymptotes as full equations, never as bare numbers.
当分母可能为零时就会出现渐近线。对于 y = 1 ÷ (x − 3)，垂直渐近线是 x = 3，水平渐近线是 y = 0。在平移下，两条渐近线随曲线一同移动，所以 y = 1 ÷ (x − 3) + 2 的渐近线是 x = 3 与 y = 2。渐近线一律写成完整方程，不要只写一个数字。
7. Sketching Polynomials: Factorised Form and Multiplicity | 多项式图像绘制：因式分解形式与重根
ouches it at x = −2. The exponent on each factor, called the multiplicity of the root, controls whether the curve crosses the axis or merely touches it.
当多项式以因式分解形式给出时，各因式直接告诉你 x 轴截距。若 y = (x − 1)(x + 2)²，曲线在 x = 1 处穿过 x 轴，在 x = −2 处与 x 轴相切。每个因式上的指数称为该根的重数，它决定曲线是穿过坐标轴还是仅仅相切。
- Odd multiplicity (1, 3, 5 …): the curve crosses the x-axis | 奇重数（1、3、5…）：曲线穿过 x 轴
- Even multiplicity (2, 4, …): the curve touches the x-axis and turns back | 偶重数（2、4…）：曲线与 x 轴相切后折返
- Degree of the polynomial: controls the end behaviour of the curve | 多项式的次数：决定曲线两端的走向
End behaviour depends on the degree and on the sign of the leading coefficient. An even degree sends both ends the same way: up if the leading coefficient is positive, down if it is negative. An odd degree sends the two ends in opposite directions, which is why every cubic has at least one real root.
两端走向由次数和首项系数的符号决定。偶次多项式两端同向：首项系数为正则两端向上，为负则两端向下。奇次多项式两端反向，这正是每个三次方程至少有一个实根的原因。
A good sketch is built in a fixed order: find the intercepts, note the multiplicities, find the y-intercept, decide the end behaviour, then join the points smoothly with the correct number of turning points. A polynomial of degree n has at most n − 1 turning points, which stops you drawing a cubic with two humps and a wiggle in the middle.
好的草图按固定顺序完成：先求各截距，标注重数，再求 y 轴截距，判断两端走向，最后用正确数量的极值点把这些点平滑连接。n 次多项式最多有 n − 1 个极值点，这提醒你不要把一个三次曲线画成两个峰还带中间小波动。
8. Intersections of Graphs and Graphical Solutions | 图像交点与图解方程
alence converts an algebra problem into a graph problem, and it works in both directions: a question may ask you to solve an equation graphically, or to use a graph to state how many roots an equation has.
方程 f(x) = g(x) 的解的个数，等于图像 y = f(x) 与 y = g(x) 的交点个数。这一等价关系把代数问题转化为图像问题，而且双向可用：题目可能要求你图解方程，也可能要求你利用图像判断方程有多少个根。
A typical higher-tier question reads: ' ing a suitable straight line, show that the equation x³ − 3x + 1 = 0 has three real roots.' e x³ = 3x − 1, you draw y = 3x − 1 on the same axes as y = x³ and count the intersections.
一道典型的进阶题会这样问："通过画一条合适的直线，说明方程 x³ − 3x + 1 = 0 有三个实根。"技巧在于把方程整理成一边与已知曲线相同的形式。由 x³ = 3x − 1 可知，在 y = x³ 的同一坐标系中画出直线 y = 3x − 1，再数交点即可。
