📚 Linear Functions and Graphs | 线性函数与图像
Linear functions are among the most fundamental topics in IGCSE Mathematics. A linear function produces a straight-line graph, and mastery of its equation, slope, and intercepts lays the foundation for coordinate geometry, simultaneous equations, and real-world modelling.
线性函数是 IGCSE 数学中最基础也最重要的主题之一。线性函数对应的图像是一条直线，掌握其方程、斜率和截距，是学习坐标几何、联立方程和实际建模的基础。
1. The Standard Form y = mx + c | 标准形式 y = mx + c
The equation of any straight line can be written in the form y = mx + c, where m represents the gradient (slope) and c represents the y-intercept. This is the most widely used form in IGCSE examinations.
任何一条直线的方程都可以写成 y = mx + c 的形式，其中 m 表示斜率（梯度），c 表示 y 轴截距。这是 IGCSE 考试中最常用的形式。
For example, the line y = 2x + 3 has gradient 2 and crosses the y-axis at (0, 3). Every unit increase in x causes y to increase by 2 units.
例如，直线 y = 2x + 3 的斜率为 2，与 y 轴相交于点 (0, 3)。x 每增加 1 个单位，y 就增加 2 个单位。
2. Understanding the Gradient m | 理解斜率 m
The gradient measures how steep a line is. It is defined as the change in y divided by the change in x between any two points on the line:
斜率衡量一条直线的倾斜程度。它定义为直线上任意两点之间 y 的变化量除以 x 的变化量：
m = (y₂ − y₁) / (x₂ − x₁)
A positive gradient means the line slopes upward from left to right. A negative gradient means the line slopes downward. A horizontal line has gradient 0, while a vertical line has an undefined gradient.
斜率为正表示直线从左到右上升；斜率为负表示直线从左到右下降。水平线的斜率为 0，而垂直线的斜率不存在。
3. The y-Intercept c | y 轴截距 c
The y-intercept is the point where the line crosses the y-axis. At this point, x = 0, so substituting x = 0 into y = mx + c gives y = c. The coordinates of the y-intercept are therefore (0, c).
y 轴截距是直线与 y 轴交点的纵坐标。在该点处 x = 0，代入 y = mx + c 得到 y = c。因此 y 轴截距点的坐标为 (0, c)。
For instance, in y = −3x + 5, the line crosses the y-axis at (0, 5). If c = 0, the line passes through the origin.
例如，在 y = −3x + 5 中，直线与 y 轴相交于点 (0, 5)。如果 c = 0，则直线经过原点。
4. Drawing a Straight-Line Graph | 绘制直线图像
To draw the graph of a linear function, you only need two points. A common method is to choose two convenient x-values, substitute them into the equation, and plot the resulting coordinates.
绘制一次函数图像只需要两个点。常用方法是选取两个方便的 x 值，代入方程求出对应的 y 值，然后标出坐标并连线。
- Choose x = 0 to find the y-intercept.
- Choose x = 1 (or another simple value) to find a second point.
- Plot both points and draw a straight line through them.
- 选取 x = 0，求出 y 轴截距。
- 选取 x = 1（或其他简单数值）求第二个点。
- 标出两点，用直尺画直线穿过它们。
Always extend the line across the full grid, and label the line with its equation.
画图时务必让直线延伸到整个坐标系，并在直线旁标注其方程。
5. Finding the Equation from a Graph | 从图像求方程
Given a straight-line graph, you can determine its equation by finding the gradient and the y-intercept. Read the y-intercept directly from the graph where the line crosses the y-axis.
给定一条直线图像，可以通过求斜率和 y 轴截距来确定其方程。直接从图上看直线与 y 轴的交点，即可得到截距 c。
To find the gradient, select two points on the line with integer coordinates. Calculate the vertical change divided by the horizontal change:
求斜率时，选取直线上两个坐标为整数的点，计算纵向变化量除以横向变化量：
m = rise / run = Δy / Δx
Substitute m and c into y = mx + c to obtain the final equation. Always check that a third point on the line satisfies the equation.
将 m 和 c 代入 y = mx + c 即得最终方程。务必用直线上的第三个点检验方程是否成立。
6. Parallel and Perpendicular Lines | 平行线与垂直线
Two lines are parallel if and only if they have the same gradient. For example, y = 2x + 1 and y = 2x − 7 are parallel because both have m = 2.
两条直线平行当且仅当它们的斜率相同。例如 y = 2x + 1 与 y = 2x − 7 平行，因为二者的 m 都等于 2。
Two lines are perpendicular if the product of their gradients is −1. In other words, if one line has gradient m, a perpendicular line has gradient −1/m.
两条直线垂直当且仅当它们的斜率之积为 −1。换句话说，若一条直线斜率为 m，则其垂线的斜率为 −1/m。
m₁ × m₂ = −1 (perpendicular)
For instance, y = 3x + 2 and y = −(1/3)x + 4 are perpendicular because 3 × (−1/3) = −1.
例如，y = 3x + 2 与 y = −(1/3)x + 4 相互垂直，因为 3 × (−1/3) = −1。
7. Horizontal and Vertical Lines | 水平线与垂直线
A horizontal line has the equation y = k, where k is a constant. Its gradient is 0 because there is no change in y regardless of x.
水平线的方程为 y = k，其中 k 是常数。它的斜率为 0，因为无论 x 如何变化，y 都不变。
A vertical line has the equation x = h, where h is a constant. Its gradient is undefined because Δx = 0, and division by zero is impossible.
垂直线的方程为 x = h，其中 h 是常数。它的斜率不存在，因为 Δx = 0，不能除以零。
Do not confuse these with the line y = x, which has gradient 1 and passes through the origin at 45°.
不要将这两类直线与 y = x 混淆。y = x 的斜率为 1，经过原点，倾斜角为 45°。
8. Solving Simultaneous Equations Graphically | 联立方程的图解法
When two linear equations are plotted on the same axes, their point of intersection gives the simultaneous solution. This is because the intersection point satisfies both equations at once.
当两条直线画在同一坐标系中时，它们的交点就是联立方程的解。这是因为交点同时满足两个方程。
For example, solve y = 2x + 1 and y = −x + 4 graphically. Plot both lines; they intersect at (1, 3), so the solution is x = 1, y = 3.
例如，图解 y = 2x + 1 和 y = −x + 4。画出两条直线后，它们相交于点 (1, 3)，因此解为 x = 1，y = 3。
If the lines are parallel, there is no solution. If the lines are identical, there are infinitely many solutions. The graphical method is especially useful for checking algebraic results.
若两条直线平行，则方程无解；若两条直线重合，则有无穷多组解。图解法特别适合用来检验代数结果。
9. Real-World Applications | 实际应用
Linear functions model many real-life situations where one quantity changes at a constant rate. Speed-time graphs, currency conversions, and taxi fare calculations are all common examples.
线性函数可以建模很多实际情境，只要一个量以恒定速率随另一个量变化即可。速度-时间图像、货币换算、出租车计费都是常见的例子。
- A taxi charges a fixed fee of $3 plus $2 per kilometre: C = 2d + 3.
- Converting dollars to euros at a fixed rate: E = 0.85D.
- A bathtub filling at 5 litres per minute: V = 5t.
- 出租车收费为固定 3 美元外加每公里 2 美元：C = 2d + 3。
- 按固定汇率将美元换算为欧元：E = 0.85D。
- 浴缸以每分钟 5 升的速度注水：V = 5t。
In each case, the coefficient of the variable is the rate of change, and the constant term is the initial value.
在每种情况中，变量的系数就是变化率，常数项就是初始值。
10. Common Pitfalls and Exam Tips | 常见错误与考点提示
Students frequently make mistakes when reading the gradient from a graph or when substituting coordinates. Be careful with negative gradients and fractional values.
学生在从图中读取斜率或代入坐标时经常出错。要特别注意负斜率和分数值。
| Error | 错误 | Correction | 正确做法 |
|---|---|
| Confusing x and y intercepts | y-intercept: set x = 0; x-intercept: set y = 0 |
| Swapping rise and run | m = Δy / Δx, always vertical over horizontal |
| Forgetting to extend the line | Use a ruler and extend across the entire grid |
| Misidentifying the y-intercept in a table | Look for the row where x = 0 |
| 错误 | 正确做法 |
|---|---|
| 混淆 x 轴和 y 轴截距 | y 轴截距令 x = 0；x 轴截距令 y = 0 |
| 把纵向与横向变化量弄反 | m = Δy / Δx，永远是纵向除以横向 |
| 忘记延长直线 | 用直尺画并延伸到整个坐标系 |
| 在表格中找错 y 轴截距 | 找到 x = 0 的那一行 |
In the exam, always show your method when finding m and c. Even if the final answer is wrong, you may still earn method marks.
考试中，求 m 和 c 时一定要写出过程。即使最终答案有误，仍可能获得步骤分。
11. Worked Example | 完整例题
Find the equation of the line passing through the points A(2, 5) and B(4, 9).
求经过点 A(2, 5) 和 B(4, 9) 的直线方程。
Write down the intermediate results step by step:
逐步写出中间结果：
| Step | 步骤 | Calculation | 计算 |
|---|---|
| 1. Find the gradient | m = (9 − 5) / (4 − 2) = 4 / 2 = 2 |
| 2. Substitute into y = mx + c | y = 2x + c |
| 3. Use point A(2, 5) | 5 = 2(2) + c → c = 1 |
| 4. Write the final equation | y = 2x + 1 |
| 步骤 | 计算 |
|---|---|
| 1. 求斜率 | m = (9 − 5) / (4 − 2) = 4 / 2 = 2 |
| 2. 代入 y = mx + c | y = 2x + c |
| 3. 代入点 A(2, 5) | 5 = 2(2) + c → c = 1 |
| 4. 写出最终方程 | y = 2x + 1 |
Check with point B: substituting x = 4 gives y = 2(4) + 1 = 9, which matches B(4, 9). The equation is correct.
用点 B 检验：代入 x = 4 得 y = 2(4) + 1 = 9，与 B(4, 9) 一致。方程正确。
12. Practice Questions | 巩固练习
Attempt the following questions on your own before checking the answers. They cover the full range of skills discussed above.
先独立完成以下练习再对照答案。这些题目覆盖了上文讨论的所有技能。
- Find the gradient of the line passing through (−1, 3) and (2, 9).
- Write down the equation of the line with gradient −2 and y-intercept 7.
- Determine whether y = 4x − 1 and y = −(1/4)x + 2 are perpendicular.
- A line passes through (0, −3) and (5, 7). Find its equation.
- 求经过 (−1, 3) 和 (2, 9) 两点的直线斜率。
- 写出斜率为 −2 且 y 轴截距为 7 的直线方程。
- 判断 y = 4x − 1 与 y = −(1/4)x + 2 是否垂直。
- 一条直线经过 (0, −3) 和 (5, 7)，求其方程。
Answers: (1) m = 2; (2) y = −2x + 7; (3) Yes, since 4 × (−1/4) = −1; (4) m = 2, so y = 2x − 3.
答案：(1) m = 2；(2) y = −2x + 7；(3) 垂直，因为 4 × (−1/4) = −1；(4) m = 2，所以 y = 2x − 3。
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