📚 Solving the Intersections of Function Graphs | 函数图像交点的求解方法
Finding the intersections of two function graphs is a central skill in A-Level mathematics. It links algebra, graph sketching and equation solving, and it appears in many exam questions involving curves, lines, trigonometric functions, exponentials and modulus functions.
求两个函数图像的交点是 A-Level 数学中的核心技能。它将代数、画图与解方程联系在一起，并且在涉及曲线、直线、三角函数、指数函数和绝对值函数的许多考试题中都会出现。
1. The Meaning of an Intersection Point | 交点的含义
An intersection point is a coordinate (x, y) that lies on both graphs simultaneously. For two functions y = f(x) and y = g(x), any common point must satisfy y = f(x) and y = g(x) at the same time. Therefore, the x-coordinate of every intersection must solve the equation f(x) = g(x).
交点就是同时位于两个图像上的坐标 (x, y)。对于两个函数 y = f(x) 和 y = g(x)，任何公共点都必须同时满足 y = f(x) 与 y = g(x)。因此，每个交点的横坐标 x 都必须满足方程 f(x) = g(x)。
After solving for x, substitute each x-value into either function to find the corresponding y-value. The result is an ordered pair (x, y). The number of intersection points may be zero, one, two, or even infinitely many, depending on the types of functions involved.
解出 x 后，将每一个 x 值代入任意一个函数，求出对应的 y 值。结果写成有序数对 (x, y)。交点的个数可以是零个、一个、两个甚至无穷多个，这取决于函数的类型。
2. The Algebraic Method: Setting Equal | 代数方法：令两式相等
To find intersections algebraically, follow this general procedure:
用代数方法求交点时，通常按以下步骤操作：
Step 1: Write down the equations as y = f(x) and y = g(x).
步骤一：将两个函数写成 y = f(x) 和 y = g(x)。
Step 2: Set f(x) = g(x) and solve for x.
步骤二：令 f(x) = g(x)，解出 x。
Step 3: Substitute each x back into either original equation to find y.
步骤三：将每个 x 代回原方程中的任意一个，求出相应的 y。
Step 4: State the coordinates as (x₁, y₁), (x₂, y₂), etc.
步骤四：将坐标写为 (x₁, y₁)、(x₂, y₂) 等形式。
For example, find the intersection of y = 2x + 1 and y = x + 3. Setting them equal gives 2x + 1 = x + 3, so x = 2. Substituting into y = 2x + 1 gives y = 5. The intersection point is (2, 5).
例如，求直线 y = 2x + 1 与 y = x + 3 的交点。令两式相等：2x + 1 = x + 3，得到 x = 2。代入 y = 2x + 1，得 y = 5。交点就是 (2, 5)。
3. Quadratic and Linear Curves: Discriminant | 二次函数与一次函数：判别式
When one graph is a quadratic and the other is a line, substituting the line into the quadratic produces a quadratic equation. Suppose y = ax² + bx + c and the line is y = mx + k. Then the intersection condition becomes ax² + (b − m)x + (c − k) = 0.
当其中一个图像是二次函数，另一个是一次函数时，将一次函数代入二次函数会得到一个二次方程。设二次函数为 y = ax² + bx + c，直线为 y = mx + k，则交点条件变为 ax² + (b − m)x + (c − k) = 0。
This equation is only valid when a ≠ 0. If a = 0, the original graph is actually a line, not a quadratic. For a genuine quadratic, the discriminant Δ = (b − m)² − 4a(c − k) decides the number of intersections.
这个方程仅在 a ≠ 0 时有效。如果 a = 0，那么原来的图像实际上是一条直线，而不是二次函数。对于真正的二次曲线，判别式 Δ = (b − m)² − 4a(c − k) 决定了交点个数。
| Discriminant | Graphical Meaning | Number of Intersections |
| Δ > 0 | The line cuts the curve | Two distinct points |
| Δ = 0 | The line is tangent to the curve | One point of tangency |
| Δ < 0 | The line does not meet the curve | No real intersection |
A common example is y = x² and y = 2x + 3. Setting x² = 2x + 3 gives x² − 2x − 3 = 0, which factors as (x − 3)(x + 1) = 0. Hence x = 3 or x = −1. The corresponding y-values are 9 and 1, so the intersections are (3, 9) and (−1, 1).
一个常见例子是 y = x² 与 y = 2x + 3。令 x² = 2x + 3，得 x² − 2x − 3 = 0，因式分解为 (x − 3)(x + 1) = 0。所以 x = 3 或 x = −1。对应的 y 值分别是 9 和 1，因此交点为 (3, 9) 和 (−1, 1)。
4. Quadratic and Quadratic | 二次函数与二次函数
When both graphs are quadratic functions, the equation f(x) = g(x) still gives a quadratic equation after simplification. For two quadratics y = a₁x² + b₁x + c₁ and y = a₂x² + b₂x + c₂, the intersection equation is (a₁ − a₂)x² + (b₁ − b₂)x + (c₁ − c₂) = 0.
当两个图像都是二次函数时，令 f(x) = g(x) 仍会在化简后得到一个二次方程。对于两个二次函数 y = a₁x² + b₁x + c₁ 和 y = a₂x² + b₂x + c₂，交点方程为 (a₁ − a₂)x² + (b₁ − b₂)x + (c₁ − c₂) = 0。
If a₁ ≠ a₂, this is a quadratic equation, and the discriminant tells us how many times the two parabolas meet. If a₁ = a₂, the quadratic terms cancel, leaving a linear or constant equation; then there can be exactly one intersection, no intersection, or infinitely many if the two graphs are identical.
如果 a₁ ≠ a₂，这就是一个二次方程，判别式可以告诉我们两条抛物线相交的次数。如果 a₁ = a₂，二次项相互抵消，剩下一次或常数方程；此时可能恰好有一个交点、没有交点，或者当两条抛物线完全相同时有无限多个交点。
Consider y = x² and y = x² − 4x + 4. Setting them equal produces x² = x² − 4x + 4, so 4x = 4 and x = 1. Thus there is one intersection at (1, 1). Geometrically, the two parabolas have the same shape and touch at a single point here.
考虑 y = x² 和 y = x² − 4x + 4。令它们相等：x² = x² − 4x + 4，得到 4x = 4，所以 x = 1。因此只有一个交点 (1, 1)。从几何上看，这两条抛物线形状相同，在这里恰好相切于一点。
5. Polynomials and the Factor Theorem | 多项式与因式定理
For higher-degree polynomial graphs, the strategy is the same: set y₁ = y₂ and simplify to a single equation. However, the resulting polynomial is often cubic or higher, so solving it requires factorisation, the factor theorem or polynomial division.
对于更高次的多项式图像，策略相同：令 y₁ = y₂，化简为一个方程。但得到的多项式常常是三次或更高次的，因此需要用因式定理、因式分解或多项式除法来求解。
For example, find the intersections of y = x³ − 3x + 2 and y = x + 2. Setting them equal gives x³ − 3x + 2 = x + 2, which simplifies to x³ − 4x = 0. Factoring out x, we get x(x² − 4) = 0, so x = 0, x = 2 and x = −2.
例如，求 y = x³ − 3x + 2 与 y = x + 2 的交点。令两式相等：x³ − 3x + 2 = x + 2，化简得 x³ − 4x = 0。提出公因式 x，得到 x(x² − 4) = 0，于是 x = 0、x = 2 和 x = −2。
Substituting each x-value into y = x + 2 gives the points (0, 2), (2, 4) and (−2, 0). Notice that all three points lie on the straight line and also on the cubic curve.
将每个 x 值代入 y = x + 2，得到交点 (0, 2)、(2, 4) 和 (−2, 0)。注意这三个点同时位于直线和三次曲线上。
6. Trigonometric Functions and Periodicity | 三角函数与周期性
When one or both functions are trigonometric, the intersection equation may have infinitely many solutions because sine, cosine and tangent are periodic. In A-Level questions, the domain is usually restricted, for example 0 ≤ x ≤ 2π, so we must find all solutions within that interval.
当一个或两个函数是三角函数时，交点方程可能有无穷多解，因为正弦、余弦和正切都是周期函数。在 A-Level 考试中，通常会限定定义域，例如 0 ≤ x ≤ 2π，因此要在该区间内求出所有解。
Consider y = sin x and y = 0.5, with the domain 0 ≤ x ≤ 2π. Setting sin x = 0.5 gives two principal solutions: x = π/6 and x = 5π/6. The corresponding intersection points are (π/6, 0.5) and (5π/6, 0.5).
考虑 y = sin x 与 y = 0.5，定义域为 0 ≤ x ≤ 2π。令 sin x = 0.5，得到两个基本解：x = π/6 和 x = 5π/6。对应的交点为 (π/6, 0.5) 和 (5π/6, 0.5)。
Another useful example is y = sin x and y = cos x. Setting sin x = cos x leads
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