📚 A-Level Maths: Solving Intersections of Function Graphs | A-Level数学：函数图像交点求解方法
The points at which two function graphs meet are called intersection points. At these points, the two functions share the same x-coordinate and the same y-coordinate. Finding these points is a key skill in Edexcel A-Level Mathematics, as it connects algebra, coordinate geometry, and numerical methods.
两个函数图像相交的点称为交点。在这些点上，两个函数具有相同的 x 坐标和相同的 y 坐标。求交点是 Edexcel A-Level 数学中的核心技能，它把代数、坐标几何与数值方法联系在一起。
1. The Core Principle | 核心原理
Suppose two functions are given by y = f(x) and y = g(x). At any intersection point, the y-values are equal, so we can write f(x) = g(x). Solving this equation gives the x-coordinates of all intersection points.
设有两个函数 y = f(x) 和 y = g(x)。在任意交点处，两个函数的 y 值相等，因此可以列出 f(x) = g(x)。解这个方程就能得到所有交点的 x 坐标。
Example: find the intersection points of f(x) = x² and g(x) = 2x + 3.
例：求 f(x) = x² 与 g(x) = 2x + 3 的交点。
x² = 2x + 3
Rearrange and factorise:
移项并因式分解：
x² − 2x − 3 = 0
(x − 3)(x + 1) = 0
So x = 3 or x = −1. Substituting into either function gives y = 9 or y = 1. The intersection points are (3, 9) and (−1, 1).
因此 x = 3 或 x = −1。代入任一函数可得 y = 9 或 y = 1，所以交点为 (3, 9) 和 (−1, 1)。
2. Linear-Linear Intersections | 线性函数与线性函数的交点
For two straight lines, the equation f(x) = g(x) is linear. Solving it gives a single x-value, provided the lines are not parallel.
对于两条直线，方程 f(x) = g(x) 是一次方程。只要两直线不平行，解方程就能得到唯一的 x 值。
Example: find the intersection of y = 2x + 1 and y = −x + 4.
例：求 y = 2x + 1 与 y = −x + 4 的交点。
2x + 1 = −x + 4
Then 3x = 3, so x = 1. Substituting x = 1 into either line gives y = 3. The intersection point is (1, 3).
于是 3x = 3，即 x = 1。将 x = 1 代入任一直线得 y = 3，所以交点坐标为 (1, 3)。
If the gradients of two lines are equal, they are parallel. Parallel lines have no intersection unless they are the same line, in which case there are infinitely many intersections.
若两条直线的斜率相等，则两直线平行。平行线没有交点；若两直线完全重合，则有无数个交点。
3. Quadratic Equations and the Discriminant | 二次方程与判别式
When both functions are quadratic, subtracting one equation from the other gives a quadratic equation in x. This equation can be solved by factorising, completing the square, or using the quadratic formula.
当两个函数都是二次函数时，将两个方程相减可以得到一个关于 x 的一元二次方程。这个方程可以用因式分解、配方法或求根公式求解。
The quadratic formula is:
求根公式为：
x = [−b ± √(b² − 4ac)] / (2a)
The discriminant Δ = b² − 4ac determines the number of real intersection points. If Δ > 0, there are two distinct intersections; if Δ = 0, there is one repeated intersection; if Δ < 0, there are no real intersections.
判别式 Δ = b² − 4ac 决定实数交点的个数：若 Δ > 0，有两个不同交点；若 Δ = 0，有一个重根交点；若 Δ < 0，没有实数交点。
Example: find the x-coordinates of the intersections of y = x² − 2x + 1 and y = −x² + 3.
例：求 y = x² − 2x + 1 与 y = −x² + 3 交点的 x 坐标。
x² − 2x + 1 = −x² + 3
2x² − 2x − 2 = 0
Dividing by 2 gives x² − x − 1 = 0. Using the quadratic formula:
两边除以 2 得 x² − x − 1 = 0。使用求根公式：
x = (1 ± √5) / 2
Substitute each value into either original function to find the corresponding y-coordinate.
将每个 x 值代入原函数中的任一个，即可求出对应的 y 坐标。
4. A Line and a Quadratic: Tangent or Secant | 直线与二次函数：相切还是相交
A common exam question asks whether a straight line meets a quadratic curve, and if so, how many times. Substitute y = mx + c into y = ax² + bx + d. The resulting quadratic determines the number of intersection points.
常见考试题型是判断一条直线与一条二次曲线是否相交，以及交点的个数。将 y = mx + c 代入 y = ax² + bx + d，得到的二次方程决定了交点的个数。
ax² + bx + d = mx + c
ax² + (b − m)x + (d − c) = 0
| Discriminant | 判别式 | Number of intersections | 交点个数 |
| Δ > 0 | Two distinct points | 两个不同交点 |
| Δ = 0 | One point, the line is a tangent | 一个交点，直线与曲线相切 |
| Δ < 0 | 更多咨询请联系16621398022（同微信）
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