📚 Rational Functions | 有理函数
Rational functions are among the most important types of functions in the IB Mathematics curriculum, appearing in both Analysis & Approaches and Applications & Interpretation courses. A rational function is formed by the quotient of two polynomials. Understanding their key features — domain restrictions, asymptotes, holes, and intercepts — enables students to sketch graphs accurately, solve equations and inequalities, and apply these functions to real-world modelling. This comprehensive revision guide walks through every essential topic step by step, pairing explanations in English and Chinese to support bilingual learners.
有理函数是 IB 数学课程中最重要的函数类型之一，出现在分析与方法和应用与解释两个方向的课程中。有理函数由两个多项式的商构成。理解其关键特征——定义域限制、渐近线、可去间断点和截距——能让学生准确地绘制图像、解方程和不等式，并将这些函数应用于现实建模。本综合复习指南逐步讲解每一个核心主题，并用中英双语配对解释，以支持双语学习者。
1. Definition and Basic Form | 定义与基本形式
ision by zero is undefined, the domain of a rational function excludes any x-values that make the denominator zero. The behaviour of the function near these excluded values leads to vertical asymptotes or removable discontinuities.
有理函数是可以写成 R(x) = P(x)/Q(x) 形式的函数，其中 P(x) 和 Q(x) 是多项式，且 Q(x) 不是零多项式。最简单的例子包括 f(x) = 1/x、g(x) = (2x+1)/(x-3) 和 h(x) = (x²-4)/(x²+x-6)。由于除以零没有定义，有理函数的定义域排除所有使分母为零的 x 值。函数在这些被排除值附近的行为会导致垂直渐近线或可去间断点。
- A rational function is defined wherever the denominator is non-zero.
- Rational functions can model reciprocal relationships, rates, and many physical situations.
- 只要分母不为零，有理函数就有定义。
- 有理函数可以模拟倒数关系、速率和许多物理情形。
2. Domain and Holes | 定义域与可去间断点
find the domain of R(x) = P(x)/Q(x), solve Q(x) = 0 and exclude those real solutions from ℝ s in both the numerator and the denominator, it can be cancelled algebraically, creating a "hole" The y-coordinate of the hole is found by substituting x = a into the simplified function.
要求出 R(x) = P(x)/Q(x) 的定义域，解方程 Q(x) = 0，并从 ℝ 中排除那些实数解。如果因式 (x - a) 同时出现在分子和分母中，它可以通过代数约分而被消去，从而在 x = a 处产生一个“孔”（可去间断点），前提是约分后的分母中不再含有该因式。孔的 y 坐标通过将 x = a 代入化简后的函数来求得。
eal x except x = 2 and x = -2, with a hole at (2, 1/4) and a vertical asymptote at x = -2.
示例：f(x) = (x-2)/(x²-4) = (x-2)/[(x-2)(x+2)] = 1/(x+2)（x ≠ 2）。因此，定义域为除了 x = 2 和 x = -2 的所有实数，在 (2, 1/4) 处有一个孔，在 x = -2 处有一条垂直渐近线。
3. Vertical Asymptotes | 垂直渐近线
ned by testing the sign of the function on each side of the asymptote. Vertical asymptotes are essential boundaries that the graph never crosses.
如果在某点 x = a 处，化简后的分母为零而分子不为零，则此处存在一条垂直渐近线。当 x 从左侧或右侧趋近 a 时，函数值趋向 +∞ 或 -∞。具体方向可以通过在渐近线两侧检验函数的符号来确定。垂直渐近线是图像永远不能跨越的重要边界。
(x-1) after cancelling (x+1), there is a vertical asymptote at x = 1 and a hole at x = -1.
如何确定垂直渐近线：将分子和分母因式分解，约去公因式，然后令剩余的分母因式等于零。每一个单根给出一个垂直渐近线。例如，f(x) = (x+1)/(x²-1) = 1/(x-1) 约去 (x+1) 后，在 x = 1 处有一条垂直渐近线，在 x = -1 处有一个孔。
4. Horizontal Asymptotes | 水平渐近线
s based on the degrees of the numerator polynomial (n) and the denominator polynomial (d):
水平渐近线描述了当 x → +∞ 或 x → -∞ 时有理函数的末端行为。根据分子多项式的次数 n 和分母多项式的次数 d，有以下三种情形：
- If n < d, the horizontal asymptote is y = 0.
- e is y = a/b, where a and b are the leading coefficients of the numerator and denominator.
- ; d, there is no horizontal asymptote (but an oblique asymptote may exist when n = d + 1).
- 若 n < d，水平渐近线为 y = 0。
- 若 n = d，水平渐近线为 y = a/b，其中 a 和 b 分别为分子和分母的首项系数。
- 若 n > d，不存在水平渐近线（但当 n = d + 1 时可能存在斜渐近线）。
ents 3 and 1, so horizontal asymptote y = 3. For f(x) = (2x)/(x²+1), n=1, d=2, giving y=0.
示例：f(x) = (3x²+2)/(x²-1)，n = d = 2，首项系数 3 和 1，因此水平渐近线为 y = 3。对于 f(x) = (2x)/(x²+1)，n = 1，d = 2，得出 y = 0。
5. Oblique (Slant) Asymptotes | 斜渐近线
r) gives the equation of the slant asymptote. The graph will approach this line as x → ±∞.
当分子的次数刚好比分母的次数大 1（n = d + 1）时，函数有一条斜渐近线。它可以通过多项式长除法求得。商式（不含余项）给出斜渐近线的方程。当 x → ±∞ 时，图像将趋近这条直线。
)/(x+1) = x+2 + 0/(x+1), so the oblique asymptote is y = x+2. Notice that here x+1 cancels? = (x²+x+1)/(x-1). Long division yields quotient x+2 and remainder 3, so asymptote y = x+2.
示例：f(x) = (x²+x+1)/(x-1)。长除法得到商 x+2 和余数 3，因此渐近线为 y = x+2。严格说，前例 f(x) = (x²+3x+2)/(x+1) 约分后成 x+2（x≠-1），是一条带孔的直线，并非通常意义上的斜渐近线。斜渐近线是针对不能完全约分的情形而言的。
6. Intercepts | 截距
ion at x = 0, provided 0 is in the domain. Intercepts provide anchor points for sketching.
要找到有理函数的 x 截距（根），令分子等于零并求解，同时确保这些 x 值在定义域内（即在这些点处分母不为零）。y 截距通过计算 f(0) 得到，前提是 0 在定义域内。截距为绘制图像提供了锚点。
o x-intercepts. For example, f(x) = 1/(x²+1) has no x-intercepts and y-intercept at (0,1).
对于 f(x) = (x-2)/(x+3)，x 截距为 (2, 0)，y 截距为 f(0) = -2/3。若分子无实数根，则不存在 x 截距。例如，f(x) = 1/(x²+1) 无 x 截距，y 截距为 (0,1)。
7. Graphing Rational Functions | 绘制有理函数图像
Sketching a rational function involves a systematic approach:
绘制有理函数图像需要系统的方法：
- Factorise numerator and denominator fully. Identify any common factors (holes) and remaining denominator factors (vertical asymptotes).
- Determine horizontal or oblique asymptotes using degree comparisons.
- Find x- and y-intercepts.
- Construct a sign chart to analyse intervals where the function is positive or negative, guided by the vertical asymptotes, holes, and x-intercepts.
- Plot key points and draw branches approaching asymptotes correctly.
- 将分子和分母完全因式分解。识别所有公因式（孔）和保留的分母因式（垂直渐近线）。
- 通过次数比较确定水平或斜渐近线。
- 求出 x 截距和 y 截距。
- 构建符号表，分析函数在各区间上的正负，并以垂直渐近线、孔和 x 截距为分界。
- 标出关键点并正确绘制趋近渐近线的分支。
on (-∞,-2), negative on (-2,1), positive on (1,2), negative on (2,∞). Sketch accordingly.
示例：f(x) = (x-1)/(x²-4)。分母因式分解为 (x-2)(x+2)，无公因式。垂直渐近线 x=2, x=-2；水平渐近线 y=0。x 截距 (1,0)；y 截距 1/4。符号表显示在 (-∞,-2) 为正，(-2,1) 为负，(1,2) 为正，(2,∞) 为负。据此绘制图像。
8. Solving Rational Equations | 解有理方程
To solve an equation involving rational functions, multiply both sides by the least common denominator (LCD) to eliminate fractions. This yields a polynomial equation. Always check for extraneous solutions — values that make any original denominator zero. Extraneous roots must be rejected.
要解含有有理函数的方程，两边同时乘以最小公分母 (LCD) 以消去分母，得到一个多项式方程。务必检验增根——即任何会使原方程分母为零的值。增根必须舍去。
→ x=2. But x=2 makes denominators zero, so it is extraneous. The equation has no solution.
示例：解方程 1/(x-2) + 1/(x+2) = 4/(x²-4)。LCD 为 (x-2)(x+2)。相乘得 (x+2)+(x-2)=4 → 2x=4 → x=2。但 x=2 使分母为零，故为增根。该方程无解。
9. Solving Rational Inequalities | 解有理不等式
ne side and combine into a single rational expression. Then find all critical values: zeros of the numerator and zeros of the denominator. Place these on a number line and test the sign of the expression in each interval. Exclude any values that make the denominator zero. Express the solution using interval notation.
要解形如 R(x) > 0 或 R(x) ≤ 0 的有理不等式，将所有项移到一边并合并成一个有理式。接着找出所有临界值：分子的零点和分母的零点。将这些点标在数轴上，并在每一区间检验表达式的符号。排除所有使分母为零的值。用区间表示法写出解集。
values: numerator zero x=1, denominator zero x=-2. Test intervals: (-∞,-2) gives positive? includes equality and numerator can be zero, include x=1; exclude x=-2. Solution: (-∞,-2) ∪ [1,∞).
示例：(x-1)/(x+2) ≥ 0。临界值：分子零点 x=1，分母零点 x=-2。检验区间：(-∞,-2) 选 x=-3 得正；(-2,1) 选 x=0 得负；(1,∞) 选 x=2 得正。不等式包含等号且分子可为零，故包含 x=1；排除 x=-2。解集为 (-∞,-2) ∪ [1,∞)。
10. Partial Fractions | 部分分式
Partial fraction decomposition is the process of expressing a rational function as a sum of simpler rational expressions. It is particularly useful for integration and for analysing function behaviour. The method depends on the factorisation of the denominator:
部分分式分解是将有理函数表示为较简单有理式之和的过程。它对于积分和分析函数行为特别有用。分解方法取决于分母的因式分解形式：
- Distinct linear factors: A/(ax+b) + B/(cx+d) + …
- Repeated linear factors: A/(ax+b) + B/(ax+b)² + …
- Irreducible quadratic factors: (Ax+B)/(ax²+bx+c).
- 不同的一次因式：A/(ax+b) + B/(cx+d) + …
- 重复的一次因式：A/(ax+b) + B/(ax+b)² + …
- 不可约的二次因式：(Ax+B)/(ax²+bx+c)。
A → A=11/3; let x=-1 → -4 = -3B → B=4/3. Thus, decomposition = (11/3)/(x-2) + (4/3)/(x+1).
示例：将 (5x+1)/(x²-x-2) 分解为部分分式。因式分解分母为 (x-2)(x+1)。设 (5x+1)/[(x-2)(x+1)] = A/(x-2) + B/(x+1)。两边乘 LCD：5x+1 = A(x+1) + B(x-2)。求解 A 和 B：令 x=2 得 11=3A → A=11/3；令 x=-1 得 -4=-3B → B=4/3。因此分解为 (11/3)/(x-2) + (4/3)/(x+1)。
11. Modelling with Rational Functions | 有理函数的建模应用
Rational functions appear in many real-world contexts: average cost functions in economics (total cost divided by number of items), concentration of a solution after mixing, time required to travel a fixed distance at varying speed, and inverse variation relationships. In IB problems, students may be asked to construct a model, find asymptotes, interpret their meaning, and solve related equations.
有理函数出现在许多现实情境中：经济学中的平均成本函数（总成本除以产品数量）、混合后溶液的浓度、以不同速度行驶固定距离所需的时间，以及反比例关系。在 IB 问题中，学生可能需要建立模型、找出渐近线、解释其含义并求解相关方程。
production must be positive. Solving C(x) < 25 gives 20+500/x < 25 → 500/x < 5 → x > 100.
示例：生产 x 件产品时的单位平均成本为 C(x) = (500+20x)/x，化简为 C(x) = 20 + 500/x。水平渐近线 y = 20 表明长期最低平均成本。垂直渐近线 x=0 反映了产量必须为正。求解 C(x) < 25 得 20+500/x < 25 → 500/x < 5 → x > 100。
12. Key Points for Exam Success | 考试成功要点
When tackling rational function questions in IB exams, keep the following in mind:
在 IB 考试中解答有理函数题目时，请牢记以下要点：
- Always fully factorise first; this reveals holes and asymptotes.
- Label asymptotes with their equations clearly on graphs.
- Check for extraneous solutions when solving equations.
- Use a sign chart for inequalities — never simply clear denominators without considering sign changes.
- Remember that horizontal and oblique asymptotes can be crossed, but vertical asymptotes cannot.
- Simplify rational expressions before differentiating or integrating.
- When using partial fractions, verify by combining the fractions to see if you get back the original expression.
- 务必先完全因式分解；这能揭示孔和渐近线。
- 在图像上清楚地用方程标注渐近线。
- 解方程时检验增根。
- 对于不等式，使用符号表——不要不经考虑符号变化就简单地消去分母。
- 切记水平渐近线和斜渐近线可以被穿过，但垂直渐近线绝不可穿过。
- 在求导或积分前化简有理式。
- 使用部分分式时，通过合并分式来检验是否能还原为原式。
Mastering these techniques will build confidence in tackling both standard problems and more complex multi-step exam questions.
掌握这些技巧将有助于自信地应对标准问题以及更复杂的多步考试题。
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