Unit B3: Functions
with Asymptotes单元 B3：含渐近线的函数
Where B1 introduces functions in the abstract and B2 develops polynomial functions, B3 studies the family that misbehaves at isolated inputs: rational functions. A rational function is a quotient of polynomials. The denominator zeros drive the geometry. Some create vertical asymptotes, some create removable holes. Far from the origin, the function settles toward a horizontal or oblique line. The same algebraic toolkit that built the function also lets us solve rational inequalities by sign analysis.B1 抽象地介绍函数，B2 发展多项式函数；B3 研究在某些孤立输入处"失常"的函数族：有理函数。有理函数就是两个多项式的商。分母的零点决定几何形态：有的产生竖直渐近线，有的产生可去间断点。在原点远处，函数趋向某条水平线或斜线。构造函数所用的代数工具同样可以通过符号分析解有理不等式。
How to use this guide本指南使用说明
B3 is short on theory and long on diagnostic moves. Almost every question reduces to three sub-tasks: factor the numerator and denominator, locate the denominator zeros, then compare degrees. The sections below isolate each move so that, when you face an unfamiliar rational expression, you know which tool to reach for.B3 理论不多，但"诊断动作"很多。几乎每道题都可拆为三步：把分子分母分别因式分解、定位分母零点、再比较次数。下文按动作分节，遇到陌生的有理表达式时知道用哪一招。
Memorise the linear-over-linear sketch: $f(x) = \frac{ax + b}{cx + d}$ has a vertical asymptote at $x = -d/c$ and a horizontal asymptote at $y = a/c$. Hole-spotting: a cancelled factor in numerator and denominator is a hole, not an asymptote. For inequalities, build a sign chart on a number line and read off the intervals.
背熟一次比一次的草图：$f(x) = \frac{ax + b}{cx + d}$ 的竖直渐近线为 $x = -d/c$、水平渐近线为 $y = a/c$。识别可去间断点：分子分母同有一个被约掉的因式，就是"洞"而不是渐近线。解不等式：在数轴上画符号表，从图上读出区间。
For HL rational functions, master the degree comparison rule (lower / equal / one higher) and execute polynomial long division for the slant asymptote case. State the hole as an ordered pair, not just a value of $x$. When solving rational inequalities, never multiply both sides by an expression of unknown sign; always rearrange to a single fraction and analyse signs.
HL 题型要熟练应用次数比较规则（分子次数低、等、高一）并在斜渐近线情形熟练做多项式长除法。写"洞"时给出有序对 $(a, y)$，不要只写 $x = a$。解有理不等式时永远不要两边同乘符号未知的表达式；要先化为一个分式再做符号分析。
Linear over Linear Rational Functions一次比一次有理函数 SL 2.8
- Default domain. $\{ x \in \mathbb{R} : x \ne -d/c \}$. The single excluded input is where the denominator vanishes.
- Default range. $\{ y \in \mathbb{R} : y \ne a/c \}$. The horizontal asymptote value is the one $y$-coordinate the curve never reaches.
- Why the condition $a d - b c \ne 0$. If $a d = b c$, the numerator is a scalar multiple of the denominator and $f$ collapses to the constant $a/c$. The graph would degenerate to a horizontal line.
- 默认定义域。$\{ x \in \mathbb{R} : x \ne -d/c \}$。被剔除的唯一输入即分母零点。
- 默认值域。$\{ y \in \mathbb{R} : y \ne a/c \}$。水平渐近线对应的纵坐标是曲线永远到达不了的那个值。
- 为何要求 $a d - b c \ne 0$。若 $a d = b c$，分子是分母的标量倍，$f$ 退化为常数 $a/c$，图像退化为一条水平线。
Find the domain and range of $f(x) = \dfrac{2x + 1}{x - 3}$.求 $f(x) = \dfrac{2x + 1}{x - 3}$ 的定义域与值域。
Domain. The denominator $x - 3$ vanishes only at $x = 3$. Therefore the domain is $\{ x \in \mathbb{R} : x \ne 3 \}$, often written $\mathbb{R} \setminus \{ 3 \}$.
定义域。分母 $x - 3$ 仅在 $x = 3$ 处为零。故定义域为 $\{ x \in \mathbb{R} : x \ne 3 \}$，也写作 $\mathbb{R} \setminus \{ 3 \}$。
Range. Solve $y = \dfrac{2x + 1}{x - 3}$ for $x$ in terms of $y$. Clearing the denominator gives $y(x - 3) = 2x + 1$, then $y x - 3 y = 2 x + 1$, so $x (y - 2) = 3 y + 1$. Provided $y \ne 2$:
值域。把 $y = \dfrac{2x + 1}{x - 3}$ 反解为 $x$ 关于 $y$ 的表达式。去分母：$y(x - 3) = 2x + 1$，整理 $y x - 3 y = 2 x + 1$，故 $x (y - 2) = 3 y + 1$。当 $y \ne 2$ 时：
$$ x \;=\; \frac{3 y + 1}{y - 2}. $$For every $y \ne 2$ there is exactly one $x$ in the domain mapping to it. Therefore the range is $\{ y \in \mathbb{R} : y \ne 2 \}$.
对每个 $y \ne 2$，定义域内恰有一个 $x$ 映射过去。故值域为 $\{ y \in \mathbb{R} : y \ne 2 \}$。
Sanity check. The "forbidden $y$" turned out to be $2$, which equals the ratio of leading coefficients $a / c = 2 / 1$. That ratio always identifies the missing horizontal value for a linear-over-linear function (the horizontal asymptote, formalised in B3.2).
合理性检查。"被禁止的 $y$"是 $2$，恰好等于首项系数比 $a / c = 2 / 1$。对任何"一次比一次"函数，此比值就是缺失的水平值（即水平渐近线，B3.2 正式给出）。
Going deeper: why every linear-over-linear is a translated reciprocal深入：为何"一次比一次"皆为平移过的倒数函数
Start with $f(x) = \dfrac{a x + b}{c x + d}$ and divide numerator by denominator (polynomial long division of degree-$1$ by degree-$1$ leaves a constant quotient and a constant remainder):
由 $f(x) = \dfrac{a x + b}{c x + d}$ 出发，对分子做长除（一次除以一次得常数商加常数余）：
$$ \frac{a x + b}{c x + d} \;=\; \frac{a}{c} \;+\; \frac{b - \tfrac{a d}{c}}{c x + d} \;=\; \frac{a}{c} \;+\; \frac{b c - a d}{c \,(c x + d)}. $$Writing $k = (b c - a d)/c^{2}$ and $h = -d/c$, this becomes
令 $k = (b c - a d)/c^{2}$、$h = -d/c$，可改写为
$$ f(x) \;=\; \frac{a}{c} \;+\; \frac{k}{x - h}. $$Therefore every linear-over-linear function is a translation of the reciprocal $y = 1/x$ horizontally by $h$, vertically by $a/c$, and scaled by $k$. The two asymptotes ($x = h$ and $y = a/c$) are now visible directly. The condition $a d - b c \ne 0$ becomes $k \ne 0$, which prevents the curve from collapsing to a horizontal line.
故每条"一次比一次"曲线都是 $y = 1/x$ 经水平平移 $h$、竖直平移 $a/c$、按 $k$ 缩放而成。两条渐近线（$x = h$ 与 $y = a/c$）现在一目了然。$a d - b c \ne 0$ 等价于 $k \ne 0$，避免曲线退化为水平线。
Vertical and Horizontal Asymptotes竖直渐近线与水平渐近线 SL 2.8
- Vertical asymptote at $x = a$. Occurs when $D(a) = 0$ and $N(a) \ne 0$. The function blows up: $|f(x)| \to \infty$ as $x \to a$.
- Horizontal asymptote of $f(x) = \dfrac{a x + b}{c x + d}$. The line $y = a / c$. This is the ratio of leading coefficients of numerator and denominator.
- Sign of approach to the vertical asymptote. Read off the sign of $f$ on each side from a sign chart. The graph may approach $+\infty$ on one side and $-\infty$ on the other.
- 竖直渐近线 $x = a$。发生于 $D(a) = 0$ 且 $N(a) \ne 0$。函数发散：$x \to a$ 时 $|f(x)| \to \infty$。
- $f(x) = \dfrac{a x + b}{c x + d}$ 的水平渐近线。直线 $y = a / c$，即分子与分母首项系数之比。
- 逼近竖直渐近线的方向。由符号表读 $f$ 在渐近线两侧的符号。曲线可能一侧趋 $+\infty$，另一侧趋 $-\infty$。
Find the vertical and horizontal asymptotes of $f(x) = \dfrac{2 x + 1}{x - 3}$.求 $f(x) = \dfrac{2 x + 1}{x - 3}$ 的竖直与水平渐近线。
Vertical asymptote. Set the denominator to zero: $x - 3 = 0 \Rightarrow x = 3$. Check that the numerator is non-zero there: $2 (3) + 1 = 7 \ne 0$. Therefore $x = 3$ is a vertical asymptote.
竖直渐近线。令分母为零：$x - 3 = 0 \Rightarrow x = 3$。检验此处分子不为零：$2 (3) + 1 = 7 \ne 0$。故 $x = 3$ 是竖直渐近线。
Horizontal asymptote. Leading coefficient of numerator is $2$, of denominator is $1$. The horizontal asymptote is $y = 2 / 1 = 2$.
水平渐近线。分子首项系数 $2$，分母首项系数 $1$。水平渐近线为 $y = 2 / 1 = 2$。
Direction of approach. Just to the left of $x = 3$, say $x = 2.9$, the numerator is positive and the denominator is a small negative, so $f \to -\infty$. Just to the right, say $x = 3.1$, both numerator and denominator are positive small, so $f \to +\infty$. The two horizontal arms approach $y = 2$ from below on the left and from above on the right.
逼近方向。取 $x = 2.9$（$x = 3$ 左侧），分子正、分母为很小的负数，故 $f \to -\infty$。取 $x = 3.1$（右侧），分子分母都是很小的正数，故 $f \to +\infty$。左半支从下方逼近 $y = 2$，右半支从上方逼近 $y = 2$。
Removable Holes可去间断点（"洞"） SL 2.8
Procedure.
- Factor numerator and denominator fully.
- Identify each shared factor $(x - a)$. Each contributes a hole at $x = a$.
- Cancel the shared factor and substitute $x = a$ into the simplified expression to find the $y$-coordinate.
- Report the hole as an ordered pair $(a, y_{\text{hole}})$.
步骤。
- 把分子分母完全因式分解。
- 找出每个公共因式 $(x - a)$，每个对应 $x = a$ 处一个"洞"。
- 约去公共因式，把 $x = a$ 代入化简式求 $y$ 坐标。
- 把"洞"写成有序对 $(a, y_{\text{洞}})$。
Locate any vertical asymptotes and removable holes of $f(x) = \dfrac{x^{2} - 4}{x - 2}$, and give the simplified form valid where the function is defined.找出 $f(x) = \dfrac{x^{2} - 4}{x - 2}$ 的竖直渐近线与可去间断点，并给出函数有定义处的化简式。
Factor. Numerator: $x^{2} - 4 = (x - 2)(x + 2)$. Denominator: $x - 2$.
因式分解。分子 $x^{2} - 4 = (x - 2)(x + 2)$；分母 $x - 2$。
Cancel the shared factor.
约去公共因式。
$$ f(x) \;=\; \frac{(x - 2)(x + 2)}{x - 2} \;=\; x + 2, \qquad x \ne 2. $$No vertical asymptote. Both numerator and denominator vanish at $x = 2$, so the candidate $x = 2$ is a hole, not an asymptote.
不存在竖直渐近线。$x = 2$ 处分子分母同为零，候选点 $x = 2$ 是"洞"，不是渐近线。
Locate the hole. Substitute $x = 2$ into the simplified expression $x + 2$: $y = 2 + 2 = 4$. Therefore $f$ has a removable hole at $(2, 4)$.
定位"洞"。把 $x = 2$ 代入化简式 $x + 2$：$y = 2 + 2 = 4$。故 $f$ 在 $(2, 4)$ 处有一个"洞"。
Graphical description. The graph is the line $y = x + 2$ with the single point $(2, 4)$ deleted. The function is still strictly undefined at $x = 2$, even though "the limit as $x \to 2$" of the expression is $4$.
几何描述。图像即直线 $y = x + 2$ 挖去点 $(2, 4)$。函数本身在 $x = 2$ 仍严格无定义，尽管表达式当 $x \to 2$ 时"极限"为 $4$。
Rational Functions with Higher-Degree Polynomials含高次多项式的有理函数 HL AHL 2.13
- $\deg N < \deg D$. Horizontal asymptote $y = 0$. Denominator outgrows numerator.
- $\deg N = \deg D$. Horizontal asymptote $y = \dfrac{\text{leading coeff of } N}{\text{leading coeff of } D}$.
- $\deg N = \deg D + 1$. No horizontal asymptote, but there is an oblique (slant) asymptote $y = q(x)$, where $q(x)$ is the quotient from polynomial long division of $N$ by $D$. Treated in B3.5.
- $\deg N \ge \deg D + 2$. Neither horizontal nor slant asymptote. The end behaviour follows the polynomial quotient, which is itself non-linear.
- $\deg N < \deg D$。水平渐近线 $y = 0$。分母比分子"长得快"。
- $\deg N = \deg D$。水平渐近线 $y = \dfrac{N \text{ 首项系数}}{D \text{ 首项系数}}$。
- $\deg N = \deg D + 1$。无水平渐近线，但存在斜（倾斜）渐近线 $y = q(x)$，其中 $q(x)$ 为 $N$ 除以 $D$ 的多项式商。B3.5 详述。
- $\deg N \ge \deg D + 2$。既无水平也无斜渐近线。末端行为由多项式商决定，该商本身非线性。
Describe the end behaviour and vertical asymptotes of (a) $f(x) = \dfrac{x^{2} - 1}{x + 3}$ and (b) $g(x) = \dfrac{x + 4}{x^{2} - 1}$.分别描述 (a) $f(x) = \dfrac{x^{2} - 1}{x + 3}$ 与 (b) $g(x) = \dfrac{x + 4}{x^{2} - 1}$ 的末端行为与竖直渐近线。
(a) Quadratic over linear. Degrees: $\deg N = 2$, $\deg D = 1$. Numerator degree exceeds denominator degree by exactly one. Therefore no horizontal asymptote; there is a slant asymptote (treated formally in B3.5). Vertical asymptote candidate: $x + 3 = 0 \Rightarrow x = -3$. Numerator at $x = -3$ is $9 - 1 = 8 \ne 0$. Hence $x = -3$ is a vertical asymptote.
(a) 二次比一次。$\deg N = 2$、$\deg D = 1$。分子次数比分母高一。故无水平渐近线，但有斜渐近线（B3.5 详述）。竖直渐近线候选：$x + 3 = 0 \Rightarrow x = -3$。$x = -3$ 处分子 $9 - 1 = 8 \ne 0$。故 $x = -3$ 是竖直渐近线。
(b) Linear over quadratic. Degrees: $\deg N = 1$, $\deg D = 2$. Numerator degree is less, so horizontal asymptote $y = 0$. Vertical asymptote candidates: $x^{2} - 1 = (x - 1)(x + 1) = 0 \Rightarrow x = \pm 1$. Numerator at $x = 1$ is $5 \ne 0$ and at $x = -1$ is $3 \ne 0$. Both candidates are genuine vertical asymptotes. The graph has two vertical asymptotes and a horizontal asymptote $y = 0$.
(b) 一次比二次。$\deg N = 1$、$\deg D = 2$。分子次数较低，水平渐近线 $y = 0$。竖直渐近线候选：$x^{2} - 1 = (x - 1)(x + 1) = 0 \Rightarrow x = \pm 1$。$x = 1$ 处分子 $5 \ne 0$；$x = -1$ 处分子 $3 \ne 0$。两候选皆为真正的竖直渐近线。图像有两条竖直渐近线及水平渐近线 $y = 0$。
Going deeper: why "leading terms dominate" at infinity深入：为何无穷远处"首项主导"
Suppose $N(x) = a_{n} x^{n} + (\text{lower order})$ and $D(x) = b_{m} x^{m} + (\text{lower order})$. Divide numerator and denominator by $x^{m}$:
设 $N(x) = a_{n} x^{n} + (\text{低次项})$、$D(x) = b_{m} x^{m} + (\text{低次项})$。分子分母同除以 $x^{m}$：
$$ \frac{N(x)}{D(x)} \;=\; \frac{a_{n} x^{n - m} + O(x^{n - m - 1})}{b_{m} + O(x^{-1})}. $$As $|x| \to \infty$, every $O(x^{-1})$ term vanishes. The limit depends only on the sign of $n - m$:
当 $|x| \to \infty$ 时，每个 $O(x^{-1})$ 项消失。极限只依赖 $n - m$ 的符号：
- $n < m$: $x^{n - m} \to 0$, so the ratio $\to 0 / b_{m} = 0$.
- $n < m$：$x^{n - m} \to 0$，比值 $\to 0 / b_{m} = 0$。
- $n = m$: $x^{n - m} = 1$, so the ratio $\to a_{n} / b_{m}$.
- $n = m$：$x^{n - m} = 1$，比值 $\to a_{n} / b_{m}$。
- $n > m$: $x^{n - m} \to \pm \infty$ depending on sign of $x$, so no finite horizontal asymptote.
- $n > m$：$x^{n - m} \to \pm \infty$（视 $x$ 的符号而定），无有限水平渐近线。
When $n = m + 1$, the polynomial-division quotient is linear and gives the slant asymptote of B3.5.
当 $n = m + 1$ 时，多项式商为一次，给出 B3.5 的斜渐近线。
Oblique (Slant) Asymptotes斜（倾斜）渐近线 HL AHL 2.13
Procedure.
- Confirm $\deg N = \deg D + 1$.
- Perform polynomial long division of $N$ by $D$.
- The quotient is the slant asymptote.
- The remainder over $D$ is the "wiggle" that decays to zero at infinity.
步骤。
- 确认 $\deg N = \deg D + 1$。
- 对 $N$ 关于 $D$ 做多项式长除。
- 商即斜渐近线。
- 余项除以 $D$ 是无穷远处衰减至零的"扰动"。
Find the slant asymptote of $f(x) = \dfrac{x^{2} + 1}{x - 1}$, and describe the end behaviour.求 $f(x) = \dfrac{x^{2} + 1}{x - 1}$ 的斜渐近线，并描述末端行为。
Degree check. $\deg N = 2$, $\deg D = 1$. Numerator degree exceeds denominator degree by one, so a slant asymptote exists.
次数检验。$\deg N = 2$、$\deg D = 1$。分子次数比分母高一，存在斜渐近线。
Polynomial long division. Divide $x^{2} + 1$ by $x - 1$.
多项式长除。用 $x - 1$ 除 $x^{2} + 1$。
First step: $x^{2} \div x = x$. Multiply back: $x \cdot (x - 1) = x^{2} - x$. Subtract: $(x^{2} + 1) - (x^{2} - x) = x + 1$.
第一步：$x^{2} \div x = x$。回乘：$x \cdot (x - 1) = x^{2} - x$。相减：$(x^{2} + 1) - (x^{2} - x) = x + 1$。
Second step: $x \div x = 1$. Multiply back: $1 \cdot (x - 1) = x - 1$. Subtract: $(x + 1) - (x - 1) = 2$.
第二步：$x \div x = 1$。回乘：$1 \cdot (x - 1) = x - 1$。相减：$(x + 1) - (x - 1) = 2$。
Result of division. Quotient $q(x) = x + 1$, remainder $r = 2$. Therefore
长除结果。商 $q(x) = x + 1$、余数 $r = 2$。故
$$ f(x) \;=\; x + 1 \;+\; \frac{2}{x - 1}. $$Slant asymptote. $y = x + 1$. As $x \to \pm \infty$, the remainder term $2 / (x - 1) \to 0$, so $f(x)$ approaches the line $y = x + 1$ from above (when $x > 1$) and from below (when $x < 1$).
斜渐近线。$y = x + 1$。当 $x \to \pm \infty$ 时，余项 $2 / (x - 1) \to 0$，故 $f(x)$ 在 $x > 1$ 一侧自上方、$x < 1$ 一侧自下方逼近 $y = x + 1$。
Vertical asymptote. The denominator zero is $x = 1$; the numerator at $x = 1$ is $1 + 1 = 2 \ne 0$. Therefore $x = 1$ is a vertical asymptote (no cancellation, no hole).
竖直渐近线。分母零点 $x = 1$；$x = 1$ 处分子 $1 + 1 = 2 \ne 0$。故 $x = 1$ 是竖直渐近线（无约分、无"洞"）。
Sign Analysis and Rational Inequalities符号分析与有理不等式 SL 2.8 · AHL 2.13
- Zeros of the numerator $N(x) = 0$ (where $f$ itself equals zero).
- Vertical asymptotes $D(x) = 0$ with $N(x) \ne 0$ (where $f$ is undefined and may flip sign).
- Move everything to one side; collect over a common denominator. You should have a single fraction compared to zero.
- Factor numerator and denominator.
- List all boundaries on a number line: numerator zeros and denominator zeros.
- Test one point in each interval and record the sign of $f$.
- Read off the solution intervals. Include numerator zeros if the inequality is $\ge 0$; never include denominator zeros (function undefined).
- 分子零点 $N(x) = 0$（$f$ 本身为零）。
- 竖直渐近线 $D(x) = 0$ 且 $N(x) \ne 0$（$f$ 无定义，且可能变号）。
- 所有项移到一侧；通分。化为"单个分式与零比较"。
- 把分子分母分别因式分解。
- 在数轴上列出所有边界：分子零点与分母零点。
- 每段取一个测试点，记录 $f$ 的符号。
- 读出解集。若不等式为 $\ge 0$，把分子零点纳入；分母零点永不纳入（函数无定义）。
Solve $\dfrac{x - 1}{x + 2} > 0$ using a sign chart.用符号表解 $\dfrac{x - 1}{x + 2} > 0$。
One fraction, already. The expression is already in the form "single fraction $> 0$".
已为单个分式。表达式已为"单个分式 $> 0$"。
Boundaries. Numerator zero: $x - 1 = 0 \Rightarrow x = 1$. Denominator zero (vertical asymptote): $x + 2 = 0 \Rightarrow x = -2$.
边界。分子零点：$x - 1 = 0 \Rightarrow x = 1$。分母零点（竖直渐近线）：$x + 2 = 0 \Rightarrow x = -2$。
Sign chart. The two boundaries split $\mathbb{R}$ into three intervals.
符号表。两条边界把 $\mathbb{R}$ 分成三段。
| $x$ | $x < -2$ | $-2 < x < 1$ | $x > 1$ |
|---|---|---|---|
| $x - 1$ (numerator) | $-$ | $-$ | $+$ |
| $x + 2$ (denominator) | $-$ | $+$ | $+$ |
| quotient sign | $+$ | $-$ | $+$ |
Read off the solution. The quotient is positive on $x < -2$ and on $x > 1$. Strict inequality means we exclude the boundaries:
读取解集。商在 $x < -2$ 与 $x > 1$ 上为正。严格不等式意味着不取边界：
$$ \boxed{\; x < -2 \quad \text{or} \quad x > 1 \;} \qquad \text{equivalently} \quad (-\infty, -2) \cup (1, \infty). $$Why not multiply through by $x + 2$? The sign of $x + 2$ depends on whether $x > -2$ or $x < -2$. Multiplying through without splitting cases would force you to keep the inequality on one half-line and reverse it on the other, doubling the work and the error rate. The sign chart handles both halves in one pass.
为何不能两边同乘 $x + 2$？$x + 2$ 的符号取决于 $x > -2$ 还是 $x < -2$。若不分情况就乘，必须在一段保持不等号、另一段翻转，工作量与出错率翻倍。符号表一次性处理两段。
Going deeper: sign analysis with repeated factors深入：含重因式时的符号分析
If a factor appears with even multiplicity (for example $(x - 3)^{2}$ in the numerator), then the sign of that factor does not change at $x = 3$. The factor squared is always non-negative. Mark the boundary on the number line but record "no sign change" at that boundary. The function value at the boundary is still zero (in this example $f(3) = 0$), so for an inequality $\ge 0$ the boundary is included; for an inequality $> 0$ it is excluded.
若某因式以偶重数出现（例如分子中的 $(x - 3)^{2}$），则该因式的符号在 $x = 3$ 处不变。平方项恒非负。仍把该边界标在数轴上，但记录"不变号"。函数在边界处的值仍为零（此例中 $f(3) = 0$），故对 $\ge 0$ 取此边界，对 $> 0$ 不取。
Conversely, an odd-multiplicity factor (such as $(x - 3)^{3}$) does flip sign at $x = 3$, exactly like the simple factor $(x - 3)$. The cube has the same sign as $(x - 3)$ itself. The rule of thumb: only the parity of the multiplicity matters for sign analysis.
反之，奇重因式（如 $(x - 3)^{3}$）在 $x = 3$ 处确实变号，规律与一次因式 $(x - 3)$ 一致。立方与 $(x - 3)$ 同号。经验法则：符号分析只看重数的奇偶。
B3 Exam StrategyB3 考试策略
- Factor first, always. Before computing anything, factor the numerator and denominator. A cancelled factor signals a hole, not a vertical asymptote.
- 永远先因式分解。动笔算之前，把分子分母分别分解。被约掉的因式提示"洞"而非竖直渐近线。
- Compare degrees second. The degree comparison decides the end behaviour: $y = 0$, ratio of leading coefficients, slant, or no asymptote at all.
- 其次比较次数。次数比较决定末端行为：$y = 0$、首项系数之比、斜渐近线或无渐近线。
- Only then do long division. Long division is needed only in the slant case ($\deg N = \deg D + 1$). Skipping it elsewhere saves time.
- 最后才做长除。只在斜渐近线情形（$\deg N = \deg D + 1$）需要长除。其它情形跳过可节省时间。
- Mark axes intercepts first. $x$-intercepts from numerator zeros (after cancellation); $y$-intercept from $f(0)$.
- 先标轴上交点。$x$ 截距由约分后的分子零点给出；$y$ 截距即 $f(0)$。
- Dash in the asymptotes. Draw each vertical asymptote as a dashed vertical line; the horizontal or slant asymptote as a dashed line of the appropriate slope.
- 虚线画渐近线。每条竖直渐近线画为虚直线；水平或斜渐近线画为对应斜率的虚线。
- Mark holes with open circles. Every cancelled factor becomes an open dot on the simplified curve at $(a, y_{\text{hole}})$.
- 用空心圆标"洞"。每个约掉的因式在化简曲线上 $(a, y_{\text{洞}})$ 处画空心圆。
- Use the sign chart to decide which side of each asymptote the graph lives on. The signs determine whether each branch leaves the asymptote upward or downward.
- 用符号表判定每条渐近线两侧曲线的位置。符号决定每支曲线"向上"或"向下"远离渐近线。
- For slant asymptotes, lay out long division neatly. Each step shows: "this divided by leading term equals quotient piece", multiply back, subtract, bring down. Mark errors immediately by checking that the remainder degree drops at every step.
- 斜渐近线题：长除排版要工整。每步显示"被除式÷首项=商的一段"，回乘、相减、落下一项。每步余式次数必须降低；若没有降低，立即返工。
- For rational inequalities, never multiply by an expression of unknown sign. Always rearrange to "single fraction $\gtrless 0$" and use a sign chart. Multiplying through is the single most-penalised slip in HL inequality questions.
- 有理不等式：永不两边同乘符号未知的表达式。务必化为"单个分式 $\gtrless 0$"并用符号表。两边同乘是 HL 不等式题最常被扣分的错误。
- State holes as ordered pairs. A hole is a point in the plane: write "$(a, y_{\text{hole}})$", not just "$x = a$". Markers reward the coordinates.
- "洞"要写成有序对。"洞"是平面上的一点：写"$(a, y_{\text{洞}})$"，不要只写"$x = a$"。评分给坐标。
Flashcards闪卡
Unit B3 Practice Quiz单元 B3 练习测验
$x < -1$: $(-)(-) / (-) = -$.
$-1 < x < 2$: $(-)(+) / (-) = +$.
$2 < x < 5$: $(+)(+) / (-) = -$.
$x > 5$: $(+)(+) / (+) = +$.
Quotient $\ge 0$ on $[-1, 2]$ (include numerator zeros) and on $(5, \infty)$ (exclude denominator zero).边界：$x = -1, 2, 5$。符号表：
$x < -1$：$(-)(-) / (-) = -$。
$-1 < x < 2$：$(-)(+) / (-) = +$。
$2 < x < 5$：$(+)(+) / (-) = -$。
$x > 5$：$(+)(+) / (+) = +$。
商 $\ge 0$ 的解集为 $[-1, 2]$（含分子零点）与 $(5, \infty)$（不含分母零点）。
Readiness Checklist备考清单
Tick each item when you can do it cold, without notes, on your first attempt.
每一条都要"裸做"做对（不看笔记、一次过）才打勾。
- Write the default domain and range of a linear-over-linear rational function写出"一次比一次"有理函数的默认定义域与值域
- Identify a vertical asymptote from a denominator zero with a non-zero numerator由分母零点（分子非零）识别竖直渐近线
- Identify a horizontal asymptote of a linear-over-linear function as the ratio of leading coefficients把"一次比一次"的水平渐近线写为首项系数之比
- Distinguish a removable hole from a vertical asymptote by checking shared factors通过公共因式区分可去间断点与竖直渐近线
- Report a hole as an ordered pair $(a, y_{\text{hole}})$ using the simplified expression用化简式把"洞"写成有序对 $(a, y_{\text{洞}})$
- HL Apply the degree-comparison rule to decide between $y = 0$, matched ratio, slant, or no asymptote用次数比较规则在 $y = 0$、首项系数之比、斜渐近线、无渐近线之间做出判断
- HL Find the vertical asymptotes of a rational function with quadratic denominator求分母为二次的有理函数的竖直渐近线
- HL Perform polynomial long division of a quadratic by a linear to extract a slant asymptote对二次除一次做多项式长除以提取斜渐近线
- HL Write a rational function in the form $q(x) + r(x) / D(x)$ and describe how each piece governs the geometry把有理函数写为 $q(x) + r(x) / D(x)$ 并说明各部分如何决定几何形态
- Build a sign chart for a rational function from numerator zeros and vertical asymptotes由分子零点与竖直渐近线建有理函数的符号表
- Solve a rational inequality without multiplying through by an expression of unknown sign解有理不等式时不两边同乘符号未知的表达式
- Include numerator zeros and exclude denominator zeros at boundary points when writing the solution set写解集时在边界处纳入分子零点、排除分母零点
- Sketch a rational function by combining asymptotes, intercepts, holes, and sign analysis综合渐近线、截距、"洞"与符号分析徒手作图
IB Paper-Style PracticeIB 试卷风格练习
B3 Practice and Solutions are on the roadmap. They will ship under Practice Questions/Unit_B3_*.html with the bilingual built-in pattern.
B3 配套的 Practice 与 Solutions 已在排期，上线后位于 Practice Questions/Unit_B3_*.html，采用双语内嵌格式。