Axiom Learning·AP PrecalculusChapter 3 · Rational Functions
有理函数 — Asymptotes, Holes, and Graphing Strategies
- 理解有理函数的定义与性质 · Understand properties of rational functions
- 用多项式长除法求等价形式 · Rewrite using polynomial long division
- 求零点、截距、渐近线与间断点 · Find zeros, intercepts, asymptotes, holes
- 用七步法绘制有理函数图像 · Graph rational functions step-by-step
- 求解有理不等式 · Solve rational inequalities with sign charts
- 用单侧极限描述垂直渐近线两侧、用极限描述洞 · Use limit notation near VAs and holes
- 求值域，并在情境中解释渐近线 · Find range and interpret asymptotes in context
| Term 术语 | Definition 定义 | Chinese 中文 |
|---|---|---|
| Rational Function | Quotient of two polynomials $f(x)=\frac{p(x)}{q(x)}$ | 有理函数 |
| Vertical Asymptote | Line $x=a$ where $a$ is a zero of $q$ and either $p(a)\neq 0$, or $a$ has greater multiplicity in $q$ than in $p$ | 垂直渐近线 |
| Horizontal Asymptote | Line $y=L$ describing end behavior | 水平渐近线 |
| Slant Asymptote | Linear end-behavior asymptote when deg $p$ = deg $q$+1 | 斜渐近线 |
| Hole (Removable Discontinuity) | $a$ is a zero of both $p$ and $q$, with multiplicity in $p$ ≥ multiplicity in $q$ | 洞（可去间断点） |
| Sign Chart | Number line diagram for solving inequalities | 符号表（数轴分析） |
| Long Division | Algorithm to divide polynomials | 多项式长除法 |
3.1 Rational Functions 有理函数
关键：先因式分解分子和分母，再判断各种特征。
含义（CED 1.7.A.1）：$\dfrac{p(x)}{q(x)}$ 衡量的是分子相对于分母有多大。分母接近 0 → 函数值很大（垂直渐近线）；分母比分子"长得快" → 函数值趋于 0；两者一样快 → 趋于一个常数（水平渐近线）。
✓ $\dfrac{x^2-4}{x+1}$ — Yes, both polynomial
✓ $\dfrac{3}{x^2-9}$ — Yes
✗ $\dfrac{\sqrt{x}+1}{x-2}$ — No, $\sqrt{x}$ is not polynomial
✗ $\dfrac{x^2}{2^x}$ — No, $2^x$ is not polynomial
3.2 Long Division 多项式长除法
Equivalently: $\dfrac{p(x)}{d(x)} = q(x) + \dfrac{r(x)}{d(x)}$
①用被除式最高次项 ÷ 除式最高次项，得商的当前项
②将该商项乘以除式，与被除式相减
③重复，直到余式次数低于除式次数
Find the quotient and remainder when $x^3$ is divided by $x^2-9$ using long division.
Divide leading term: $x^3 \div x^2 = x$. So first term of quotient is $x$.
$x\cdot(x^2-9) = x^3 - 9x$
Subtract: $x^3 - (x^3 - 9x) = 9x$
$9x \div x^2$ — degree of $9x$ (1) < degree of divisor (2), so stop.
$$x^3 = x\cdot(x^2-9) + 9x$$ Quotient: $x$, Remainder: $9x$ $$\frac{x^3}{x^2-9} = x + \frac{9x}{x^2-9}$$
Using long division, write an equivalent form of $f(x)=\dfrac{2x^3+5x^2-3x+7}{x^2+6}$.
$2x^3 \div x^2 = 2x$ → multiply: $2x(x^2+6)=2x^3+12x$ → subtract:
$(2x^3+5x^2-3x+7)-(2x^3+12x) = 5x^2-15x+7$
$5x^2 \div x^2 = 5$ → multiply: $5(x^2+6)=5x^2+30$ → subtract:
$(5x^2-15x+7)-(5x^2+30) = -15x-23$
Degree of remainder $-15x-23$ (1) < degree of divisor $x^2+6$ (2) → stop. $$f(x) = 2x+5+\frac{-15x-23}{x^2+6}$$
Use long division to write $g(x)=\dfrac{3x^3-x^2+2x-5}{x^2+1}$ in equivalent form.
Then $-x^2\div x^2=-1$; subtract to get $-x-4$.
$$g(x) = 3x-1+\frac{-x-4}{x^2+1}$$
3.3 Zeros & Intercepts 零点与截距
- x-intercepts / zeros: real zeros of $p(x)$ that are in the domain
- y-intercept: $(0, f(0))$, if $x=0$ is in the domain
⚠️ If a zero of $p(x)$ is also a zero of $q(x)$, that input is not in the domain, so it is not an x-intercept. It is a hole or a vertical asymptote; compare multiplicities (see 3.5).
y截距 = $f(0)$（若 $x=0$ 属于定义域）
注意：若分子分母有公因式，该 $x$ 不在定义域内，不是 x 截距；它是洞还是垂直渐近线，要比较重数（见 3.5）。
Find the x- and y-intercepts of $f(x)=\dfrac{x^2-64}{x+2}$.
$x^2-64=0 \Rightarrow (x-8)(x+8)=0 \Rightarrow x=8$ or $x=-8$
Check domain: $q(x)=x+2=0$ when $x=-2$; neither $8$ nor $-8$ equals $-2$. ✓
x-intercepts: $(8,0)$ and $(-8,0)$
$f(0)=\dfrac{0^2-64}{0+2}=\dfrac{-64}{2}=-32$
y-intercept: $(0,-32)$
Find the x- and y-intercepts of $h(x)=\dfrac{x^2-x-6}{x^2-4}$.
Denominator: $x^2-4=(x-2)(x+2)$, zeros at $x=\pm2$.
$x=-2$ is a common factor → hole, not an intercept.
x-intercept: $(3,0)$ only.
y-intercept: $h(0)=\dfrac{-6}{-4}=\dfrac{3}{2}$, so $(0,\tfrac{3}{2})$.
3.4 Vertical Asymptotes 垂直渐近线
Without simplifying (CED 1.9.A.1): $x=a$ is a vertical asymptote if $a$ is a zero of $q$ and either (i) $a$ is not a zero of $p$, or (ii) the multiplicity of $a$ in $q$ is greater than its multiplicity in $p$.
约去公因式的零点是间断点（洞），不是垂直渐近线。
Find and describe the vertical asymptotes of $f(x)=\dfrac{x+8}{x^2-5x+6}$.
$x^2-5x+6=(x-2)(x-3)$
Numerator: $x+8$
At $x=2$: $2+8=10\neq 0$ ✓ → vertical asymptote at $x=2$
At $x=3$: $3+8=11\neq 0$ ✓ → vertical asymptote at $x=3$
Near $x=2$: numerator $\approx 10>0$, $(x-3)\approx-1<0$.
$x\to2^-$: $(x-2)<0$, so $f\approx\dfrac{+}{(-)(-)}>0$ → $\displaystyle\lim_{x\to2^-}f(x)=+\infty$; $x\to2^+$: $(x-2)>0$ → $\displaystyle\lim_{x\to2^+}f(x)=-\infty$.
Near $x=3$: numerator $\approx 11>0$, $(x-2)\approx1>0$.
$x\to3^-$: $(x-3)<0$ → $\displaystyle\lim_{x\to3^-}f(x)=-\infty$; $x\to3^+$: $(x-3)>0$ → $\displaystyle\lim_{x\to3^+}f(x)=+\infty$.
Find the vertical asymptotes of $g(x)=\dfrac{x^2-1}{x^2-3x-4}$.
Common factor $(x+1)$ → hole at $x=-1$, not a VA.
Remaining denominator factor: $(x-4)$, $x=4$ → check numerator: $(4-1)(4+1)=15\neq 0$ ✓
Vertical asymptote: $x=4$ only.
Behavior Near a Vertical Asymptote · One-Sided Limits 垂直渐近线两侧：单侧极限
$\lim_{x\to a^-}=+\infty$
$\lim_{x\to a^+}=+\infty$
$\lim_{x\to a^-}=-\infty$
$\lim_{x\to a^+}=-\infty$
$\lim_{x\to a^-}=-\infty$
$\lim_{x\to a^+}=+\infty$
$\lim_{x\to a^-}=+\infty$
$\lim_{x\to a^+}=-\infty$
- Plug $x=a$ into every factor except $(x-a)$ to get their signs. 除了 $(x-a)$，其余因式直接代入 $a$ 看正负。
- Just left of $a$, $(x-a)$ is a tiny negative number; just right, a tiny positive number. $a$ 左侧 $(x-a)$ 是很小的负数，右侧是很小的正数。
- Combine the signs → $+\infty$ or $-\infty$. (With a calculator you can also test $x=a\pm0.01$.) 合并符号得到 $\pm\infty$，也可以用计算器代入 $a\pm0.01$ 验证。
Let $h(x)=\dfrac{x+1}{(x-1)(x-3)}$. Find each vertical asymptote and the one-sided limits there, then find $\displaystyle\lim_{x\to\pm\infty}h(x)$ and sketch the graph.
No common factors. Denominator zeros $x=1$ and $x=3$ (both multiplicity 1 → the sign changes across each).
$x+1\approx2>0$, $x-3\approx-2<0$.
$x\to1^-$: $\dfrac{+}{(-)(-)}>0$ → $\displaystyle\lim_{x\to1^-}h(x)=+\infty$. $x\to1^+$: $\dfrac{+}{(+)(-)}<0$ → $\displaystyle\lim_{x\to1^+}h(x)=-\infty$.
$x+1\approx4>0$, $x-1\approx2>0$.
$x\to3^-$: $\dfrac{+}{(+)(-)}<0$ → $\displaystyle\lim_{x\to3^-}h(x)=-\infty$. $x\to3^+$: → $\displaystyle\lim_{x\to3^+}h(x)=+\infty$.
Degree 1 < degree 2 → $\displaystyle\lim_{x\to\pm\infty}h(x)=0$ (HA $y=0$). Zero $x=-1$; $y$-intercept $h(0)=\tfrac13$.
(a) $r(x)=\dfrac{x+1}{(x-2)^2}$: find $\displaystyle\lim_{x\to2^-}r(x)$, $\displaystyle\lim_{x\to2^+}r(x)$, and $\displaystyle\lim_{x\to\pm\infty}r(x)$.
(b) $s(x)=\dfrac{x-4}{x+1}$: find $\displaystyle\lim_{x\to-1^-}s(x)$ and $\displaystyle\lim_{x\to-1^+}s(x)$.
3.5 Holes (Removable Discontinuities) 间断点
- Cancel the common factor to get the simplified function $f_{\text{simp}}(x)$
- Hole coordinates: $\big(a,\; L\big)$ where $L=\displaystyle\lim_{x\to a}f(x)=f_{\text{simp}}(a)$
约去公因式后，将 $x=a$ 代入简化函数，得到洞的坐标 $(a, L)$。
用极限的语言（CED 1.10.A.2）：$x$ 从两侧靠近 $a$ 时，函数值都趋近 $L$，即 $\displaystyle\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=\lim_{x\to a}f(x)=L$，但 $f(a)$ 本身无定义。
图像上该点画一个空心圆。
Hole or Vertical Asymptote? Compare Multiplicities 洞还是渐近线？比较重数
| Function | Mult. of $(x-1)$ in numerator | Mult. in denominator | Result at $x=1$ |
|---|---|---|---|
| $\dfrac{x^2-1}{x-1}=\dfrac{(x-1)(x+1)}{x-1}$ | 1 | 1 | Hole at $(1,2)$ |
| $\dfrac{(x-1)^2(x+3)}{x-1}$ | 2 | 1 | Hole at $(1,0)$ (not an $x$-intercept!) |
| $\dfrac{x-1}{(x-1)^2}$ | 1 | 2 | Vertical asymptote $x=1$ |
Find the discontinuities for $f(x)=\dfrac{x^2+9x+8}{x+8}$.
$x^2+9x+8=(x+8)(x+1)$
$f(x)=\dfrac{(x+8)(x+1)}{x+8}=x+1$, with $x\neq -8$
At $x=-8$: $f_{\text{simp}}(-8)=-8+1=-7$
Hole at $(-8,-7)$.
No vertical asymptotes (denominator fully cancelled).
The graph looks like $y=x+1$ with a hole at $(-8,-7)$. In limit notation: $\displaystyle\lim_{x\to-8}f(x)=-7$.
For each function, decide what happens at $x=1$. If it is a hole, give its coordinates using a limit.
(a) $a(x)=\dfrac{x^2-1}{x-1}$ (b) $b(x)=\dfrac{x-1}{x^2-2x+1}$ (c) $c(x)=\dfrac{(x-1)^2(x+3)}{x-1}$
$a(x)=\dfrac{(x-1)(x+1)}{x-1}=x+1$ for $x\neq1$. $\displaystyle\lim_{x\to1}a(x)=2$ → hole at $(1,2)$.
$b(x)=\dfrac{x-1}{(x-1)^2}=\dfrac{1}{x-1}$ for $x\neq1$. The factor $(x-1)$ is still in the denominator, so $x=1$ is a VA: $\displaystyle\lim_{x\to1^-}b(x)=-\infty$, $\displaystyle\lim_{x\to1^+}b(x)=+\infty$.
$c(x)=(x-1)(x+3)$ for $x\neq1$. $\displaystyle\lim_{x\to1}c(x)=0$ → hole at $(1,0)$.
For $h(x)=\dfrac{x^2-x-6}{x^2-4}$, identify all holes and vertical asymptotes, with exact coordinates of holes.
Common factor $(x+2)$: hole at $x=-2$. Simplified: $h_s(x)=\dfrac{x-3}{x-2}$.
$h_s(-2)=\dfrac{-5}{-4}=\dfrac{5}{4}$ → Hole at $\left(-2,\dfrac{5}{4}\right)$.
Remaining denominator: $(x-2)$, $h_s(2)=\dfrac{-1}{0}$ undefined, numerator $\neq 0$ → VA at $x=2$.
3.6 Horizontal Asymptotes & End Behavior 水平渐近线与终端行为
分子次数更低 → 水平渐近线 $y=0$
次数相等 → 最高次系数之比
(slant if $m=n+1$)
分子次数更高 → 无水平渐近线（可能有斜渐近线）
余式趋于0，所以图像趋向于该直线。
Quotient of Leading Terms 首项之商（CED 1.7.A.2–A.5）
| Quotient of leading terms is… | Example | End behavior |
|---|---|---|
| constant ÷ nonconstant ($m<n$) | $\dfrac{5}{x^2-9}\to\dfrac{5}{x^2}$ | HA $y=0$ |
| a nonzero constant ($m=n$) | $\dfrac{3x^2-4}{2x^2+5}\to\dfrac32$ | HA $y=\tfrac32$ |
| linear ($m=n+1$) | $\dfrac{2x^3-x+4}{x^2+3}\to 2x$ | Slant asymptote parallel to $y=2x$ (exact line from long division) |
| higher-degree polynomial ($m\ge n+2$) | $\dfrac{3x^3-x^2+3x+1}{x-1}\to 3x^2$ | Behaves like $3x^2$: $\displaystyle\lim_{x\to\pm\infty}=+\infty$ |
Find horizontal asymptotes (if any) for each function:
- $f(x)=\dfrac{5}{x^2-9}$
- $f(x)=\dfrac{3x^2-4}{2x^2+5}$
- $f(x)=\dfrac{3x^3-x^2+3x+1}{x-1}$
- $f(x)=\dfrac{3x}{x^2-9}$ — evaluate $\displaystyle\lim_{x\to\pm\infty}f(x)$
$m<n$ → Horizontal asymptote: $y=0$
$m=n$ → HA at $y=\dfrac{3}{2}$ (ratio of leading coefficients)
$m>n$ → No horizontal asymptote.
Since $m=n+2$, not even a slant asymptote. Quotient of leading terms: $\dfrac{3x^3}{x}=3x^2$, so $f$ has the end behavior of $3x^2$: $\displaystyle\lim_{x\to-\infty}f(x)=+\infty$ and $\displaystyle\lim_{x\to+\infty}f(x)=+\infty$.
$f(x)=\dfrac{3x}{x^2-9}$: $m(1)<n(2)$, HA at $y=0$.
$\displaystyle\lim_{x\to-\infty}f(x)=0$ and $\displaystyle\lim_{x\to+\infty}f(x)=0$.
(For $x<-3$, $f(x)<0$, so the graph approaches $y=0$ from below; for $x>3$, $f(x)>0$, so it approaches from above. 左端从下方、右端从上方趋近 $y=0$。)
Evaluate $f(x)=\dfrac{10x^4-2x^2+3x+7}{5x^4+x^3-2}$:
(a) $\displaystyle\lim_{x\to-\infty}f(x)$ (b) $\displaystyle\lim_{x\to+\infty}f(x)$
Divide numerator and denominator by $x^4$ (highest power in denominator): $$f(x)=\frac{10-\frac{2}{x^2}+\frac{3}{x^3}+\frac{7}{x^4}}{5+\frac{1}{x}-\frac{2}{x^4}}$$
All terms with $x$ in denominator → 0: $$\lim_{x\to\pm\infty}f(x)=\frac{10}{5}=2$$ Both (a) and (b): $\boxed{2}$
Horizontal asymptote: $y=2$
For $r(x)=\dfrac{2x^3-x+4}{x^2+3}$, determine whether there is a horizontal asymptote, slant asymptote, or neither. Find it if it exists.
Long division: $2x^3-x+4 \div (x^2+3)$:
$2x^3\div x^2=2x$; subtract $2x(x^2+3)=2x^3+6x$; remainder $-7x+4$.
$-7x\div x^2$ has lower degree → stop.
Slant asymptote: $y=2x$.
Use the quotient of leading terms to describe the end behavior with limits. For (c), also find the slant asymptote.
(a) $\dfrac{x^4-1}{x-3}$ (b) $\dfrac{-2x^3+x}{x+1}$ (c) $\dfrac{x^2+3x+1}{x-1}$
(b) $\dfrac{-2x^3}{x}=-2x^2$: $\displaystyle\lim_{x\to\pm\infty}=-\infty$.
(c) $\dfrac{x^2}{x}=x$ → slant asymptote parallel to $y=x$. Long division: $x^2+3x+1=(x-1)(x+4)+5$, so $\dfrac{x^2+3x+1}{x-1}=x+4+\dfrac{5}{x-1}$. Slant asymptote: $y=x+4$. $\displaystyle\lim_{x\to-\infty}=-\infty$, $\displaystyle\lim_{x\to+\infty}=+\infty$.
3.7 七步绘图策略 Seven-Step Graphing Strategy
- Factor numerator and denominator; identify and remove holes
- Find x- and y-intercepts
- Determine degrees; find end-behavior asymptote (HA or slant)
- Find vertical asymptotes from remaining denominator factors
- Check symmetry via $f(-x)$ — even/odd?
- Use test points in regions divided by VAs to determine sign
- Sketch the graph using all information
Find key features and sketch $f(x)=\dfrac{x^2-64}{x^2-16}$.
Numerator: $x^2-64=(x-8)(x+8)$
Denominator: $x^2-16=(x-4)(x+4)$
No common factors → no holes.
x-intercepts: $(8,0)$ and $(-8,0)$
y-intercept: $f(0)=\dfrac{-64}{-16}=4$ → $(0,4)$
$\deg p = \deg q = 2$, leading coefficients both 1 → HA: $y=1$
Denominator zeros: $x=4$ and $x=-4$
Check numerator: $(4)^2-64=-48\neq 0$; $(-4)^2-64=-48\neq 0$ ✓
VAs: $x=4$ and $x=-4$
$f(-x)=\dfrac{(-x)^2-64}{(-x)^2-16}=\dfrac{x^2-64}{x^2-16}=f(x)$ → Even function, symmetric about y-axis ✓
Regions: $x<-8$, $-8<x<-4$, $-4<x<0$, $0<x<4$, $4<x<8$, $x>8$
Test $x=5$: $f(5)=\dfrac{25-64}{25-16}=\dfrac{-39}{9}<0$
Test $x=10$: $f(10)=\dfrac{36}{84}\approx0.43>0$ (but below the HA $y=1$)
• Even function (symmetric)
• x-intercepts at $\pm 8$; y-intercept $(0,4)$
• VAs at $x=\pm 4$; HA at $y=1$
• Approaches $y=1$ from below as $x\to\pm\infty$, because $f(x)=1-\dfrac{48}{x^2-16}$ and $\dfrac{48}{x^2-16}>0$ when $|x|>4$.
• One-sided limits: $\displaystyle\lim_{x\to4^-}f=+\infty$, $\displaystyle\lim_{x\to4^+}f=-\infty$ (and the mirror image at $x=-4$).
Find the key features of $$g(x)=\frac{(x-1)(x+2)}{(x-1)(x-2)}$$ and sketch the graph.
The original denominator is zero at $x=1$ and $x=2$, so $$x\neq1,\qquad x\neq2.$$ Cancel the common factor $(x-1)$: $$g(x)=\frac{(x-1)(x+2)}{(x-1)(x-2)} =\frac{x+2}{x-2},\qquad x\neq1,2.$$ 约去公因式以后，$x=1$ 仍不能放回定义域。
The canceled factor gives a hole at $x=1$. Substitute $x=1$ into the simplified function: $$g_s(1)=\frac{1+2}{1-2}=\frac{3}{-1}=-3.$$ Therefore, the hole is the open point $$\boxed{(1,-3)}.$$
x-intercept: set the simplified numerator equal to zero: $$x+2=0\Rightarrow x=-2,$$ so the x-intercept is $\boxed{(-2,0)}$.
y-intercept: $$g_s(0)=\frac{0+2}{0-2}=-1,$$ so the y-intercept is $\boxed{(0,-1)}$.
After cancellation, the remaining denominator is zero at $x=2$, so $$\boxed{\text{Vertical asymptote: }x=2}.$$ The numerator and denominator have the same degree and the same leading coefficient, so $$\boxed{\text{Horizontal asymptote: }y=1}.$$ There is no slant asymptote.
Rewrite: $$\frac{x+2}{x-2}=\frac{(x-2)+4}{x-2}=1+\frac{4}{x-2}.$$ Therefore:
- When $x<2$, $\dfrac{4}{x-2}<0$, so the graph lies below $y=1$.
- When $x>2$, $\dfrac{4}{x-2}>0$, so the graph lies above $y=1$.
- Draw an open circle at $(1,-3)$.
- The graph passes through $(-2,0)$ and $(0,-1)$.
VA: $x=2$ | HA: $y=1$ | Domain: $x\neq1,2$
Use the seven-step strategy to analyze $r(x)=\dfrac{x^2-1}{x^2-x-2}$ and describe all key features.
Hole at $x=-1$: $r_s(x)=\dfrac{x-1}{x-2}$, $r_s(-1)=\dfrac{-2}{-3}=\dfrac{2}{3}$ → hole at $(-1, \tfrac{2}{3})$
x-intercept: $x=1$ → $(1,0)$
y-intercept: $r_s(0)=\dfrac{-1}{-2}=\dfrac{1}{2}$ → $(0,\tfrac{1}{2})$
HA: $\deg p=\deg q=2$, leading coefficients both 1 → $y=1$
VA: $x=2$ ($\lim_{x\to2^-}r=-\infty$, $\lim_{x\to2^+}r=+\infty$)
Symmetry: neither
3.8 Rational Inequalities 有理不等式
- Rewrite as $f(x)>0$, $\geq 0$, $<0$, or $\leq 0$
- Find critical values: zeros of numerator & denominator (VAs)
- Plot critical values on a number line
- Use test points in each interval to find the sign of $f(x)$
- Write solution intervals (include/exclude endpoints based on $\geq$ or $>$)
①把所有东西移到一边（$f(x)$ 与 0 比较）
②找临界值：分子零点（可以等于0）和分母零点（不能等于0，即VA）
③在数轴上标出这些点，逐区间用测试点判断正负
⚠️ VA处的临界值用开区间（不含），分子零点用闭/开根据 $\geq$ 或 $>$
Solve: $\dfrac{3}{x+2}\geq 2$
$\dfrac{3}{x+2}-2\geq 0 \Rightarrow \dfrac{3-2(x+2)}{x+2}\geq 0 \Rightarrow \dfrac{3-2x-4}{x+2}\geq 0 \Rightarrow \dfrac{-2x-1}{x+2}\geq 0$
Numerator $-2x-1=0$: $x=-\dfrac{1}{2}$ (can equal 0 since $\geq$)
Denominator $x+2=0$: $x=-2$ (excluded — VA)
| Interval | Test point | $-2x-1$ | $x+2$ | $f(x)$ |
|---|---|---|---|---|
| $x<-2$ | $x=-3$ | $+5$ | $-1$ | $-$ |
| $-2<x<-\frac{1}{2}$ | $x=-1$ | $+1$ | $+1$ | $+$ |
| $x>-\frac{1}{2}$ | $x=0$ | $-1$ | $+2$ | $-$ |
$f(x)\geq 0$ on $\left(-2,\,-\dfrac{1}{2}\right]$ (include $-\tfrac{1}{2}$, exclude $-2$) $$\boxed{-2 < x \leq -\tfrac{1}{2}}$$
Solve for $x$: $\dfrac{5}{x-2}\leq 3$
$\dfrac{5}{x-2}-3\leq 0 \Rightarrow \dfrac{5-3(x-2)}{x-2}\leq 0 \Rightarrow \dfrac{11-3x}{x-2}\leq 0$
Numerator: $11-3x=0 \Rightarrow x=\dfrac{11}{3}$
Denominator: $x=2$ (excluded)
| Interval | Test | $11-3x$ | $x-2$ | $f(x)$ |
|---|---|---|---|---|
| $x<2$ | $x=0$ | $+11$ | $-2$ | $-$ |
| $2<x<\frac{11}{3}$ | $x=3$ | $+2$ | $+1$ | $+$ |
| $x>\frac{11}{3}$ | $x=4$ | $-1$ | $+2$ | $-$ |
$f(x)\leq 0$ on $(-\infty, 2)\cup\left[\dfrac{11}{3},+\infty\right)$ $$\boxed{x<2 \text{ or } x\geq\dfrac{11}{3}}$$
Solve: $\dfrac{x+1}{x-3} > 2$
Critical values: $x=7$ (num zero, excluded since $>$) and $x=3$ (VA, excluded).
Sign chart: $(-\infty,3)$: test $x=0$: $\dfrac{7}{-3}<0$ ✗; $(3,7)$: test $x=5$: $\dfrac{2}{2}>0$ ✓; $(7,\infty)$: test $x=8$: $\dfrac{-1}{5}<0$ ✗
Solution: $\boxed{3 < x < 7}$
Solve: $\dfrac{x^2-4}{x+1}\geq 0$
Critical values: $x=-2, x=2$ (numerator zeros, included since $\geq$); $x=-1$ (VA, excluded).
Number line: $\ldots,-2,-1,\ldots,2,\ldots$
Test $x=-3$: $\dfrac{(-)(-)}{(-)}=\dfrac{(+)}{(-)} < 0$ ✗
Test $x=-1.5$: $\dfrac{(-)(+)}{(-)}>0$ ✓
Test $x=0$: $\dfrac{(-2)(2)}{1}=-4<0$ ✗
Test $x=3$: $\dfrac{(1)(5)}{4}>0$ ✓
Solution: $\boxed{[-2,-1)\cup[2,+\infty)}$
3.9 Range & Rational Models in Context 值域与情境应用
- Rewrite in a useful equivalent form, e.g. $\dfrac{x+2}{x-2}=1+\dfrac{4}{x-2}$. The added fraction is never 0, so $y=1$ (the HA) is never reached here. 改写成"常数 + 分式"，分式不为 0 → 取不到水平渐近线的值（但并非所有函数都如此，见上方"可以穿过 HA"）。
- A hole removes one $y$-value unless that value is reached at another $x$. 洞会让值域少一个值，除非这个值在别的 $x$ 处也能取到。
- Use the graph (or a calculator) to find any maximum or minimum values. 借助图像找最大值、最小值。
(a) Find the range of $g(x)=\dfrac{(x-1)(x+2)}{(x-1)(x-2)}$ from T-11. (b) Find the range of $k(x)=\dfrac{3}{x^2+1}$.
For $x\neq1,2$: $g(x)=1+\dfrac{4}{x-2}$. Since $\dfrac{4}{x-2}$ takes every nonzero value, $g$ takes every value except 1.
The hole: $g$ would equal $-3$ only when $1+\dfrac{4}{x-2}=-3 \Rightarrow x-2=-1 \Rightarrow x=1$, which is excluded. So $-3$ is also missing.
Range: $(-\infty,-3)\cup(-3,1)\cup(1,+\infty)$.
$x^2+1\ge1$, so $0<\dfrac{3}{x^2+1}\le3$, with the maximum 3 at $x=0$ and values approaching 0 (HA $y=0$) but never reaching it. Range: $(0,3]$.
(a) Find the range of $r(x)=\dfrac{x^2-1}{x^2-x-2}$ from E-8. (b) Find the range of $h(x)=\dfrac{1}{x^2-4}$.
(b) $x^2-4\ge-4$ and $x^2-4\neq0$. If $-4\le x^2-4<0$, then $\dfrac{1}{x^2-4}\le-\tfrac14$; if $x^2-4>0$, then $\dfrac{1}{x^2-4}>0$. Range: $\left(-\infty,-\tfrac14\right]\cup(0,+\infty)$.
Rational Models in Context 情境中的有理函数
A 10-liter drink is 20% pure juice (so it contains 2 L of juice). Pure juice is added, $x$ liters at a time. The concentration of juice is
$$C(x)=\frac{2+x}{10+x},\qquad x\ge0.$$
(a) Find $C(0)$ and interpret it. (b) How much pure juice must be added to reach 50% concentration?
(c) Find $\displaystyle\lim_{x\to\infty}C(x)$ and interpret it. (d) $C$ has a vertical asymptote at $x=-10$. Is it meaningful here? (e) State the range of $C$ in this context.
A company's cost to make $x$ items is $1500+4x$ dollars, so the average cost per item is $A(x)=\dfrac{1500+4x}{x}$, $x>0$.
(a) Find $A(100)$ and interpret it. (b) Find $\displaystyle\lim_{x\to\infty}A(x)$ and interpret it.
(c) Find $\displaystyle\lim_{x\to0^+}A(x)$ and interpret it. (d) How many items must be made for the average cost to be at most \$5?
(b) $\displaystyle\lim_{x\to\infty}A(x)=4$ (HA $y=4$): for very large production, the average cost approaches the \$4 cost of materials per item, because the fixed \$1500 is spread over more items.
(c) $\displaystyle\lim_{x\to0^+}A(x)=+\infty$ (VA $x=0$): making very few items makes the average cost extremely large.
(d) $\dfrac{1500+4x}{x}\le5 \Rightarrow 1500+4x\le5x$ (since $x>0$) $\Rightarrow x\ge1500$ items.
Chapter 3 Summary 本章总结
Key Concepts 核心知识点
| Feature | How to Find 求法 | Notes 备注 |
|---|---|---|
| Holes 洞 | Common zero of $p$ & $q$ with mult. in $p$ ≥ mult. in $q$ | Hole at $(a,L)$, $L=\lim_{x\to a}f(x)$ (cancel, then plug in) |
| VA 垂直渐近线 | Zeros of simplified denominator (mult. in $q$ > mult. in $p$) | One-sided limits $\pm\infty$: check signs; odd mult. → sign changes, even → same |
| x-intercepts 零点 | Zeros of simplified numerator in domain | Not at holes |
| y-intercept | $f(0)$ if $0$ in domain | Check VA at $x=0$ |
| HA 水平渐近线 | Compare degrees $m$ vs $n$ | $m<n$: $y=0$; $m=n$: ratio; $m>n$: none. General: quotient of leading terms |
| End behavior, $m\ge n+2$ | Quotient of leading terms is a polynomial | Same end behavior as that polynomial ($\pm\infty$) |
| Slant asymptote 斜渐近线 | Long division when $m=n+1$ | Quotient is the asymptote |
| Rational inequalities 有理不等式 | Sign chart method | Exclude VA from solution |
| Range 值域 | Equivalent form + holes + graph | A hole removes a $y$-value unless reached elsewhere |
| Context 情境 | HA = long-run value; VA = unbounded growth | Check the context domain; use units |
- 🕳️ 公因式且分子重数 ≥ 分母重数 → 洞（hole），约去后代入求 $y$ 坐标，即 $\lim_{x\to a}f(x)$
- 📏 约分后分母仍为 0 → 垂直渐近线；两侧看符号：奇数重异号，偶数重同号
- 🔝 首项之商决定端点行为：常数 → HA，一次式 → 斜渐近线，更高次 → $\pm\infty$
- 📐 次数比较 → 水平渐近线：低↓0，等↓比，高↓斜/无
- 📊 有理不等式 → 符号表法，找临界值分区间
- ⚠️ VA处的 $x$ 值永远不包含在解集中
①因式分解+洞 → ②截距 → ③端行为渐近线 → ④垂直渐近线 → ⑤对称性 → ⑥测试点 → ⑦描图
Chapter 3 Review 章末习题
Let $r$ be the function given by $r(x)=\dfrac{3(x-2)(x+4)(x+1)}{(x+4)(x+1)^2(x-5)}$. Which of the following is true about the graph of $r$ in the $xy$-plane?
答案：(A) 比较每个公共零点在分子、分母中的重数：
$x=-4$：分子 1 次、分母 1 次（分子 $\ge$ 分母）→ 洞。$x=-1$：分子 1 次、分母 2 次（分母更多）→ 约分后分母仍有 $(x+1)$ → 垂直渐近线。$x=5$ 只在分母 → 垂直渐近线。$x=2$ 只在分子 → 零点。
In the $xy$-plane, the graph of a rational function $f$ has a hole at $x=3$. Input values sufficiently close to $3$ correspond to output values arbitrarily close to $8$. Which of the following could define $f(x)$?
答案：(B) 约去 $(x-3)$ 后代入 $x=3$：(B) $2(x+1)\to 2\cdot4=8$ ✓。
(A) 约分后 $\dfrac{x+5}{x-1}\to\dfrac82=4$，洞的位置对但极限是 4；(C) 分母 $(x-3)$ 是 2 次，约分后仍在分母 → 垂直渐近线，不是洞；(D) $(x-3)$ 只在分子 → $x=3$ 是零点。
The rational function $r$ is given by $r(x)=\dfrac{4x^3-x+2}{2x-1}$. Which of the following describes the end behavior of $r$?
答案：(B) 首项之商：$\dfrac{4x^3}{2x}=2x^2$，所以 $r$ 的端点行为和 $2x^2$ 一样，两端都趋向 $+\infty$。分子次数比分母高 2，没有水平渐近线，也没有斜渐近线。
The rational function $f$ is given by $f(x)=\dfrac{2x^3+x^2-5x}{x^2+x}$. Which of the following is an equation of the slant asymptote of the graph of $f$?
答案：(A) 先约去公因式 $x$（$x=0$ 处是洞）：$f(x)=\dfrac{2x^2+x-5}{x+1}$。长除法：$2x^2+x-5=(x+1)(2x-1)-4$，所以 $f(x)=2x-1-\dfrac{4}{x+1}$，斜渐近线 $y=2x-1$。
(B) 只看了首项之商 $2x$ —— 它只给出斜率，截距要用长除法求。
The function $f$ is given by $f(x)=\dfrac{x+1}{(x-2)^2(x+3)}$. Which of the following statements is true?
答案：(B) $x\to2$：$x+1\to3$，$x+3\to5$，$(x-2)^2\to0^+$（偶次，两侧都为正）→ 两侧都是 $+\infty$。
$x\to-3$：$x+1\to-2$，$(x-2)^2\to25$；左侧 $x+3\to0^-$ → $\dfrac{-2}{25\cdot0^-}\to+\infty$，右侧 → $-\infty$。所以 (C)、(D) 都不对。
The rational functions $f$ and $g$ are given by $f(x)=\dfrac{12x}{x^2+6}$ and $g(x)=\dfrac{x+3}{x^2+5}$. If $h(x)=f(x)-g(x)$, what are all real zeros of $h$?
答案：(B) 两个分母恒为正，$h$ 的定义域是全体实数。在计算器上画 $Y_1=f(x)-g(x)$ 求零点，或求 $f$ 与 $g$ 的交点：只有 $x\approx0.332$。
（通分后分子 $11x^3-3x^2+54x-18$ 只有一个实根。）(A) 是 $f$ 和 $g$ 各自的零点，不是 $h$ 的零点。
The concentration of a medicine in a patient's blood $t$ hours after it is taken is modeled by $C(t)=\dfrac{5t}{t^2+4}$ milligrams per liter. For how many hours is the concentration at least $1$ milligram per liter?
答案：(C) 解 $\dfrac{5t}{t^2+4}\ge1$：$t^2+4>0$，两边同乘得 $t^2-5t+4\le0$，即 $(t-1)(t-4)\le0$，$1\le t\le4$，共 $3$ 小时。（也可以用计算器求 $Y_1=C(t)$ 与 $Y_2=1$ 的交点。）
(D) 是第二次降到 $1$ 的时刻，不是持续时间。
| $x$ | $-5.1$ | $-5.01$ | $-5.001$ | $-4.999$ | $-4.99$ | $-4.9$ |
|---|---|---|---|---|---|---|
| $f(x)$ | $10$ | $100$ | $1000$ | $-1000$ | $-100$ | $-10$ |
The table gives values of a rational function $f$ at selected values of $x$. The polynomials in the numerator and denominator of $f$ have no zeros in common. Based on the table, which of the following is a possible conclusion about the graph of $f$?
答案：(C) $x$ 从左边靠近 $-5$ 时输出无限增大，从右边靠近时无限减小 → $x=-5$ 处是垂直渐近线：$\displaystyle\lim_{x\to-5^-}f(x)=\infty$，$\displaystyle\lim_{x\to-5^+}f(x)=-\infty$。
分子分母没有公共零点，所以不可能是洞；输出不趋近 0，也不是零点。
The function $r$ is given by $r(x)=\dfrac{2(x-3)(x+2)(x-1)}{(x-2)(x+1)(x-1)}$.
(a) State the domain of $r$, the zeros of $r$, and the $y$-intercept of the graph.
(b) Find the vertical asymptotes. For each one, describe the behavior of $r$ on both sides using limit notation.
(c) Find the horizontal asymptote, and express the end behavior of $r$ using limit notation.
(d) The graph has a hole. Find its coordinates, and write a limit statement that describes the behavior of $r$ near the hole.
(e) Does the graph of $r$ ever cross its horizontal asymptote? Justify your answer algebraically.
(f) Sketch the graph of $r$.
(a) 定义域：$x\ne-1,\ 1,\ 2$。零点：$x=3$ 和 $x=-2$（$x=1$ 是洞，不是零点）。$y$ 截距：$r(0)=\dfrac{2(-3)(2)}{(-2)(1)}=6$，即 $(0,6)$。
(b) 约分后 $r(x)=\dfrac{2(x-3)(x+2)}{(x-2)(x+1)}$，垂直渐近线 $x=2$ 和 $x=-1$。
$x\to2$：分子 $\to2(-1)(4)=-8$，$x+1\to3$：$\displaystyle\lim_{x\to2^-}r(x)=\infty$，$\displaystyle\lim_{x\to2^+}r(x)=-\infty$。
$x\to-1$：分子 $\to2(-4)(1)=-8$，$x-2\to-3$：$\displaystyle\lim_{x\to-1^-}r(x)=-\infty$，$\displaystyle\lim_{x\to-1^+}r(x)=\infty$。
(c) 分子分母都是 2 次（约分后），首项系数之比 $\dfrac21$：水平渐近线 $y=2$，$\displaystyle\lim_{x\to\pm\infty}r(x)=2$。
(d) 约分后代入 $x=1$：$\dfrac{2(-2)(3)}{(-1)(2)}=6$，洞在 $(1,6)$；$\displaystyle\lim_{x\to1}r(x)=6$，但 $r(1)$ 无定义。
(e) 令 $\dfrac{2(x-3)(x+2)}{(x-2)(x+1)}=2$：$x^2-x-6=x^2-x-2\Rightarrow-6=-2$，无解，所以图像从不穿过 $y=2$。（也可以写成 $r(x)=2-\dfrac{8}{(x-2)(x+1)}$，后一项永远不为 0。）
(f) 参考图：
Directions: Solve each part without a calculator. Show the work that leads to your answers.
(a) Use long division to write $g(x)=\dfrac{2x^3-x^2+3}{x^2+1}$ in the form $q(x)+\dfrac{r(x)}{x^2+1}$. Then state the equation of the slant asymptote of the graph of $g$.
(b) Solve $\dfrac{x-4}{x+1}\ge2$. Write your answer in interval notation.
(c) The function $k$ is given by $k(x)=\dfrac{x^2-9}{x^2+2x-15}$. Find all values of $x$ for which $k(x)=2$.
(a) $2x(x^2+1)=2x^3+2x$，余 $-x^2-2x+3$；$-1\cdot(x^2+1)$，余 $-2x+4$。所以 $g(x)=2x-1+\dfrac{-2x+4}{x^2+1}$，斜渐近线 $y=2x-1$。
(b) 移项通分（不能两边同乘 $x+1$，它的符号未知）：$\dfrac{x-4}{x+1}-2=\dfrac{-x-6}{x+1}\ge0\iff\dfrac{x+6}{x+1}\le0$。临界值 $-6$（可取）和 $-1$（不可取）；符号表：$x<-6$ 为正，$-6<x<-1$ 为负，$x>-1$ 为正。
解集：$[-6,-1)$
(c) $k(x)=\dfrac{(x-3)(x+3)}{(x-3)(x+5)}=\dfrac{x+3}{x+5}$，且 $x\ne3,\,-5$。令 $\dfrac{x+3}{x+5}=2$：$x+3=2x+10\Rightarrow x=-7$。检查：$-7$ 在定义域内，$k(-7)=\dfrac{40}{20}=2$ ✓。所以只有 $x=-7$。
A tank initially contains $100$ liters of water with $2$ kilograms of salt dissolved in it. Salt water containing $0.3$ kilogram of salt per liter is pumped into the tank at $5$ liters per minute, and nothing leaves the tank. The concentration of salt in the tank, in kilograms per liter, $t$ minutes after the pumping starts is $C(t)=\dfrac{2+1.5t}{100+5t}$. The tank holds at most $500$ liters. A calculator is allowed.
(a) Find $C(0)$ and interpret it in context.
(b) Find the average rate of change of $C$ from $t=0$ to $t=20$. Interpret your answer in context, with units.
(c) Find $\displaystyle\lim_{t\to\infty}C(t)$. Explain what this value means in terms of the salt water being added.
(d) Find the domain of $C$ in this context. Will the concentration in the tank ever reach $0.25$ kilogram per liter? Justify your answer.
(a) $C(0)=\dfrac{2}{100}=0.02$：开始注水前，水箱里每升含盐 $0.02$ 千克。
(b) $C(20)=\dfrac{32}{200}=0.16$，$\dfrac{0.16-0.02}{20}=0.007$ kg/L per minute：前 20 分钟，浓度平均每分钟升高 $0.007$ 千克/升。
(c) 首项之商 $\dfrac{1.5t}{5t}=0.3$，$\displaystyle\lim_{t\to\infty}C(t)=0.3$：时间越长，水箱里的浓度越接近注入盐水的浓度 $0.3$ kg/L，但永远达不到（原来的 100 升清水一直在稀释）。
(d) 水量 $100+5t\le500\Rightarrow t\le80$，情境定义域 $0\le t\le80$。令 $C(t)=0.25$：$2+1.5t=25+1.25t\Rightarrow t=92$，超出定义域。$C$ 在定义域内递增，最大值 $C(80)=\dfrac{122}{500}=0.244<0.25$，所以水箱满之前浓度达不到 $0.25$ kg/L。
The graph of a rational function $f$ is shown. The dashed lines $x=-2$ and $y=-1$ are asymptotes. The graph crosses the $x$-axis at $(1,0)$ and has a hole at $x=3$. The numerator and denominator of $f$ are each the product of two linear factors.
(a) Write a possible expression for $f(x)$. Explain how each feature of the graph is used.
(b) Find the $y$-coordinate of the hole.
(c) Describe the behavior of $f$ near $x=-2$ and as $x\to\pm\infty$ using limit notation.
(d) Find the range of $f$.
(a) 零点 $x=1$ → 分子有 $(x-1)$；垂直渐近线 $x=-2$ → 分母有 $(x+2)$；洞在 $x=3$ → 分子分母各有一个 $(x-3)$。分子分母同次，水平渐近线 $y=-1$ → 首项系数之比为 $-1$。所以 $f(x)=\dfrac{-(x-1)(x-3)}{(x+2)(x-3)}$。
(b) 约分后 $f(x)=-\dfrac{x-1}{x+2}$，代入 $x=3$：$-\dfrac25$，洞在 $\left(3,-\dfrac25\right)$。
(c) $\displaystyle\lim_{x\to-2^-}f(x)=-\infty$，$\displaystyle\lim_{x\to-2^+}f(x)=\infty$；$\displaystyle\lim_{x\to\pm\infty}f(x)=-1$。
(d) $f(x)=-1+\dfrac{3}{x+2}$，约分后的函数能取到除 $-1$ 以外的所有值；而 $-\dfrac25$ 只在 $x=3$ 处取到，那里是洞，所以也取不到。值域：$(-\infty,-1)\cup\left(-1,-\dfrac25\right)\cup\left(-\dfrac25,\infty\right)$。
