有人知道为什么最后一项（以及第 7 项之后添加的项）与前面的项不一致吗？我不明白……
\documentclass[12pt]{article}
\usepackage{graphicx}
\usepackage{float}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage[margin=0.5in]{geometry}
\title{Unit 1 Assessment, Part 2}
\date{May 2022}
\begin{document}
\maketitle
\begin{enumerate}
\item x - intercepts of the quadratic function $f(x)=-x^{2}+7x-6$ is at: \\
$0 = -x^{2} + 7x - 6 = (x - 1)(-x + 6) \\
x = 1,x = 6$ \\
The vertical asymptotes of the reciprocal function $g(x)=\frac{1}{-x^{2}+7x-6}$: \\
$x = 1,x = 6$ \\
\\
The horizontal asymptote of the reciprocal is y = 0 since all reciprocal functions have a horizontal asymptote at y = 0. \\
\\
The interval of increase of the quadratic function is $(-\infty,3.5)$, and the interval of decrease of the quadratic function is $(3.5,\infty)$. Therefore, the interval of decrease of the reciprocal function is $(-\infty,1)\cup(1,3.5)$, and the interval of increase of the reciprocal function is $(3.5,6)\cup(6,\infty)$.\\
\\
The quadratic function has a maximum point at x = 3.5, therefore the reciprocal has a minimum point at x = 3.5. \\
\\
The positive interval of the quadratic function is (1,6), and the negative interval of the quadratic function is $(-\infty,1)\cup(6,\infty)$. Therefore, the positive interval of the reciprocal function is (1,6), and the negative interval of the quadratic function is $(-\infty,1)\cup(6,\infty)$.\\
\begin{figure}[H]
\centering
\includegraphics[scale=0.5]{unit1part2a.png}
\caption{Graph: $f(x)=-x^{2}+7x-6$ \& $g(x)=\frac{1}{-x^{2}+7x-6}$.}
\end{figure}
\item \begin{enumerate}
\item $f(x) = \frac{-2x - 5}{3x + 18}$
\begin{center}
\begin{tabular}{|c|c|c|c|c|}
\hline
Vertical asymptotes & Horizontal asymptotes & x - intercept & y - intercept & Domain \\
\hline
x = -6 & $y = -\frac{2}{3}$ & $(-\frac{5}{2},0)$ & $(0,-\frac{5}{18})$ & $D = \{x\in\mathbb{R}|x\neq-6\}$ \\
\hline
\end{tabular}
\end{center}
\begin{figure}[H]
\centering
\includegraphics[scale=0.5]{unit1part2b.png}
\caption{Graph: $f(x) = \frac{-2x - 5}{3x + 18}$.}
\end{figure}
\item Positive interval: $(-6,-2.5)$ \\
Negative intervals: $(-\infty,-6)\cup(-2.5,\infty)$
\end{enumerate}
\item Find the real roots of the following rational equations.
\begin{enumerate}
\item $\frac{-7x}{9x + 11} - 12 = \frac{1}{x} \\
\frac{-7x}{9x + 11} = \frac{1 + 12x}{x} \\
(-7x)(x) = (9x + 11)(1 + 12x) \\
-7x^2 = 9x + 108x^2 + 11 + 132x \\
115x^2 + 141x + 11 = 0 \\
x = \frac{-141\pm\sqrt{141^2 - (4)(115)(11)}}{(2)(115)} \\
x \approx -0.08,x \approx -1.14$
\item $\frac{x - 1}{x + 2} = \frac{3x + 8}{5x - 1} \\
(x - 1)(5x - 1) = (x + 2)(3x + 8) \\
5x^2 - x - 5x + 1 = 3x^2 + 8x + 6x + 16 \\
2x^2 - 20x - 15 = 0 \\
x = \frac{-(-20)\pm\sqrt{(-20)^2 - (4)(2)(-15)}}{(2)(2)} \\
x = \frac{10\pm\sqrt{130}}{2}$
\end{enumerate}
\item $8x - 3 \leq 2x+1 \leq 17x - 8 \\
8x - 3 \leq 2x+1 \\
6x \leq 4 \\
x \leq \frac{2}{3} \\
2x+1 \leq 17x - 8 \\
9 \leq 15x \\
x \geq \frac{3}{5} \\
\frac{3}{5} \leq x \leq \frac{2}{3}$
\item $\frac{5x + 4}{x - 11} < \frac{5x - 7}{x + 13} \\
\frac{5x + 4}{x - 11} - \frac{5x - 7}{x + 13} < 0 \\
\frac{(5x + 4)(x + 13) - (5x - 7)(x - 11)}{(x - 11)(x + 13)} < 0 \\
\frac{5x^2 + 65x + 4x + 52 - 5x^2 + 55x + 7x - 77}{(x - 11)(x + 13)} < 0 \\
\frac{121x - 25}{(x - 11)(x + 13)} < 0 \\
\text{Critical numbers:} \\
121x - 25 = 0 \\
x = \frac{25}{121} \\
x - 11 = 0 \\
x = 11 \\
x + 13 = 0 \\
x = -13$ \\
\begin{tabular}{|c|c|c|c|c|c|}
\hline
Intervals & Test value x & 121x - 25 & x - 11 & x + 13 & $\frac{121x - 25}{(x - 11)(x + 13)}$ \\
\hline
$x < -13$ & -14 & - & - & - & - \\
\hline
$-13 < x < \frac{25}{121}$ & 0 & - & - & + & + \\
\hline
$ \frac{25}{121} < x < 11$ & 10 & + & - & + & - \\
\hline
$x > 11$ & 12 & + & + & + & + \\
\hline
\end{tabular} \\
\\
Therefore, the solution is $(-\infty,-13)\cup(\frac{25}{121},11)$.
\item $(3 + x)(5 + x)(7 + x) = 693 \\
(15 + 3x + 5x + x^2)(7 + x) = 693 \\
105 + 15x + 56x + 8x^2 + 7x^2 + x^3 - 693 = 0 \\
x^3 + 15x^2 + 71x - 588 = 0 \\
\because 4^3 + 15(4)^2 + 71(4) - 588 = 0 \\
\therefore x - 4$ is a factor. \\
Devide $x^3 + 15x^2 + 71x - 588$ by x - 4: \\
$(x - 4)(x^2 + 19x + 127) = 0 \\
\because 19^2 - (4)(1)(127) < 0 \\
\therefore x^2 + 19x + 127 = 0$ has no real solution. \\
When x - 4 = 0, x = 4 \\
The value of x is 4 will produce a box with a volume of 693 cm^3.
\item Let x represent the width in meters. \\
$(3x + 1)(2x - 5)x \geq 8436 \\
6x^3 - 13x^2 - 5x - 8436 \geq 0 \\
\because 6(12)^3 - 13(12)^2 - 5(12) - 8436 = 0 \\
\therefore x - 12$ is a factor. \\
Devide $6x^3 - 13x^2 - 5x - 8436$ by x - 12: \\
$(x - 12)(6x^2 + 59x + 703) \geq 0 $\\
Critical number: \\
$\because 59^2 - (4)(6)(703) < 0 \\
\therefore 6x^2 + 59x + 703 = 0$ has no real solution. \\
When x - 12 = 0, x = 12 \\
\begin{tabular}{|c|c|c|c|c|}
\hline
Intervals & Test value x & x - 12 & 6x^2 + 59x + 703 & (x - 12)(6x^2 + 59x + 703) \\
\hline
$x < 12$ & 11 & - & + & - \\
\hline
$x > 12$ & 13 & + & + & + \\
\hline
\end{tabular} \\
$D = \{x\in\mathbb{R}|x\geq12\} \\
3(12) + 1 = 37 \\
2(12) - 5 = 19 \\$
When the length is greater or equal to 37 m, the height is greater or equal to 19 m, and the width is greater or equal to 12 m, the volume of the container is at least 8436 m^3.
\end{enumerate}
\end{document}
答案1
您的文档中有错误，因此实际上它无法编译。这就是为什么在输出中（无论如何您都不应该相信它，因为有错误）枚举的最后一项没有正确对齐。
- 在第 122 行，您是
cm^3在文本模式下编写的，但用 编写的指数^3必须在数学模式下。要排版单位，最好的方法是使用siunitx：您可以简单地cm^3用替换\unit{cm^3}。 - 在第 137 行，文本模式下有带指数的多项式。这应该是数学模式。
- 在第 147 行，您
m^3在文本模式下写入，因此发生的错误与 相同cm^3。
这是您示例的更正版本。
\documentclass[12pt]{article}
\usepackage{graphicx}
\usepackage{float}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage[margin=0.5in]{geometry}
\usepackage{siunitx}
\title{Unit 1 Assessment, Part 2}
\date{May 2022}
\begin{document}
\maketitle
\begin{enumerate}
\item x - intercepts of the quadratic function $f(x)=-x^{2}+7x-6$ is at: \\
$0 = -x^{2} + 7x - 6 = (x - 1)(-x + 6) \\
x = 1,x = 6$ \\
The vertical asymptotes of the reciprocal function $g(x)=\frac{1}{-x^{2}+7x-6}$: \\
$x = 1,x = 6$ \\
\\
The horizontal asymptote of the reciprocal is y = 0 since all reciprocal functions have a horizontal asymptote at y = 0. \\
\\
The interval of increase of the quadratic function is $(-\infty,3.5)$, and the interval of decrease of the quadratic function is $(3.5,\infty)$. Therefore, the interval of decrease of the reciprocal function is $(-\infty,1)\cup(1,3.5)$, and the interval of increase of the reciprocal function is $(3.5,6)\cup(6,\infty)$.\\
\\
The quadratic function has a maximum point at x = 3.5, therefore the reciprocal has a minimum point at x = 3.5. \\
\\
The positive interval of the quadratic function is (1,6), and the negative interval of the quadratic function is $(-\infty,1)\cup(6,\infty)$. Therefore, the positive interval of the reciprocal function is (1,6), and the negative interval of the quadratic function is $(-\infty,1)\cup(6,\infty)$.\\
\begin{figure}[H]
\centering
\includegraphics[scale=0.5]{example-image-a}
\caption{Graph: $f(x)=-x^{2}+7x-6$ \& $g(x)=\frac{1}{-x^{2}+7x-6}$.}
\end{figure}
\item \begin{enumerate}
\item $f(x) = \frac{-2x - 5}{3x + 18}$
\begin{center}
\begin{tabular}{|c|c|c|c|c|}
\hline
Vertical asymptotes & Horizontal asymptotes & x - intercept & y - intercept & Domain \\
\hline
x = -6 & $y = -\frac{2}{3}$ & $(-\frac{5}{2},0)$ & $(0,-\frac{5}{18})$ & $D = \{x\in\mathbb{R}|x\neq-6\}$ \\
\hline
\end{tabular}
\end{center}
\begin{figure}[H]
\centering
\includegraphics[scale=0.5]{example-image-b}
\caption{Graph: $f(x) = \frac{-2x - 5}{3x + 18}$.}
\end{figure}
\item Positive interval: $(-6,-2.5)$ \\
Negative intervals: $(-\infty,-6)\cup(-2.5,\infty)$
\end{enumerate}
\item Find the real roots of the following rational equations.
\begin{enumerate}
\item $\frac{-7x}{9x + 11} - 12 = \frac{1}{x} \\
\frac{-7x}{9x + 11} = \frac{1 + 12x}{x} \\
(-7x)(x) = (9x + 11)(1 + 12x) \\
-7x^2 = 9x + 108x^2 + 11 + 132x \\
115x^2 + 141x + 11 = 0 \\
x = \frac{-141\pm\sqrt{141^2 - (4)(115)(11)}}{(2)(115)} \\
x \approx -0.08,x \approx -1.14$
\item $\frac{x - 1}{x + 2} = \frac{3x + 8}{5x - 1} \\
(x - 1)(5x - 1) = (x + 2)(3x + 8) \\
5x^2 - x - 5x + 1 = 3x^2 + 8x + 6x + 16 \\
2x^2 - 20x - 15 = 0 \\
x = \frac{-(-20)\pm\sqrt{(-20)^2 - (4)(2)(-15)}}{(2)(2)} \\
x = \frac{10\pm\sqrt{130}}{2}$
\end{enumerate}
\item $8x - 3 \leq 2x+1 \leq 17x - 8 \\
8x - 3 \leq 2x+1 \\
6x \leq 4 \\
x \leq \frac{2}{3} \\
2x+1 \leq 17x - 8 \\
9 \leq 15x \\
x \geq \frac{3}{5} \\
\frac{3}{5} \leq x \leq \frac{2}{3}$
\item $\frac{5x + 4}{x - 11} < \frac{5x - 7}{x + 13} \\
\frac{5x + 4}{x - 11} - \frac{5x - 7}{x + 13} < 0 \\
\frac{(5x + 4)(x + 13) - (5x - 7)(x - 11)}{(x - 11)(x + 13)} < 0 \\
\frac{5x^2 + 65x + 4x + 52 - 5x^2 + 55x + 7x - 77}{(x - 11)(x + 13)} < 0 \\
\frac{121x - 25}{(x - 11)(x + 13)} < 0 \\
\text{Critical numbers:} \\
121x - 25 = 0 \\
x = \frac{25}{121} \\
x - 11 = 0 \\
x = 11 \\
x + 13 = 0 \\
x = -13$ \\
\begin{tabular}{|c|c|c|c|c|c|}
\hline
Intervals & Test value x & 121x - 25 & x - 11 & x + 13 & $\frac{121x - 25}{(x - 11)(x + 13)}$ \\
\hline
$x < -13$ & -14 & - & - & - & - \\
\hline
$-13 < x < \frac{25}{121}$ & 0 & - & - & + & + \\
\hline
$ \frac{25}{121} < x < 11$ & 10 & + & - & + & - \\
\hline
$x > 11$ & 12 & + & + & + & + \\
\hline
\end{tabular} \\
\\
Therefore, the solution is $(-\infty,-13)\cup(\frac{25}{121},11)$.
\item $(3 + x)(5 + x)(7 + x) = 693 \\
(15 + 3x + 5x + x^2)(7 + x) = 693 \\
105 + 15x + 56x + 8x^2 + 7x^2 + x^3 - 693 = 0 \\
x^3 + 15x^2 + 71x - 588 = 0 \\
\because 4^3 + 15(4)^2 + 71(4) - 588 = 0 \\
\therefore x - 4$ is a factor. \\
Devide $x^3 + 15x^2 + 71x - 588$ by x - 4: \\
$(x - 4)(x^2 + 19x + 127) = 0 \\
\because 19^2 - (4)(1)(127) < 0 \\
\therefore x^2 + 19x + 127 = 0$ has no real solution. \\
When x - 4 = 0, x = 4 \\
The value of x is 4 will produce a box with a volume of 693 \unit{cm^3}.
\item Let x represent the width in meters. \\
$(3x + 1)(2x - 5)x \geq 8436 \\
6x^3 - 13x^2 - 5x - 8436 \geq 0 \\
\because 6(12)^3 - 13(12)^2 - 5(12) - 8436 = 0 \\
\therefore x - 12$ is a factor. \\
Devide $6x^3 - 13x^2 - 5x - 8436$ by x - 12: \\
$(x - 12)(6x^2 + 59x + 703) \geq 0 $\\
Critical number: \\
$\because 59^2 - (4)(6)(703) < 0 \\
\therefore 6x^2 + 59x + 703 = 0$ has no real solution. \\
When x - 12 = 0, x = 12 \\
\begin{tabular}{|c|c|c|c|c|}
\hline
Intervals & Test value x & x - 12 & $6x^2 + 59x + 703$ & $(x - 12)(6x^2 + 59x + 703)$ \\
\hline
$x < 12$ & 11 & - & + & - \\
\hline
$x > 12$ & 13 & + & + & + \\
\hline
\end{tabular} \\
$D = \{x\in\mathbb{R}|x\geq12\} \\
3(12) + 1 = 37 \\
2(12) - 5 = 19 \\$
When the length is greater or equal to 37 m, the height is greater or equal to 19 m, and the width is greater or equal to 12 m, the volume of the container is at least 8436 \unit{m^3}.
\end{enumerate}
\end{document}
现在，它确实可以编译，并且枚举的最后一项在输出中正确对齐。
我觉得我还应该指出，即使不会导致 LaTeX 错误，您的文档中仍然有印刷错误。以下是两条建议。
x有时，当要将其作为数学变量时，会以文本模式书写，例如在枚举中的第 7 项的第一行。当“x”表示数学量时，应始终以数学模式书写。equation如果使用显示数学环境（例如或）来编写代数运算和其他数学内容align，许多行将更容易阅读。gather
答案2
问题是由第七项中的 m^3 引起的。这是个错误；输入m^3m 立方仅在数学模式下有效。请m\textsuperscript{3}改用。