曲线 $C$ 的方程为 $y=\mathrm{f}(x)$，其中 $\mathrm{f}(x)=\dfrac{x^2+2}{x^2-x-2}$。

(a) 求 $C$ 的渐近线方程。

(b) 求 $C$ 上所有驻点的坐标，答案精确到小数点后一位。

(c) 画出 $C$ 的草图，并写出它与坐标轴的所有交点的坐标。

(d) 画出方程为 $y=\dfrac1{\mathrm{f}(x)}$ 的曲线的草图。

(e) 求使 $\dfrac1{\mathrm{f}(x)}<\mathrm{f}(x)$ 成立的取值集合。

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

分母因式分解且分子恒正，给两条竖直渐近线；最高次项系数之比为1。

Since $x^2-x-2=(x+1)(x-2)$ and $x^2+2>0$,

are the vertical asymptotes. The numerator and denominator have the same degree and leading coefficient, so the horizontal asymptote is $\boxed{y=1}$.

商法则求导后，分子化成负的二次式。先保留精确横坐标再代回求纵坐标，最后一起保留一位小数。

Thus $x^2+8x-2=0$, giving $x=-4\pm3\sqrt2$, neither excluded from the domain. Substitution gives

Therefore, to one decimal place, the stationary points are

分子恒正，所以没有横轴交点；注意左支会在x=−4穿过水平渐近线，不能画成永远在其上方。

At $x=0$, $f(0)=-1$, so the only axis intersection is $\boxed{(0,-1)}$. There are no $x$-intercepts since $x^2+2>0$.

Also

so the curve crosses $y=1$ at $\boxed{(-4,1)}$.

Sketch three branches with asymptotes $x=-1$, $x=2$ and $y=1$:

• For $x<-1$, approach $y=1$ from below as $x\to-\infty$, decrease to the shallow minimum $(-4-3\sqrt2,2\sqrt2/3)$, then rise through $(-4,1)$ to $+\infty$ as $x\to-1^-$.
• For $-1<x<2$, the curve is negative, tends to $-\infty$ at both ends and has maximum $(-4+3\sqrt2,-2\sqrt2/3)$, passing through $(0,-1)$.
• For $x>2$, decrease from $+\infty$ to $y=1$ from above.

共同定义域必须排除−1和2。把不等式移到同一边后因式分解，分子零点和分母零点一同列入符号表，不能直接乘符号不定的分母。

The common domain is $x\ne-1,2$; $f(x)$ has no zeros because $x^2+2>0$. On this domain,

We require this expression to be positive. Since $x^2+2>0$, consider the sign of

The critical values in order are $-4,-1,0,1/2,2$.

The zeros are excluded by the strict inequality, and $-1,2$ are outside the domain. Hence