曲线 $C$ 的方程为 $y=\dfrac{x^2-x}{x+1}$。

(a) 求 $C$ 的渐近线方程。

(b) 求 $C$ 上驻点的精确坐标。

(c) 画出 $C$ 的草图，并写出曲线与坐标轴所有交点的坐标。

(d) 画出方程为 $y=\Big|\dfrac{x^2-x}{x+1}\Big|$ 的曲线的草图，并用精确形式求出满足 $\Big|\dfrac{x^2-x}{x+1}\Big|<6$ 的 $x$ 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。

多项式除法分出斜渐近线与真分式，分母零点不约消。

Therefore the asymptotes are $\boxed{x=-1}$ and $\boxed{y=x-2}$.

求导后令x+1的平方等于2，先保留精确横坐标再代回，利用2除以正负根号2化简。

Thus $(x+1)^2=2$, giving $x=-1\pm\sqrt2$. For these values,

Hence the exact stationary coordinates are

The left point is a maximum and the right point a minimum, from the derivative sign changes.

先标明两个横轴交点，右支最低点在横轴稍下方。两侧都按真分式符号趋近斜渐近线。

The axis intersections are $\boxed{(0,0)}$ and $\boxed{(1,0)}$, since the numerator is $x(x-1)$.

Label $x=-1$ and $y=x-2$. For $x<-1$, the curve lies below the oblique asymptote, rises to the maximum $(-1-\sqrt2,-3-2\sqrt2)$ and falls to $-\infty$ as $x\to-1^-$.

For $x>-1$, it lies above the oblique asymptote, falls from $+\infty$ through $(0,0)$ to the minimum $(-1+\sqrt2,-3+2\sqrt2)$, then rises through $(1,0)$ towards $+\infty$. Both outer tails approach the oblique asymptote.

将原图横轴以下部分反射，0与1处都形成尖点。解不等式时左支和右支的限制不同，左支由f大于−6控制，右支只需f小于6。

For the absolute-value graph, reflect the branch with $x<-1$ and the segment $0<x<1$ in the $x$-axis; keep the other parts unchanged.

There are sharp minima at $(0,0)$ and $(1,0)$, a small intervening maximum at $(-1+\sqrt2,3-2\sqrt2)$, and a left minimum at $(-1-\sqrt2,3+2\sqrt2)$. Both sides tend to $+\infty$ at $x=-1$.

Let $f(x)=(x^2-x)/(x+1)$. We require $-6<f(x)<6$.

For $x<-1$, $f(x)<0$, so only $f(x)>-6$ is restrictive. Since the denominator is negative,

Thus $-3<x<-2$.

For $x>-1$, the minimum of $f$ is $-3+2\sqrt2>-6$, so only $f(x)<6$ is restrictive. Since the denominator is positive,

The roots are $(7\pm\sqrt{73})/2$, both greater than $-1$. Therefore the required set is