曲线 $C$ 的方程为 $y=\dfrac{x^2-x+2}{4x+4}$。

(a) 求 $C$ 的渐近线方程。

(b) 求 $C$ 上所有驻点的坐标。

(c) 画出 $C$ 的草图。

(d) 画出方程为 $y=\bigg|\dfrac{x^2-x+2}{4x+4}\bigg|$ 的曲线的草图。

(e) 求满足 $\bigg|\dfrac{x^2-x+2}{4x+4}\bigg|<2$ 的 $x$ 值的集合，用精确形式表示。

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。

先用多项式除法拆出直线和趋零的分式余项；分母为零而分子非零处给出竖直渐近线。

Since

we can write

The remainder term tends to zero as $|x|\to\infty$, giving the oblique asymptote $\boxed{y=x/4-1/2}$.

At $x=-1$ the denominator is zero but the numerator is $4\ne0$, so the vertical asymptote is $\boxed{x=-1}$.

利用拆分后的形式求导，求驻点并代回原函数得到坐标。

Differentiating the expression in (a),

At a stationary point,

Substitution gives $y(1)=1/4$ and $y(-3)=-7/4$. Thus the stationary points are

The derivative is positive for $x<-3$ and $x>1$, and negative for $-3<x<-1$ and $-1<x<1$. Hence $(-3,-7/4)$ is a local maximum and $(1,1/4)$ a local minimum.

草图标清两条渐近线及驻点。分子恒正，因此左支全在横轴下方，右支全在横轴上方。

Sketch two branches with asymptotes $x=-1$ and $y=x/4-1/2$:

• For $x<-1$, the branch is below the oblique asymptote, has a smooth maximum at $(-3,-7/4)$, and tends to $-\infty$ as $x\to-1^-$.
• For $x>-1$, the branch is above the oblique asymptote, tends to $+\infty$ as $x\to-1^+$, and has a smooth minimum at $(1,1/4)$.
• Mark the $y$-intercept $(0,1/2)$. There are no $x$-intercepts, since $x^2-x+2=(x-1/2)^2+7/4>0$.

对整个函数取绝对值，保留横轴上方的右支，把横轴下方的左支反射到横轴上方。

Keep the branch for $x>-1$ unchanged. Reflect the entire branch for $x<-1$ in the $x$-axis, so its maximum $(-3,-7/4)$ becomes a minimum $(-3,7/4)$.

Both branches are now above the $x$-axis and tend to $+\infty$ on their respective sides of $x=-1$. Retain the vertical asymptote $x=-1$. At the left-hand end the reflected oblique asymptote is $y=-x/4+1/2$; at the right-hand end it remains $y=x/4-1/2$.

按分母符号分左右两支。左支为负，只需解y>−2；右支为正，只需解y<2。乘负分母时必须反向不等号。

The numerator $x^2-x+2$ is positive for all real $x$.

For $x<-1$, $y<0$, so $|y|<2$ is equivalent to $y>-2$. Since $4x+4<0$,

This gives $-5<x<-2$, entirely within $x<-1$.

For $x>-1$, $y>0$, so $|y|<2$ is equivalent to $y<2$. Multiplying by the positive denominator,

Its roots are $(9\pm\sqrt{105})/2$, so this gives the interval between them. The lower root is greater than $-1$.

Therefore the required set is