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

(a) 写出 $C$ 的渐近线方程。

(b) 证明：对所有实数 $x$，$-\dfrac{17}{7}\leqslant y<1$。

(c) 求 $C$ 上所有驻点的坐标。

(d) 画出 $C$ 的草图，并写出它与坐标轴交点的坐标。

(e) 画出方程为 $y=\dfrac{|x|^2+|x|-4}{|x|^2+|x|+2}$ 的图像，并求满足

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。

分子分母同次，最高次项系数之比为1。分母配方恒正，所以没有竖直渐近线。

The asymptote is $\boxed{y=1}$.

There are no vertical asymptotes because

for every real $x$.

把方程视作关于x的二次式，实数x要求判别式非负；最后单独排除使二次项消失的y=1。

Rearranging the equation gives

The value $y=1$ would give $6=0$, so is impossible. For any other attainable value of $y$, this quadratic has a real root, hence

Therefore $-\frac{17}{7}\leqslant y\leqslant1$. Excluding the impossible value $y=1$ gives

改写成1减去正分式。分母配方得到最小值，正分式相应得到最大值，于是直接夹出值域。

Write

where

It follows that

Subtracting from $1$ gives

Equality at the lower bound occurs at $x=-\frac12$; the upper bound is approached as $|x|\to\infty$.

对1减正分式求导更简短。分母恒正，只有导数分子为零给出驻点。

Differentiating $y=1-6/(x^2+x+2)$,

This is zero only when $x=-\frac12$, giving $y=-\frac{17}{7}$. Thus the only stationary point is

The derivative changes from negative to positive there, so it is a minimum.

先求三个截距，结合最低点、对称轴和水平渐近线画图。分母配方显示曲线关于x=−1/2对称。

At $x=0$, $y=-2$. At an $x$-intercept the numerator vanishes:

Thus the intercepts are

Sketch a smooth curve symmetric about $x=-\frac12$, decreasing to $(-\frac12,-\frac{17}{7})$ and then increasing. It approaches $y=1$ from below as $x\to\pm\infty$ and has no vertical asymptote.

把输入换成绝对值，应保留原曲线右半边并向左反射；不是把纵坐标取绝对值。解不等式时令u为x的绝对值，分母恒正，可直接交叉相乘。

Let $f(x)=(x^2+x-4)/(x^2+x+2)$. For $y=f(|x|)$, keep the part of $C$ with $x\geqslant0$ and reflect it in the $y$-axis.

The graph is symmetric about the $y$-axis, with minimum $(0,-2)$ and intercepts

It rises on each side away from the origin and tends to $y=1$ from below. Since $f'(0)=\frac32$, its one-sided gradients at $x=0$ are $-\frac32$ and $\frac32$, so there is a sharp corner at $(0,-2)$.

For the inequality, put $u=|x|\geqslant0$. Since $u^2+u+2>0$,

With $u\geqslant0$, this gives $0\leqslant u<1$. Hence