曲线 $C$ 的方程为 $y=\dfrac{4x^2+x+1}{2x^2-7x+3}$。

(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，并写出它与坐标轴的所有交点的坐标。

(d) 画出方程为 $y=\left|\dfrac{4x^2+x+1}{2x^2-7x+3}\right|$ 的曲线的草图，并写出使方程 $\left|\dfrac{4x^2+x+1}{2x^2-7x+3}\right|=k$ 有 $4$ 个不同实数解的 $k$ 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。

分母因式分解且与分子没有公因子，得到两条竖直渐近线；最高次项系数比给水平渐近线。

The denominator factors as

The numerator is nonzero at both roots, so the vertical asymptotes are $\boxed{x=\frac12}$ and $\boxed{x=3}$.

The numerator and denominator have the same degree, so the horizontal asymptote is $\boxed{y=2}$.

用商法则并整理导数分子。分母平方在定义域内恒正，驻点由分子为零决定。

The quotient rule gives

Thus stationary points have $x=-\frac13$ or $x=1$, both in the domain. Substituting into the curve gives

Hence the stationary points are

分子恒正，所以三支的正负由分母决定。用导数符号区分左支最低点与中支最高点，并标明纵轴截距。

The numerator satisfies

so there are no $x$-intercepts. At $x=0$, $y=\frac13$, giving the only axis intersection $\boxed{(0,\frac13)}$.

Sketch the three branches with asymptotes $x=\frac12$, $x=3$ and $y=2$:

• For $x<\frac12$, the curve is positive. It approaches $y=2$ from below as $x\to-\infty$, decreases to the minimum $(-\frac13,\frac15)$, then increases through $(0,\frac13)$ to $+\infty$ as $x\to\frac12^-$.
• For $\frac12<x<3$, it is negative, tending to $-\infty$ at both ends and having a maximum at $(1,-3)$.
• For $x>3$, it decreases from $+\infty$ to $y=2$ from above.

These directions follow from the sign of $-10(3x+1)(x-1)$ in (b).

把中间负支向上反射，外侧两支保持。数水平线的交点时要分别考虑三支的高度范围，等号处只有切点不能计成两个交点。

For the absolute-value graph, reflect the branch with $\frac12<x<3$ in the $x$-axis. Keep the other two branches unchanged.

The middle branch now has minimum $(1,3)$ and tends to $+\infty$ at both vertical asymptotes. The left minimum remains $(-\frac13,\frac15)$ and the horizontal asymptote remains $y=2$.

If $k>3$, the line $y=k$ meets the middle branch twice, the left branch once and the right branch once, giving four distinct real solutions.

For $k<3$ the middle branch gives no intersections, while the outer branches give at most two in total. For $k=3$ the middle branch contributes only its minimum, so there are three intersections. Therefore the required set is