曲线 $C$ 的方程为 $y=\dfrac{x^2+a}{x+a}$，其中 $a$ 为正常数。

(a) 求 $C$ 的渐近线方程。

(b) 求 $C$ 上驻点的 $x$ 坐标，以 $a$ 表示。

(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 $y=\left|\dfrac{x^2+a}{x+a}\right|$ 的曲线的草图。

(e) 求使方程 $\left|\dfrac{x^2+a}{x+a}\right|=a$ 有两个实数解的 $a$ 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。

先除出斜线与余项，利用a为正确认分母零点不被约去。

Polynomial division gives

Since $a>0$, $a^2+a>0$, so $x=-a$ is a genuine pole. The asymptotes are

对拆分形式求导，解导数为零后写成指定的a表达式。

Differentiating,

At a stationary point, $(x+a)^2=a^2+a$. Hence the required coordinates are

Neither equals $-a$ because $a^2+a>0$.

分子恒正，所以左支为负，右支为正。标明纵轴截距1及两条渐近线，并利用导数画光滑转折。

Let $h=\sqrt{a^2+a}>a$. The numerator $x^2+a$ is positive, so there are no $x$-intercepts. The $y$-intercept is $\boxed{(0,1)}$.

Sketch two branches with asymptotes $x=-a$ and $y=x-a$:

• The branch for $x<-a$ lies below the oblique asymptote, with a smooth maximum at $(-a-h,-2a-2h)$, and tends to $-\infty$ as $x\to-a^-$.
• The branch for $x>-a$ lies above the oblique asymptote, tends to $+\infty$ as $x\to-a^+$, passes through $(0,1)$, and has a smooth minimum at $(-a+h,-2a+2h)$ in the first quadrant.

Both branches approach the oblique asymptote at their outer ends.

将横轴下方的整条左支反射到上方，右支保持不动。

Reflect the branch for $x<-a$ in the $x$-axis and leave the branch for $x>-a$ unchanged. Both branches are positive and tend to $+\infty$ on their respective sides of $x=-a$.

The left branch has a smooth minimum at $(-a-h,2a+2h)$ and approaches $y=-x+a$ as $x\to-\infty$. The right branch retains its minimum and approaches $y=x-a$ as $x\to+\infty$. Mark $(0,1)$ and the vertical asymptote.

绝对值等于a要分成函数值等于正a或负a两种。先证明负a没有实解，再对正a方程要求判别式严格为正。

Since $a>0$, the equation splits into

The second gives $x^2+ax+a+a^2=0$, whose discriminant is

Thus it has no real roots.

The first gives $x^2-ax+a-a^2=0$. It has two distinct real roots exactly when

Since $a>0$, this is equivalent to $a>4/5$. Substitution of $x=-a$ into this quadratic gives $a^2+a>0$, so neither root is excluded by the original denominator.

Therefore $\boxed{a>4/5}$.