曲线 $C$ 的方程为 $y=\dfrac{x+1}{x^2+3}$。

(a) 证明 $C$ 没有竖直渐近线，并写出水平渐近线的方程。

(b) 求 $C$ 上所有驻点的坐标。

(c) 画出 $C$ 的草图，并写出它与坐标轴交点的坐标。

(d) 画出 $y^2=\dfrac{x+1}{x^2+3}$ 的草图，并写出驻点及其与坐标轴交点的坐标。

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。

分母恒大于零，所以无竖直渐近线；分母次数高于分子，水平渐近线为横轴。

Since $x^2+3>0$ for all real $x$, there are no vertical asymptotes.

Also $y\to0$ as $x\to\pm\infty$, so the horizontal asymptote is $\boxed{y=0}$.

用商法则求导并因式分解分子，两个实根都在定义域中。

Setting the numerator to zero gives $x=-3$ or $x=1$. Substitution gives $y=-1/6$ and $y=1/2$, respectively.

Thus the stationary points are

The derivative changes from negative to positive at $x=-3$ and from positive to negative at $x=1$, so these are a minimum and maximum, respectively.

分母恒正，分子决定曲线在横轴哪一边。结合驻点和无穷远方向画出完整曲线。

The intercepts are $\boxed{(-1,0)}$ and $\boxed{(0,\frac13)}$.

Sketch a smooth curve approaching the asymptote $y=0$ from below as $x\to-\infty$, decreasing to $(-3,-\frac16)$, then increasing through $(-1,0)$ and $(0,\frac13)$ to $(1,\frac12)$, and finally decreasing towards $y=0$ from above as $x\to+\infty$.

平方纵坐标要求右边非负，所以删去x小于−1的部分，剩余部分取正负平方根。两支在横轴上以竖直切线接合，不能把这个点当作水平驻点。

For real $y$, $(x+1)/(x^2+3)\geqslant0$ requires $x\geqslant-1$. The two branches are

They are reflections in the $x$-axis. The axis intersections are

Let $f(x)=(x+1)/(x^2+3)$. For $x>-1$, implicit differentiation gives

Since $y\ne0$ there, horizontal stationary points require $f'(x)=0$. Only $x=1$ lies in this interval, giving

Sketch the upper branch rising from $(-1,0)$ to $(1,1/\sqrt2)$ and then tending to $0^+$ as $x\to\infty$, with the lower branch reflected beneath it. At $(-1,0)$, $f'(-1)=1/4\ne0$, so the branches join with vertical tangent $x=-1$. There is no curve for $x<-1$.