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

(a) 求 $C$ 的渐近线方程。

(b) 求 $C$ 上转折点的坐标。

(c) 画出 $C$ 的示意图。

(d) 在同一张图上画出方程分别为

的两条曲线的示意图，并清楚标明每条曲线。

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。

多项式除法直接给斜渐近线，分母零点不可约消，给竖直渐近线。

Hence the asymptotes are $\boxed{x=3}$ and $\boxed{y=x+5}$.

从除法式求导，驻点方程是x−3的平方等于16，代回得坐标。

Setting this equal to zero gives $(x-3)^2=16$, so $x=-1$ or $x=7$. Substituting gives the turning points

The derivative changes from positive to negative at $x=-1$ and from negative to positive at $x=7$, so these are a maximum and minimum, respectively.

分子是完全平方，左支始终非正且只在−1处接触横轴，右支始终为正。

Since $y=(x+1)^2/(x-3)$, the axis intersections are $\boxed{(-1,0)}$ and $\boxed{(0,-1/3)}$.

Label $x=3$ and $y=x+5$. For $x<3$, the curve lies below the oblique asymptote, rises from $-\infty$ to touch the $x$-axis at the maximum $(-1,0)$, then falls through $(0,-1/3)$ to $-\infty$ as $x\to3^-$.

For $x>3$, the branch lies above the oblique asymptote, falls from $+\infty$ to the minimum $(7,16)$ and then rises to $+\infty$. Both tails approach the oblique asymptote.

绝对值图把左支反射，−1处原本是二重零点，所以反射后仍光滑。平方纵坐标要求右边非负，除了x大于3的两支，还保留孤立点(−1,0)，它与绝对值图的最低点重合。

Let $f(x)=(x+1)^2/(x-3)$.

For $y=|f(x)|$, reflect the entire part with $x<3$ in the $x$-axis and keep the part with $x>3$ unchanged. The left branch has a smooth minimum at $(-1,0)$ and tends to $+\infty$ at both its outer end and $x\to3^-$. The right branch keeps its minimum $(7,16)$. Its far-left and far-right oblique asymptotes are $y=-x-5$ and $y=x+5$.

For $y^2=f(x)$, the right-hand side is nonnegative exactly when $x>3$ or $x=-1$. Thus draw

together with the isolated point $(-1,0)$. This point is shared with the minimum of the absolute-value curve.

The upper square-root branch tends to $+\infty$ at $x\to3^+$, has minimum $(7,4)$ and then rises without bound; the lower branch is its reflection in the $x$-axis, with maximum $(7,-4)$. Both approach the vertical line $x=3$ at its right-hand side. There is no square-root branch for $x<3$, apart from the isolated point.

Show both curves on one diagram and label them clearly. For $x>3$, $f(x)\geqslant16$, so $f(x)>\sqrt{f(x)}$; the positive square-root branch lies below the absolute-value branch.