曲线 $C$ 的方程为 $y=\dfrac{x^2+2x-15}{x-2}$。

(a) 求 $C$ 的渐近线方程。

(b) 证明 $C$ 没有驻点。

(c) 画出 $C$ 的示意图，并写出它与坐标轴的交点坐标。

(d) 画出方程为 $y=\Big|\dfrac{x^2-2x-15}{x-2}\Big|$ 的曲线的示意图。

(e) 求满足 $\Big|\dfrac{2x^2+4x-30}{x-2}\Big|<15$ 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。

多项式除法分出直线项与真分式，读出斜渐近线和分母零点。

Division gives

The remainder is nonzero, so the asymptotes are

从除法结果求导，导数恒大于零，因此不存在驻点。

Differentiating the expression from (a),

Thus the derivative is never zero on the domain, so $C$ has no stationary points.

也可用商法则，把驻点条件化为二次方程，其判别式为负。

The quotient rule gives

A stationary point would require $x^2-4x+11=0$. Its discriminant is $(-4)^2-4(11)=-28<0$, so there are no real solutions and no stationary points.

求三处轴交点，再用余项符号判断两支相对斜渐近线的位置；两支都严格上升。

Since $x^2+2x-15=(x+5)(x-3)$, the intercepts are

Label $x=2$ and $y=x+4$. For $x<2$, sketch an increasing branch above the oblique asymptote, passing through $(-5,0)$ and $(0,15/2)$ and tending to $+\infty$ as $x\to2^-$. For $x>2$, sketch an increasing branch below the oblique asymptote, tending from $-\infty$ as $x\to2^+$ through $(3,0)$ and then to $+\infty$.

Both outer tails approach the oblique asymptote.

按本页提醒中与评分图一致的加2x解释，把(c)曲线的负部分向上反射。原横轴交点成为尖点，竖直渐近线两侧都趋于正无穷。

Under the $+2x$ interpretation identified in the warning above, sketch

Reflect the portions of $C$ below the $x$-axis in that axis, keeping the other portions unchanged.

The graph has sharp minima at $(-5,0)$ and $(3,0)$, passes through $(0,15/2)$ and tends to $+\infty$ on both sides of $x=2$. Its outer asymptotes are $y=-x-4$ as $x\to-\infty$ and $y=x+4$ as $x\to+\infty$.

先将绝对值式化成原函数在正负二分之十五之间的条件，分别求两条水平边界的交点，再按两支单调性判区间。

Let $f(x)=\dfrac{x^2+2x-15}{x-2}$ for $x\ne2$. The inequality is equivalent to

For the upper boundary,

giving $x=0,11/2$.

For the lower boundary,

giving $x=-12,5/2$.

Since $f$ is strictly increasing on each side of $x=2$, the values between the two horizontal levels occur for

All endpoints are excluded because the inequality is strict; neither interval includes $x=2$.