📚 AQA A-Level Maths Question 11: Rational Functions and Curve Sketching | AQA A-Level 数学第11题：有理函数与曲线作图
Question 11 in AQA A-Level Mathematics Paper 1 often brings together several pure topics: algebraic division, differentiation, stationary points, asymptotes and curve sketching. This article works through a typical 10-mark Question 11 involving a rational function. The same method applies to many past-paper questions where a curve is defined explicitly and you are asked to describe its behaviour fully.
AQA A-Level 数学试卷一的第11题经常综合考查代数除法、微分、驻点、渐近线与曲线作图。本文通过一道典型的有理函数第11题（约10分）进行完整解析。这些方法同样适用于历年真题中以显式函数给出的曲线行为分析题。
1. The Question | 题目
The curve C has equation y = (x² + 1)/(x - 1), where x ≠ 1.
曲线 C 的方程为 y = (x² + 1)/(x - 1)，其中 x ≠ 1。
(a) Show that y can be written as y = x + 1 + 2/(x - 1).
(a) 证明 y 可以写成 y = x + 1 + 2/(x - 1)。
(b) Show that dy/dx = (x² - 2x - 1)/(x - 1)².
(b) 证明 dy/dx = (x² - 2x - 1)/(x - 1)²。
(c) Find the exact coordinates of the stationary points of C and determine their nature.
(c) 求曲线 C 驻点的精确坐标，并判断其性质。
(d) State the equations of any asymptotes and find the coordinates of any intersections with the coordinate axes.
(d) 写出所有渐近线方程，并求曲线与坐标轴交点的坐标。
(e) Sketch the curve C.
(e) 绘制曲线 C 的图像。
2. Rewriting f(x) by Algebraic Division | 通过代数除法重写 f(x)
² + 1 = (x - 1)(x + 1) + 2. Dividing through by x - 1 gives the required form immediately.
要对 x² + 1 除以 x - 1，可发现 x² + 1 = (x - 1)(x + 1) + 2。两边同时除以 x - 1，即可得到所需形式。
(x² + 1)/(x - 1) = x + 1 + 2/(x - 1)
asymptote y = x + 1 and makes substitution of stationary point x-values much easier later.
这种重写形式非常有用，因为它直接揭示了斜渐近线 y = x + 1，也使得后续代入驻点 x 值求 y 坐标更加简便。
3. Differentiating with the Quotient Rule | 用商法则求导
Write u = x² + 1 and v = x - 1. Then u' = 2x and v' = 1. The quotient rule gives dy/dx = (v u' - u v')/v².
设 u = x² + 1，v = x - 1，则 u' = 2x，v' = 1。由商法则可得 dy/dx = (v u' - u v')/v²。
dy/dx = ((x - 1)(2x) - (x² + 1)(1))/(x - 1)²
r: 2x² - 2x - x² - 1 = x² - 2x - 1. Therefore dy/dx = (x² - 2x - 1)/(x - 1)², as required.
展开分子：2x² - 2x - x² - 1 = x² - 2x - 1。因此 dy/dx = (x² - 2x - 1)/(x - 1)²，得证。
ivative is also undefined at x = 1. This will become important when discussing asymptotes.
由于原函数定义域排除 x = 1，导数在 x = 1 处也无定义。这在后面讨论渐近线时非常重要。
4. Solving for Stationary Points | 解驻点
tor (x - 1)² is never zero on the domain, we only need to set the numerator equal to zero.
驻点出现在 dy/dx = 0 处。由于在定义域内分母 (x - 1)² 不可能为零，因此只需令分子等于零。
x² - 2x - 1 = 0
Use the quadratic formula with a = 1, b = -2 and c = -1:
使用二次公式，其中 a = 1，b = -2，c = -1：
x = (2 ± √4 + 4)/2 = 1 ± √2
th values are allowed because neither equals 1. Now substitute into y = x + 1 + 2/(x - 1).
这两个值均可取，因为都不等于 1。现在代入 y = x + 1 + 2/(x - 1)。
For x = 1 + √2, 2/(x - 1) = 2/√2 = √2, so y = 1 + √2 + 1 + √2 = 2 + 2√2.
对于 x = 1 + √2，2/(x - 1) = 2/√2 = √2，因此 y = 1 + √2 + 1 + √2 = 2 + 2√2。
For x = 1 - √2, 2/(x - 1) = 2/(-√2) = -√2, so y = 1 - √2 + 1 - √2 = 2 - 2√2.
对于 x = 1 - √2，2/(x - 1) = 2/(-√2) = -√2，因此 y = 1 - √2 + 1 - √2 = 2 - 2√2。
The stationary points are (1 + √2, 2 + 2√2) and (1 - √2, 2 - 2√2).
驻点坐标为 (1 + √2, 2 + 2√2) 和 (1 - √2, 2 - 2√2)。
5. Classifying the Stationary Points | 判断驻点性质
(x² - 2x - 1)/(x - 1)², differentiate using the quotient rule or simplify first. The result is:
再次求导得到二阶导数。从 dy/dx = (x² - 2x - 1)/(x - 1)² 出发，使用商法则或先化简后求导，结果为：
d²y/dx² = 4/(x - 1)³
At x = 1 + √2, the denominator (x - 1)³ = (√2)³ is positive, so d²y/dx² > 0. This means the point is a local minimum.
当 x = 1 + √2 时，分母 (x - 1)³ = (√2)³ 为正，所以 d²y/dx² > 0。这表明该点是局部极小值点。
At x = 1 - √2, the denominator (x - 1)³ = (-√2)³ is negative, so d²y/dx² < 0. This means the point is a local maximum.
当 x = 1 - √2 时，分母 (x - 1)³ = (-√2)³ 为负，所以 d²y/dx² < 0。这表明该点是局部极大值点。
| x | y | d²y/dx² | Nature |
| 1 + √2 | 2 + 2√2 | positive | minimum |
| 1 - √2 | 2 - 2√2 | negative | maximum |
Always give exact surd coordinates unless the question asks for decimals. AQA often awards marks for leaving answers in exact form.
除非题目要求小数，否则应始终保留根号形式的精确坐标。AQA 通常对保留精确答案给予给分。
6. Intercepts with the Axes | 与坐标轴的交点
original equation: y = (0² + 1)/(0 - 1) = -1. So the curve crosses the y-axis at (0, -1).
要求 y 轴截距，将 x = 0 代入原方程：y = (0² + 1)/(0 - 1) = -1。因此曲线与 y 轴交于 (0, -1)。
uld require x² + 1 = 0, but x² is never negative for real x, so there are no x-intercepts.
要求 x 轴截距，令 y = 0。这要求 x² + 1 = 0，但对于实数 x，x² 不可能为负，因此没有 x 轴交点。
1) cannot equal zero without producing a contradiction, but the original check is quicker.
从重写形式也能看出：y = x + 1 + 2/(x - 1) 不可能为零，但直接检查原式更快捷。
7. Vertical and Oblique Asymptotes | 垂直渐近线与斜渐近线
when x = 1, and the numerator x² + 1 is not zero there, so x = 1 is a vertical asymptote.
当 x = 1 时分母 x - 1 为零，且分子 x² + 1 不为零，因此 x = 1 是一条垂直渐近线。
pproaches 1 from the right, y tends to +∞; as x approaches 1 from the left, y tends to -∞.
当 x 从右侧趋近 1 时，y 趋向于 +∞；当 x 从左侧趋近 1 时，y 趋向于 -∞。
s to 0 as x → ±∞. Therefore y approaches the line y = x + 1. This is an oblique asymptote.
由于重写形式为 y = x + 1 + 2/(x - 1)，当 x → ±∞ 时，项 2/(x - 1) 趋于 0。因此 y 趋近于直线 y = x + 1。这是一条斜渐近线。
No horizontal asymptote exists because the degree of the numerator is one greater than the degree of the denominator.
不存在水平渐近线，因为分子的次数比分母的次数高一次。
8. Sketching the Curve | 绘制曲线图像
asymptotes: the vertical line x = 1 and the oblique line y = x + 1. Both should be dashed.
先画出两条渐近线：垂直渐近线 x = 1 和斜渐近线 y = x + 1。两条线
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