📚 Curve Sketching | 曲线绘制
Curve sketching is a fundamental skill in A-Level Mathematics that combines algebraic manipulation with geometric intuition. It requires you to analyse a function's key features — intercepts, stationary points, asymptotes, and symmetry — in order to produce an accurate graph without relying on a calculator.
曲线绘制是 A-Level 数学中的一项基本技能，它将代数运算与几何直觉相结合。它要求你通过分析函数的关键特征——截距、驻点、渐近线和对称性——在不依赖计算器的情况下绘出精确的图形。
1. Finding the Domain | 确定定义域
The domain of a function is the set of all real x-values for which the function is defined. For polynomial functions, the domain is all real numbers, written as x ∈ ℝ. For rational functions, we must exclude values that make the denominator zero. For functions involving square roots, we require the expression under the root to be non-negative.
函数的定义域是使函数有定义的所有实数 x 值的集合。对于多项式函数，定义域为全体实数，记作 x ∈ ℝ。对于有理函数，我们必须排除使分母为零的值。对于含平方根的函数，我们要求根号下的表达式非负。
(x − 2), the domain is all real numbers except x = 2, since division by zero is undefined.
例如：对于 f(x) = (x + 1) / (x − 2)，定义域为除 x = 2 之外的所有实数，因为除以零是无定义的。
Domain of f(x) = (x + 1)/(x − 2): x ∈ ℝ, x ≠ 2
When sketching, always mark the excluded x-values clearly with a vertical dashed line, as these often correspond to vertical asymptotes.
绘图时，务必用垂直虚线清楚地标出被排除的 x 值，因为这些值通常对应垂直渐近线。
2. Finding Intercepts | 求截距
) = 0 and solving for x. These are the points where the curve crosses the horizontal axis.
y 截距通过令 x = 0 并计算 f(0) 得到，这是曲线与纵轴的交点。x 截距（也称为根或零点）通过令 f(x) = 0 并解方程求得，这些是曲线与横轴的交点。
ts, solve x² − 3x + 2 = 0, which factorises as (x − 1)(x − 2) = 0, giving x = 1 and x = 2.
例如，考虑 f(x) = x² − 3x + 2。y 截距为 f(0) = 2。要求 x 截距，解 x² − 3x + 2 = 0，因式分解为 (x − 1)(x − 2) = 0，得到 x = 1 和 x = 2。
| Intercept | How to find | Example: x² − 3x + 2 |
| y-intercept | Set x = 0 | (0, 2) |
| x-intercepts | Set f(x) = 0 | (1, 0) and (2, 0) |
Always check the discriminant b² − 4ac for quadratics: if it is positive there are two distinct real roots; if zero there is one repeated root; if negative there are no real roots and the curve does not cross the x-axis.
对于二次函数，务必检查判别式 b² − 4ac：若为正则有二个不同实根；若为零则有一个重根；若为负则无实根，曲线不与 x 轴相交。
3. Symmetry | 对称性
have rotational symmetry of 180° about the origin. Examples include y = x³ and y = sin(x).
如果函数满足 f(−x) = f(x) 对定义域内所有 x 成立，则该函数为偶函数。偶函数关于 y 轴对称，例如 y = x² 和 y = cos(x)。如果 f(−x) = −f(x) 对一切 x 成立，则该函数为奇函数。奇函数关于原点具有 180° 旋转对称性，例如 y = x³ 和 y = sin(x)。
Identifying symmetry before sketching can save considerable effort: you only need to sketch half the curve and then reflect it. To test for symmetry, replace x with −x in the function and simplify.
在绘图前识别对称性可以节省大量工作：你只需画出曲线的一半，然后进行对称变换即可。测试方法是：将函数中的 x 替换为 −x 并化简。
Even: f(−x) = f(x) → symmetric about y-axis
Odd: f(−x) = −f(x) → symmetric about origin
dd. Note that many functions are neither even nor odd. For instance, f(x) = x² + x has no such symmetry, so you must analyse the full domain.
注意，许多函数既非偶函数也非奇函数。例如，f(x) = x² + x 就具有这种对称性，因此你必须分析整个定义域。
4. Stationary Points | 驻点
the tangent to the curve is horizontal. The stationary points of a curve can be classified as local maxima, local minima, or points of inflection with a horizontal tangent (stationary inflection points).
驻点出现在导数 f′(x) = 0 处。在这些点上，曲线的切线是水平的。曲线的驻点可分为局部极大值、局部极小值或具有水平切线的拐点（水平拐点）。
. The x-values found are then substituted back into f(x) to find the corresponding y-coordinates.
为求驻点，先对函数求导，然后解方程 f′(x) = 0。将求得的 x 值代回 f(x)，得到对应的 y 坐标。
− 3(−1) = 2, and f(1) = 1³ − 3(1) = −2. So the stationary points are (−1, 2) and (1, −2).
例如：对于 f(x) = x³ − 3x，有 f′(x) = 3x² − 3 = 3(x² − 1)。令 3(x² − 1) = 0 得 x = −1 和 x = 1。代回原函数：f(−1) = (−1)³ − 3(−1) = 2，而 f(1) = 1³ − 3(1) = −2。因此驻点为 (−1, 2) 和 (1, −2)。
Use the second derivative test or a sign table for f′(x) to determine whether each stationary point is a maximum or a minimum.
使用二阶导数测试或 f′(x) 的符号表来判断每个驻点是极大值还是极小值。
5. Classifying Stationary Points | 判别驻点类型
x) = 0, the test is inconclusive, and you should examine the sign of f′(x) on either side.
要判别驻点类型，需在驻点处计算二阶导数 f″(x)。若 f″(x) > 0，则该点为局部极小值；若 f″(x) < 0，则为局部极大值；若 f″(x) = 0，测试无效，此时应检查 f′(x) 在两侧的符号。
0, so (−1, 2) is a local maximum. At x = 1, f″(1) = 6 > 0, so (1, −2) is a local minimum.
继续上面的例子 f(x) = x³ − 3x，有 f″(x) = 6x。在 x = −1 处，f″(−1) = −6 < 0，故 (−1, 2) 是局部极大值。在 x = 1 处，f″(1) = 6 > 0，故 (1, −2) 是局部极小值。
The sign table method: pick test points around each stationary x-value and record the sign of f′(x). If f′(x) changes from positive to negative, it is a maximum; from negative to positive, a minimum; if no sign change, it is a stationary inflection point.
符号表法：在每个驻点 x 值周围选取测试点，记录 f′(x) 的符号。若 f′(x) 由正变负，则为极大值；由负变正，则为极小值；若符号不变，则为水平拐点。
| x | −2 | −1 | 0 | 1 | 2 |
| f′(x) = 3x² − 3 | +9 | 0 | −3 | 0 | +9 |
| Sign | + | 0 | − | 0 | + |
Here f′(x) changes from + to − at x = −1 (maximum), and from − to + at x = 1 (minimum).
这里有 f′(x) 在 x = −1 处由正变负（极大值），在 x = 1 处由负变正（极小值）。
6. Points of Inflection | 拐点
A point of inflection is a point where the curvature of the curve changes sign — . These occur where f″(x) = 0 and the sign of f″(x) changes as x passes through the point.
拐点是曲线的凹凸性发生改变的点——曲线从凹向上变为凹向下，或反之。拐点出现在 f″(x) = 0 且 f″(x) 的符号在 x 经过该点时发生变化的位置。
y testing values on both sides. Exclude any x-values that are not in the domain of the original function.
要确定拐点，先解 f″(x) = 0，然后通过测试两侧的值来验证符号变化。排除不在原函数定义域内的 x 值。
< 0 (concave down); for x > 0, f″ > 0 (concave up). Thus (0, 0) is a point of inflection.
对于 f(x) = x³，有 f″(x) = 6x。令 6x = 0 得 x = 0。当 x < 0 时，f″ < 0（凹向下）；当 x > 0 时，f″ > 0（凹向上）。因此 (0, 0) 是拐点。
= 0 but f′ ≠ 0). Both are inflection points, but only the former has a horizontal tangent.
区分水平拐点（f′ = 0 且 f″ = 0）和非水平拐点（f″ = 0 但 f′ ≠ 0）很重要。二者都是拐点，但只有前者的切线是水平的。
7. Asymptotes | 渐近线
inator of a rational function is zero. To find them, set the denominator equal to zero and check that the numerator is non-zero.
渐近线是曲线无限接近但从不触及的直线。垂直渐近线出现在函数趋于 ±∞ 的 x 值处，通常是有理函数分母为零的位置。求法：令分母等于零，并验证分子不为零。
s are equal, the asymptote is y = (leading coefficient of P) / (leading coefficient of Q).
水平渐近线描述函数在 x → ±∞ 时的行为。对于有理函数 f(x) = P(x) / Q(x)：若 P 的次数小于 Q 的次数，则水平渐近线为 y = 0。若次数相等，则渐近线为 y = P 的首项系数除以 Q 的首项系数。
case, perform polynomial long division to find the oblique asymptote equation y = mx + c.
当分子的次数比分母恰好高一次时，存在斜渐近线。此时，通过多项式长除法求出斜渐近线方程 y = mx + c。
x − 1), so the oblique asymptote is y = x + 1, and there is a vertical asymptote at x = 1.
例如：对于 f(x) = (x² + 1) / (x − 1)，除法得到 x + 1 + 2/(x − 1)，因此斜渐近线为 y = x + 1，同时在 x = 1 处有垂直渐近线。
Vertical asymptote: denominator = 0, numerator ≠ 0
Horizontal asymptote: compare degrees of numerator and denominator
8. Behaviour Near Asymptotes | 渐近线附近的行为
→ −∞) on each side. This requires taking left-hand and right-hand limits at the asymptote.
为了精确绘图，你必须确定曲线在垂直渐近线的每一侧是从上方（y → +∞）还是从下方（y → −∞）接近。这需要计算渐近线处的左极限和右极限。
→ +∞. As x → 2⁻ (from the left), the denominator is a small negative number, so f(x) → −∞.
以 f(x) = 1/(x − 2) 为例。当 x → 2⁺（从右侧接近）时，分母为小的正数，故 f(x) → +∞。当 x → 2⁻（从左侧接近）时，分母为小的负数，故 f(x) → −∞。
approaches the asymptote from above or below, which determines whether the curve lies above or below the asymptote at the extremes.
对于水平渐近线，检查 f(x) 与渐近线值之差在 x → ±∞ 时的符号。这告诉你曲线是从上方还是从下方接近渐近线，从而决定曲线在两端是位于渐近线之上还是之下。
A common error is to cross a vertical asymptote when sketching. Remember that the curve splits into separate branches on either side of a vertical asymptote — never join them across the asymptote.
一个常见错误是在绘制曲线时穿过垂直渐近线。请记住，曲线的分支在垂直渐近线的两侧是分开的——切勿跨越渐近线将它们连接起来。
9. End Behaviour of Polynomials | 多项式的端部行为
at both ends. If the leading coefficient is negative, the curve falls to −∞ at both ends.
对于多项式函数，端部行为由首项决定。若次数为偶数且首项系数为正，曲线两端都升至 +∞。若首项系数为负，曲线两端都降至 −∞。
ises to +∞ as x → +∞. If the leading coefficient is negative, the directions are reversed.
若次数为奇数且首项系数为正，曲线在 x → −∞ 时降至 −∞，在 x → +∞ 时升至 +∞。若首项系数为负，则方向相反。
| Degree | Leading coefficient | x → −∞ | x → +∞ |
| Even | + | +∞ | +∞ |
| Even | − | −∞ | −∞ |
| Odd | + | −∞ | +∞ |
| Odd | − | +∞ | −∞ |
Always state the end behaviour first when sketching polynomials — it provides the overall frame into which the detailed features (turning points and intercepts) must fit.
在绘制多项式图形时，务必先说明端部行为——它提供了整体框架，细节特征（转向点和截距）必须与这个框架吻合。
10. Transformations of Curves | 曲线变换
on y = f(x + a) shifts the graph horizontally by −a units (left if a > 0, right if a < 0).
理解图形变换对绘图大有帮助。y = f(x) + a 将图形垂直移动 a 个单位（a > 0 时向上，a < 0 时向下）。y = f(x + a) 将图形水平移动 −a 个单位（a > 0 时向左，a < 0 时向右）。
in the line y = x swaps x and y and gives the graph of the inverse function, if it exists.
y = −f(x) 将图形关于 x 轴反射，而 y = f(−x) 则关于 y 轴反射。关于直线 y = x 的反射交换 x 和 y，得到反函数的图形（如果反函数存在）。
k < 1, it compresses). The function y = f(kx) compresses horizontally by scale factor 1/k.
y = k·f(x) 将图形垂直拉伸，比例因子为 k（若 k > 1 则拉伸；若 0 < k < 1 则压缩）。y = f(kx) 将图形水平压缩，比例因子为 1/k。
y = f(x) + a → vertical shift
y = f(x + a) → horizontal shift
y = −f(x) → reflection in x-axis
y = k·f(x) → vertical stretch
When applying transformations, mark the coordinates of key points (intercepts, stationary points) before transforming, and then apply the transformation to each coordinate accordingly.
应用变换时，先标记关键点（截距、驻点）的坐标，然后相应地对每个坐标施加变换。
11. Worked Example: Sketching a Rational Function | 例题：绘制有理函数图形
x = 0: y = (−1) / [(−2)(3)] = −1/(−6) = 1/6. So the curve crosses the y-axis at (0, 1/6).
让我们来绘制曲线 y = (x − 1) / [(x − 2)(x + 3)]。首先，定义域排除 x = 2 和 x = −3。y 截距通过令 x = 0 求得：y = (−1) / [(−2)(3)] = −1/(−6) = 1/6。因此曲线在 (0, 1/6) 处穿过 y 轴。
the curve crosses the x-axis at (1, 0). There are vertical asymptotes at x = 2 and x = −3.
x 截距通过令分子为零求得：x − 1 = 0 得 x = 1。因此曲线在 (1, 0) 处穿过 x 轴。垂直渐近线位于 x = 2 和 x = −3 处。
asymptote, compare degrees: both numerator and denominator are degree 1 and 2 respectively? e numerator is less than the degree of the denominator, the horizontal asymptote is y = 0.
为求水平渐近线，比较次数：分子的次数为 1，分母的次数为 2。由于分子的次数小于分母的次数，水平渐近线为 y = 0。
/ [(−0.5)(4.5)] = 0.5/(−2.25) < 0. For x > 2, choose x = 3: f(3) = 2 / [(1)(6)] = 1/3 > 0.
现在分析 f(x) 在每个区间内的符号，区间由关键 x 值 −3、1、2 分割。当 x < −3 时，取 x = −4：f(−4) = (−5) / [(−6)(−1)] = −5/6 < 0，曲线在 x 轴下方。在 −3 与 1 之间，取 x = 0：f(0) = 1/6 > 0。在 1 与 2 之间，取 x = 1.5：f(1.5) = (0.5) / [(−0.5)(4.5)] = 0.5/(−2.25) < 0。当 x > 2 时，取 x = 3：f(3) = 2 / [(1)(6)] = 1/3 > 0。
Near x = −3: as x → −3⁻, f(x) → −(−∞) pattern gives positive? Let us check: as x → −3⁻, denominator (−)(small negative)(−3 − 3) — y (x − 2) ≈ −5, (x + 3) ≈ small negative, so denominator ≈ positive small negative product? small) = positive small. Numerator ≈ negative. So f → negative small / positive small = −∞? minator = (−4.9)(0.1) = −0.49, numerator = −3.9, so f ≈ +7.96 > 0. Thus f → +∞ as x → −3⁺.
在 x = −3 附近：当 x → −3⁻ 时，取 x = −3.1，分母 = (−5.1)(−0.1) = +0.51，分子 = −4.1，故 f ≈ −8.04 < 0。所以当 x → −3⁻ 时，f → −∞。当 x → −3⁺ 时，取 x = −2.9，分母 = (−4.9)(0.1) = −0.49，分子 = −3.9，故 f ≈ +7.96 > 0。所以当 x → −3⁺ 时，f → +∞。
= 2.1: denominator = (0.1)(5.1) = +0.51, numerator = 1.1, so f ≈ +2.16 > 0, hence f → +∞.
在 x = 2 附近：当 x → 2⁻ 时，取 x = 1.9，分母 = (−0.1)(4.9) = −0.49，分子 = 0.9，故 f ≈ −1.84 < 0，因此 f → −∞。当 x → 2⁺ 时，取 x = 2.1，分母 = (0.1)(5.1) = +0.51，分子 = 1.1，故 f ≈ +2.16 > 0，因此 f → +∞。
ve comes down from +∞, crosses nothing further, and approaches y = 0 from above as x → +∞.
完整图形：从左侧开始，曲线从下方接近 y = 0，下降，再上升穿过 y 轴于 (0, 1/6)，然后穿过 x 轴于 (1, 0)，随后在 x → 2⁻ 时降至 −∞。在 x = 2 的右侧，曲线从 +∞ 下降，不再穿过任何轴，并在 x → +∞ 时从上方接近 y = 0。
12. Exam Tips for Curve Sketching | 曲线绘制考试技巧
In AQA A-Level exams, curve sketching questions typically reward method marks for each correct feature identified. Develop a systematic checklist: domain, intercepts, symmetry, derivative analysis for stationary points, second derivative for inflection points, asymptotes, and end behaviour. Tick off each item explicitly in your working.
在 AQA A-Level 考试中，曲线绘制题通常对每个
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