Curve Sketching | 曲线绘制

📚 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。

Example: For f(x) = (x + 1) / (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

The y-intercept is found by setting x = 0 and calculating f(0). This gives the point where the curve crosses the vertical axis. The x-intercepts (also called roots or zeros) are found by setting f(x) = 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。

For example, consider f(x) = x² − 3x + 2. The y-intercept is f(0) = 2. To find x-intercepts, 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

A function is even if f(−x) = f(x) for all x in its domain. Even functions are symmetric about the y-axis. Examples include y = x² and y = cos(x). A function is odd if f(−x) = −f(x) for all x. Odd functions 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

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

A stationary point occurs where the derivative f′(x) = 0. At these points, 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）。

To find stationary points, differentiate the function and solve f′(x) = 0. 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。

Example: For f(x) = x³ − 3x, we have f′(x) = 3x² − 3 = 3(x² − 1). Setting 3(x² − 1) = 0 gives x = −1 and x = 1. Substituting back: f(−1) = (−1)³ − 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

To classify a stationary point, evaluate the second derivative f″(x) at the stationary point. If f″(x) > 0, the point is a local minimum. If f″(x) < 0, the point is a local maximum. If f″(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。

Continuing the example f(x) = x³ − 3x, we have f″(x) = 6x. At x = −1, f″(−1) = −6 < 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 — the curve changes from concave up to concave down, or vice versa. 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。

To locate inflection points, solve f″(x) = 0, then verify the sign change by 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。

For f(x) = x³, we have f″(x) = 6x. Setting 6x = 0 gives x = 0. For x < 0, f″ < 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。

It is important to distinguish between a stationary inflection point (where f′ = 0 and f″ = 0) and a non-stationary inflection point (where f″ = 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

Asymptotes are lines that the curve approaches arbitrarily closely but never touches. Vertical asymptotes occur at x-values where the function tends to ±∞, typically where the denominator 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。

Horizontal asymptotes describe the behaviour of the function as x → ±∞. For a rational function f(x) = P(x) / Q(x): if the degree of P is less than the degree of Q, the horizontal asymptote is y = 0. If the degrees 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。

An oblique (slant) asymptote occurs when the degree of the numerator is exactly one greater than the degree of the denominator. In this 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。

Example: For f(x) = (x² + 1) / (x − 1), dividing gives x + 1 + 2/(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

To sketch accurately, you must determine whether the curve approaches a vertical asymptote from above (y → +∞) or from below (y → −∞) 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。

Take f(x) = 1/(x − 2). As x → 2⁺ (from the right), the denominator is a small positive number, so f(x) → +∞. 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) → −∞。

For horizontal asymptotes, check the sign of f(x) − (asymptote value) as x → ±∞. This tells you whether the curve 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

For polynomial functions, the end behaviour is determined by the leading term. If the degree is even and the leading coefficient is positive, the curve rises to +∞ 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 −∞。

If the degree is odd and the leading coefficient is positive, the curve falls to −∞ as x → −∞ and rises 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

Sketching is greatly aided by understanding graph transformations. The function y = f(x) + a shifts the graph vertically by a units (up if a > 0, down if a < 0). The function 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）。

The function y = −f(x) reflects the graph in the x-axis, while y = f(−x) reflects it in the y-axis. A reflection 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）。

The function y = k·f(x) stretches the graph vertically by a scale factor k (if k > 1, it stretches; if 0 < 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

Let us sketch the curve y = (x − 1) / [(x − 2)(x + 3)]. First, the domain excludes x = 2 and x = −3. The y-intercept is found by setting 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 x-intercept is found by setting the numerator to zero: x − 1 = 0 gives x = 1. So 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。

To find the horizontal asymptote, compare degrees: both numerator and denominator are degree 1 and 2 respectively? The numerator has degree 1 and the denominator has degree 2. Since the degree of the 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。

Now analyse the sign of f(x) in each interval separated by the key x-values: −3, 1, 2. For x < −3, choose x = −4: f(−4) = (−5) / [(−6)(−1)] = −5/6 < 0, so the curve is below the x-axis. Between −3 and 1, choose x = 0: f(0) = 1/6 > 0. Between 1 and 2, choose x = 1.5: f(1.5) = (0.5) / [(−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) — actually (x − 2) ≈ −5, (x + 3) ≈ small negative, so denominator ≈ positive small negative product? (−5) × (negative small) = positive small. Numerator ≈ negative. So f → negative small / positive small = −∞? Let us be careful: x = −3.1, denominator = (−5.1)(−0.1) = +0.51, numerator = −4.1, so f ≈ −8.04 < 0. Thus f → −∞ as x → −3⁻. As x → −3⁺, x = −2.9: denominator = (−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 → +∞。

Near x = 2: as x → 2⁻, x = 1.9: denominator = (−0.1)(4.9) = −0.49, numerator = 0.9, so f ≈ −1.84 < 0, hence f → −∞. As x → 2⁺, x = 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 → +∞。

The complete sketch: from the left, the curve approaches y = 0 from below, dips, rises to cross the y-axis at (0, 1/6), crosses the x-axis at (1, 0), then falls to −∞ as x → 2⁻. On the right of x = 2, the curve 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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Curve Sketching | 曲线绘制