AS AQA Pure Mathematics: Quadratic Functions, Completing the Square and the Discriminant — 二次函数、配方法与判别式

一、二次函数的标准形式：ax² + bx + c 中每个系数的作用 | The Standard Form of a Quadratic: What a, b and c Control

在 AQA AS 纯数学课程中，二次函数是最基础也最常考的函数类型。它的标准形式写作 y = ax² + bx + c，其中 a、b、c 是常数，且 a 不等于 0。如果 a 恰好等于 0，那么这个式子就不再是二次函数，而是退化成一条直线 y = bx + c，这一点在判断题目类型时非常重要。

In AQA AS Pure Mathematics, the quadratic is the most fundamental and frequently examined type of function. Its standard form is written as y = ax² + bx + c, where a, b and c are constants and a is not equal to zero. If a happens to equal zero, the expression is no longer quadratic at all, but collapses to a straight line y = bx + c. Recognising this distinction is the first step in identifying what kind of problem you are facing.

系数 a 决定了抛物线的开口方向和宽窄。当 a 为正数时，抛物线开口向上，图像呈 U 形，顶点是最低点；当 a 为负数时，抛物线开口向下，图像呈倒 U 形，顶点是最高点。a 的绝对值越大，抛物线越陡峭、越窄；a 的绝对值越小，抛物线越平缓、越宽。

The coefficient a controls the direction and width of the parabola. When a is positive the parabola opens upwards in a U shape and the vertex is a minimum point; when a is negative it opens downwards in an inverted U and the vertex is a maximum point. The larger the absolute value of a, the steeper and narrower the curve; the smaller it is, the flatter and wider the curve.

系数 b 主要影响抛物线顶点的水平位置，它决定了对称轴在何处。常数项 c 则代表抛物线与 y 轴交点的纵坐标，因为当 x = 0 时代入 y = ax² + bx + c，得到 y = c。因此函数图像一定经过点 (0, c)，这一点在快速画图时非常有用。

The coefficient b mainly affects the horizontal position of the vertex, deciding where the axis of symmetry lies. The constant term c represents the y-coordinate of the point where the parabola crosses the y-axis, because substituting x = 0 into y = ax² + bx + c gives y = c. The graph therefore always passes through the point (0, c), which is extremely useful when sketching quickly.

二、因式分解法解二次方程：把二次式拆成两个括号 | Factorising Quadratics: Splitting into Two Brackets

解二次方程 ax² + bx + c = 0 的第一种常用方法是因式分解。当 a = 1 时，我们需要寻找两个数 p 和 q，使得 p + q = b 且 pq = c，然后把方程写成 (x + p)(x + q) = 0。根据”乘积为零则至少一个因子为零”的原则，得到 x = -p 或 x = -q。

The first common method for solving ax² + bx + c = 0 is factorisation. When a = 1, we look for two numbers p and q such that p + q = b and pq = c, then rewrite the equation as (x + p)(x + q) = 0. Using the principle that if a product is zero then at least one factor is zero, we obtain x = -p or x = -q.

当 a 不等于 1 时，情况稍微复杂一些。我们通常使用”十字相乘法”或分组分解法。以 2x² + 7x + 3 为例，先找两个数相乘得 2 × 3 = 6、相加得 7，即 6 和 1；接着把中间项拆开，得到 2x² + 6x + x + 3，再两两分组提取公因式，最终分解为 (2x + 1)(x + 3)。

When a is not equal to 1 the situation is slightly more involved. We usually use the cross-multiplication method or factorisation by grouping. Take 2x² + 7x + 3 as an example: first find two numbers that multiply to 2 × 3 = 6 and add to 7, namely 6 and 1; then split the middle term to get 2x² + 6x + x + 3, group the terms in pairs and take out common factors, finally factorising to (2x + 1)(x + 3).

需要注意的是，并非所有二次式都能在有理数范围内因式分解。像 x² + x + 1 这样的式子就没有整数或分数形式的因子。遇到这种情况，我们就要改用后面介绍的配方法或求根公式。因此，考试中拿到一道题时，先花几秒判断能否因式分解，能就快、不能就换方法。

It is important to note that not every quadratic can be factorised over the rational numbers. Expressions such as x² + x + 1 have no factors in integer or fractional form. In such cases we switch to completing the square or the quadratic formula, introduced below. So when you meet a question in the exam, spend a few seconds deciding whether factorisation works: if it does, it is the fastest route, and if not, move on to another method.

三、配方法：把二次式改写成完全平方形式 | Completing the Square: Rewriting as a Perfect Square

配方法是把二次式 y = ax² + bx + c 改写成 y = a(x + p)² + q 的形式。当 a = 1 时，我们取 b 的一半并平方，得到”所缺的项”。例如 x² + 6x + 2，先写成 (x + 3)²，展开后是 x² + 6x + 9，比原来多了 7，所以要减去 7，最终得到 (x + 3)² – 7。

Completing the square rewrites the quadratic y = ax² + bx + c in the form y = a(x + p)² + q. When a = 1 we take half of b and square it to find the missing term. For example, x² + 6x + 2 becomes (x + 3)², which expands to x² + 6x + 9, seven more than the original, so we subtract 7 to finish with (x + 3)² – 7.

当 a 不等于 1 时，必须先提取公因式 a，再对括号内的式子配方。例如 2x² – 8x + 5，先提出 2 得 2(x² – 4x) + 5，括号内配方得 2[(x – 2)² – 4] + 5，展开化简得到 2(x – 2)² – 3。这个过程在求顶点的题目中极其常见。

When a is not equal to 1 we must first factor out a before completing the square inside the bracket. For example 2x² – 8x + 5 first becomes 2(x² – 4x) + 5, completing the square inside gives 2[(x – 2)² – 4] + 5, which simplifies to 2(x – 2)² – 3. This process appears constantly in vertex questions.

配方法最大的价值在于能直接读出顶点坐标。对于 y = a(x + p)² + q，顶点坐标是 (-p, q)，对称轴是直线 x = -p。当 a 为正时 q 是最小值，当 a 为负时 q 是最大值。这就是为什么”用配方法求函数的最值”是 AS 纯数学试卷中的必考题型。

The greatest value of completing the square is that it reveals the vertex directly. For y = a(x + p)² + q the vertex is at (-p, q) and the axis of symmetry is the line x = -p. When a is positive, q is the minimum value, and when a is negative, q is the maximum. This is why finding the maximum or minimum by completing the square is a guaranteed question type in AS Pure Mathematics papers.

四、求根公式：通用的二次方程求解工具 | The Quadratic Formula: A Universal Solving Tool

求根公式是解任何二次方程 ax² + bx + c = 0 的万能工具：x = [-b ± √(b² – 4ac)] / 2a。这个公式由配方法直接推导而来，因此它适用于所有二次方程，包括那些无法因式分解的方程。考试中只要把 a、b、c 的值代入即可。

The quadratic formula is the universal tool for solving any quadratic equation ax² + bx + c = 0: x = [-b ± √(b² – 4ac)] / 2a. It is derived directly from completing the square, so it works for every quadratic, including those that cannot be factorised. In an exam you simply substitute the values of a, b and c.

使用求根公式时最常见的错误是把符号搞错。例如解 x² – 5x + 6 = 0 时，a = 1，b = -5，c = 6。代入公式时务必把 b = -5 连同负号一起代入，分子变成 5 ± √(25 – 24)，即 5 ± 1，再除以 2，得到 x = 3 或 x = 2。很多人在这里漏掉 b 的负号，导致答案完全错误。

The most common mistake when using the formula is getting the signs wrong. For example, to solve x² – 5x + 6 = 0 we have a = 1, b = -5 and c = 6. When substituting, remember to carry the minus sign of b = -5 into the formula, so the numerator becomes 5 ± √(25 – 24), that is 5 ± 1, divided by 2, giving x = 3 or x = 2. Many candidates drop the minus sign of b here and end up with completely wrong answers.

求根公式与因式分解、配方法本质上是相通的。同一个方程，因式分解最快，配方法能同时给出顶点，求根公式则最稳妥、最不会出错。在考试中建议：先尝试因式分解，失败就用求根公式；如果题目还要求最值或顶点，则优先用配方法。三种方法要能灵活切换。

The quadratic formula, factorisation and completing the square are fundamentally connected. For the same equation, factorisation is fastest, completing the square also delivers the vertex, and the quadratic formula is the most reliable and least error-prone. In an exam, try factorising first, and use the formula if that fails; if the question also asks for the maximum or the vertex, prefer completing the square. Learn to switch between the three methods fluently.

五、判别式 Δ = b² – 4ac：判断方程根的数量与类型 | The Discriminant Δ = b² – 4ac: Counting and Classifying Roots

判别式是求根公式中根号下面的部分，记作 Δ = b² – 4ac。它虽然只是公式的一部分，却能在不实际求解的情况下告诉我们方程根的个数和性质。这是 AS 纯数学中最常考的概念之一，也是很多学生失分的地方。

The discriminant is the part under the square root in the quadratic formula, written as Δ = b² – 4ac. Although it is just one part of the formula, it tells us the number and nature of the roots without actually solving the equation. It is one of the most frequently tested concepts in AS Pure Mathematics, and a common place for students to lose marks.

判别式有三种情况。当 Δ 大于 0 时，方程有两个不相等的实根，图像与 x 轴交于两个不同的点；当 Δ 等于 0 时，方程有两个相等的实根，图像与 x 轴恰好相切于一点（这个根也叫重根）；当 Δ 小于 0 时，方程没有实根，图像与 x 轴不相交，抛物线完全位于 x 轴的一侧。

The discriminant has three cases. When Δ is greater than 0 the equation has two distinct real roots and the graph crosses the x-axis at two different points. When Δ equals 0 the equation has two equal real roots and the graph just touches the x-axis at one point (this root is also called a repeated root). When Δ is less than 0 the equation has no real roots, the graph does not meet the x-axis, and the parabola lies entirely on one side of it.

请特别注意”两个相等的实根”这个说法。虽然严格数学上可以说它是”一个根”，但 AS 考试的标准表述是”两个相等的实根”，因为二次方程按定义总有两个根（计重数）。在答题时务必使用”two equal real roots”或”repeated root”这样的标准术语，才能拿到评分标准里对应的分数。

Pay special attention to the phrase “two equal real roots”. Although strictly we might say it is “one root”, the standard AS wording is “two equal real roots”, because a quadratic equation by definition always has two roots counting multiplicity. In your answer, always use the standard wording “two equal real roots” or “repeated root” to secure the marks listed in the mark scheme.

六、二次函数的图像：顶点、对称轴与开口方向 | The Parabola: Vertex, Axis of Symmetry and Direction of Opening

理解二次函数的图像是掌握整个章节的关键。抛物线 y = ax² + bx + c 有三个核心特征：开口方向由 a 的正负决定，对称轴是一条经过顶点的竖直直线，顶点则是图像的最高点或最低点。三者结合起来，就能快速画出草图。

Understanding the graph of a quadratic is the key to mastering the whole chapter. The parabola y = ax² + bx + c has three core features: the direction of opening is decided by the sign of a, the axis of symmetry is a vertical line through the vertex, and the vertex is the highest or lowest point of the curve. Combine these three and you can sketch the graph quickly.

顶点的横坐标可以用公式 x = -b / 2a 直接求出，也可以用配方法得到。求出横坐标后代入原函数，就得到顶点的纵坐标。对称轴方程就是 x = -b / 2a。例如 y = 2x² – 8x + 5，对称轴是 x = 2，代入得 y = -3，所以顶点是 (2, -3)，与前面配方法的结果 2(x – 2)² – 3 完全一致。

The x-coordinate of the vertex can be found directly from the formula x = -b / 2a, or by completing the square. Substitute this value back into the function to find the y-coordinate. The equation of the axis of symmetry is x = -b / 2a. For example, for y = 2x² – 8x + 5, the axis is x = 2, and substituting gives y = -3, so the vertex is (2, -3), exactly matching the completed-square form 2(x – 2)² – 3 from earlier.

画草图时还有一个常用技巧：利用与坐标轴的交点。与 y 轴的交点是 (0, c)，与 x 轴的交点（如果存在）就是方程 ax² + bx + c = 0 的实根。先标出顶点、对称轴和交点，再根据 a 的符号连出平滑的曲线，一张准确的草图就完成了。

Another useful technique when sketching is to use the axis intercepts. The y-intercept is (0, c), and the x-intercepts, if they exist, are the real roots of ax² + bx + c = 0. Mark the vertex, the axis of symmetry and the intercepts first, then draw a smooth curve following the sign of a, and an accurate sketch is complete.

七、判别式的应用：求参数范围与曲线与坐标轴的交点 | Applying the Discriminant: Parameter Ranges and Intersections

判别式最经典的应用之一是”求参数范围”类题目。题目通常会给出一个含未知参数 k 的二次方程，并告诉你”方程有两个不相等的实根”、”没有实根”或”图像与 x 轴相切”，然后要求你求出 k 的取值范围。解题思路就是把”根的个数条件”翻译成”关于 Δ 的不等式”。

One of the classic applications of the discriminant is the parameter-range question. The question typically gives a quadratic containing an unknown parameter k and states that the equation has two distinct real roots, no real roots, or that the graph touches the x-axis, then asks you to find the range of k. The strategy is to translate the condition on the number of roots into an inequality involving Δ.

举例说明：若方程 x² + kx + 4 = 0 有两个不相等的实根，则 Δ = k² – 16 大于 0，解得 k² 大于 16，即 k 小于 -4 或 k 大于 4。这个结果要写成区间形式 k < -4 或 k > 4，而不能误写成 -4 < k < 4（那是 k² 小于 16 的情况，对应的是没有实根）。分清大于号和小于号的方向是这类题的关键。

As an example, if x² + kx + 4 = 0 has two distinct real roots, then Δ = k² – 16 is greater than 0, giving k² greater than 16, so k is less than -4 or greater than 4. Write this as k < -4 or k > 4, and do not mistake it for -4 < k < 4, which is the case k² less than 16 and corresponds to no real roots. Getting the direction of the inequalities right is the key to these questions.

判别式还可以用来判断一条直线与一条抛物线是否相交、相切或相离。把直线方程代入抛物线方程，消去一个变量得到关于另一个变量的二次方程，这个二次方程的判别式就决定了交点个数：Δ 大于 0 有两个交点，Δ 等于 0 相切（一个交点），Δ 小于 0 相离（无交点）。这类”直线与曲线位置关系”题目综合性强，是 AS 考试中的高分题。

The discriminant can also decide whether a line and a parabola intersect, touch or miss each other. Substitute the line into the parabola, eliminate one variable to obtain a quadratic in the other, and its discriminant determines the number of intersection points: Δ greater than 0 gives two intersections, Δ equal to 0 gives tangency (one point), and Δ less than 0 gives no intersection. These line-and-curve position questions are rich in content and are high-mark questions in the AS exam.

八、二次函数建模应用题：最大值与最小值 | Quadratic Modelling: Maximising and Minimising in Context

AS 纯数学经常把二次函数放进实际情境中，考查建模能力。典型题目包括：求抛物线的最大高度、求面积的最大值、求利润的最大值或成本的最小值。这类题目的核心是先把文字描述转化成二次函数，再求它的顶点。

AS Pure Mathematics frequently places quadratics in real contexts to test modelling skills. Typical questions include finding the maximum height of a projectile, the maximum area, the maximum profit or the minimum cost. The core of these questions is to translate the written description into a quadratic function and then find its vertex.

解题步骤如下：第一步，定义变量并写出目标量的表达式；第二步，如果目标量涉及两个变量，用题目给出的约束关系消去其中一个，把表达式化成只含一个变量的二次函数；第三步，用配方法或公式 x = -b / 2a 求顶点，得到最值；第四步，结合题目情境检验答案是否合理（例如长度不能为负、时间不能为负）。

The steps are as follows. First, define your variables and write an expression for the target quantity. Second, if the quantity involves two variables, use the constraint given in the question to eliminate one, reducing the expression to a quadratic in a single variable. Third, find the vertex by completing the square or using x = -b / 2a to obtain the maximum or minimum. Fourth, check the answer against the context, for example lengths and times cannot be negative.

一个经典例子是”围栏问题”：用固定长度的围栏靠墙围一个矩形区域，求最大面积。设矩形的宽为 x，则长为 L – 2x，面积 A = x(L – 2x) = -2x² + Lx，这是一个开口向下的二次函数，在 x = L / 4 处取得最大值。这类问题几乎每年都会以不同形式出现，务必熟练掌握。

A classic example is the fencing problem: using a fixed length of fence against a wall to enclose a rectangular region, find the maximum area. Let the width be x, then the length is L – 2x and the area is A = x(L – 2x) = -2x² + Lx, a downward-opening quadratic whose maximum occurs at x = L / 4. A problem of this type appears almost every year in some form, so be sure to master it.

九、典型例题详解：一步一步解题示范 | Worked Examples: Step-by-Step Solutions

例 1：因式分解与求根 | Example 1: Factorising and Finding Roots

解方程 x² – 7x + 10 = 0。寻找两个数，相乘得 10、相加得 -7，即 -5 和 -2。因此 x² – 7x + 10 = (x – 5)(x – 2) = 0，解得 x = 5 或 x = 2。检验：代入 x = 5 得 25 – 35 + 10 = 0，正确。

Solve x² – 7x + 10 = 0. Look for two numbers multiplying to 10 and adding to -7, namely -5 and -2. Therefore x² – 7x + 10 = (x – 5)(x – 2) = 0, giving x = 5 or x = 2. Check: substituting x = 5 gives 25 – 35 + 10 = 0, correct.

例 2：配方法求顶点 | Example 2: Completing the Square to Find the Vertex

把 y = x² – 4x + 9 写成 a(x + p)² + q 的形式，并求顶点坐标。配方得 (x – 2)² + 5，即 a = 1，p = -2，q = 5。因此顶点是 (2, 5)，且因为 a = 1 大于 0，这是最小值点，最小值为 5。

Write y = x² – 4x + 9 in the form a(x + p)² + q and find the vertex. Completing the square gives (x – 2)² + 5, so a = 1, p = -2 and q = 5. The vertex is therefore (2, 5), and since a = 1 is positive this is a minimum point with minimum value 5.

例 3：判别式判断根的类型 | Example 3: Using the Discriminant to Classify Roots

判断方程 3x² – 2x + 4 = 0 根的情况。这里 a = 3，b = -2，c = 4，Δ = (-2)² – 4 × 3 × 4 = 4 – 48 = -44，小于 0，所以方程没有实根。图像与 x 轴不相交，且因为 a = 3 大于 0，抛物线完全位于 x 轴上方。

Classify the roots of 3x² – 2x + 4 = 0. Here a = 3, b = -2 and c = 4, so Δ = (-2)² – 4 × 3 × 4 = 4 – 48 = -44, which is less than 0, so the equation has no real roots. The graph does not cross the x-axis, and since a = 3 is positive the parabola lies entirely above the x-axis.

十、考试常见错误与评分标准提醒 | Common Exam Mistakes and Mark Scheme Tips

根据多年真题和评分标准的反馈，学生在二次函数这一章最常见的错误集中在以下几个方面。第一，求根公式中漏掉 b 的负号；第二，判别式为负时错误地写成”一个根”；第三，解不等式 k² 大于某个数时，把解集方向写反；第四，配方法中提取公因式后忘记把常数项也正确处理。

Based on years of past papers and mark scheme feedback, the most common student errors in this chapter cluster around a few areas. First, dropping the minus sign of b in the quadratic formula; second, writing “one root” when the discriminant is negative instead of “no real roots”; third, reversing the direction of the solution set when solving k² greater than a number; fourth, mishandling the constant term after factoring out a when completing the square.

评分标准还要求使用规范术语。例如”两个不相等的实根”、”两个相等的实根”、”没有实根”必须逐字使用。画图题要标出顶点坐标、对称轴方程以及与坐标轴的交点，漏标任何一个都可能被扣分。解题过程中要写出判别式的计算式，直接写结论通常拿不到过程分。

The mark scheme also demands precise terminology. For example “two distinct real roots”, “two equal real roots” and “no real roots” must be used word for word. For sketching questions, label the vertex coordinates, the equation of the axis of symmetry and the axis intercepts; missing any of these can cost marks. In your working, always write out the calculation of the discriminant, as stating the conclusion alone usually earns no method marks.

最后，一定要养成代入检验的习惯。求出方程的根后，把每个根代回原方程验证是否成立；求出顶点后，检查它的横坐标是否确实满足 x = -b / 2a。这种自我检查只需要十几秒，却能避免大量粗心错误，是拿高分的重要保障。

Finally, get into the habit of checking by substitution. After finding the roots, plug each one back into the original equation to verify; after finding the vertex, check that its x-coordinate really satisfies x = -b / 2a. This self-check takes only a dozen seconds but prevents a large number of careless errors, and is an important safeguard for scoring highly.

十一、二次不等式：结合图像确定解集 | Quadratic Inequalities: Solving with Graphs

二次不等式是 AS 纯数学中的另一个重要题型，例如解 x² – 5x + 6 > 0 或 2x² – 3x – 5 ≤ 0。解决这类问题的核心思路是”先求根、再画图、最后看区间”。因为二次函数图像是连续光滑的抛物线，它的正负号只会在根处发生变化，所以求出根后，整个数轴就被分成几段，每段内符号保持不变。

Quadratic inequalities are another important question type in AS Pure Mathematics, for example solving x² – 5x + 6 > 0 or 2x² – 3x – 5 ≤ 0. The core idea is “find the roots, sketch the graph, then read off the intervals”. Because a quadratic graph is a continuous smooth parabola, its sign can only change at the roots, so once the roots are found the whole number line is split into segments within each of which the sign stays constant.

以 x² – 5x + 6 > 0 为例，先因式分解得 (x – 2)(x – 3) > 0，根为 x = 2 和 x = 3。抛物线开口向上，所以在两根之间的区间 2 < x < 3 内函数值为负，在两根之外的区间 x < 2 或 x > 3 内函数值为正。因此不等式的解集是 x < 2 或 x > 3。

Take x² – 5x + 6 > 0 as an example. Factorising gives (x – 2)(x – 3) > 0 with roots x = 2 and x = 3. The parabola opens upwards, so between the roots, in the interval 2 < x < 3, the function is negative, and outside the roots, for x < 2 or x > 3, it is positive. The solution set is therefore x < 2 or x > 3.

需要注意的是”大于零取两边、小于零取中间”这个口诀只在 a 为正时成立。如果 a 为负，务必先把两边同乘 -1 并反转不等号，把二次项系数变成正的再套用口诀，否则极易出错。此外，解集一定要用标准区间或集合符号表示，不能只写”两边”这样含糊的文字。

Note that the rule “greater than zero takes the two outer regions, less than zero takes the middle” only holds when a is positive. If a is negative, always multiply both sides by -1 and reverse the inequality sign first, making the leading coefficient positive before applying the rule, otherwise errors are very likely. Also, the solution set must be written in proper interval or set notation, never in vague words like “the two sides”.

十二、直线与二次曲线联立方程组：代入消元法 | Simultaneous Equations: One Linear, One Quadratic

AS 纯数学还常考”一条直线与一条二次曲线”的联立方程组问题，也就是由一个一次方程和一个二次方程组成的方程组。标准解法是代入消元：从直线方程中把 y 用 x 表示出来，代入二次方程，得到一个只含 x 的二次方程，解出 x 后回代求 y。

AS Pure Mathematics also frequently tests simultaneous equations with one linear and one quadratic equation. The standard method is substitution: express y in terms of x from the linear equation, substitute it into the quadratic to get a quadratic in x alone, solve for x, then substitute back to find y.

例如解方程组 y = x + 1 与 y = x² – 3。把第一个方程代入第二个，得 x + 1 = x² – 3，整理成 x² – x – 4 = 0。用求根公式解得 x = (1 ± √17) / 2，再分别代入 y = x + 1 得到对应的 y 值。注意解是成对出现的，每个 x 值只对应一个 y 值，切勿把 x 和 y 随便搭配。

For example, solve the system y = x + 1 and y = x² – 3. Substituting the first into the second gives x + 1 = x² – 3, which rearranges to x² – x – 4 = 0. The quadratic formula gives x = (1 ± √17) / 2, and substituting each back into y = x + 1 yields the corresponding y values. Note that solutions come in pairs, each x value pairing with exactly one y value, so never mix up the pairing.

这类联立方程组与前面第七节的”直线与抛物线位置关系”在本质上是一回事：代入后得到的二次方程，其判别式决定了交点个数。Δ 大于 0 有两个交点（两组解），Δ 等于 0 相切（一组重解），Δ 小于 0 无交点（无实数解）。掌握这个联系，就能把两个看似不同的题型统一起来理解。

This kind of simultaneous equation is essentially the same as the line-and-parabola position problem from Section Seven: the quadratic obtained after substitution has a discriminant that decides the number of intersection points. Δ greater than 0 gives two intersections (two solution pairs), Δ equal to 0 gives tangency (one repeated pair), and Δ less than 0 gives no intersection (no real solutions). Grasping this link lets you unify two seemingly different question types.

Summary | 总结

二次函数是 AS AQA 纯数学的基石章节，它把代数、图像与建模能力融为一体。本文系统讲解了二次函数的标准形式、三种求解方法（因式分解、配方法、求根公式）、判别式的判断规则、图像特征、参数范围应用以及建模应用题，并配以典型例题和常见错误提醒。

The quadratic is the cornerstone chapter of AS AQA Pure Mathematics, bringing together algebra, graphs and modelling. This article has systematically covered the standard form, the three solution methods (factorising, completing the square and the quadratic formula), the rules of the discriminant, graph features, parameter-range applications and modelling problems, together with worked examples and common-error warnings.

掌握这一章的核心技巧在于三点：熟练因式分解以追求速度，掌握配方法以直接读取顶点，牢记判别式规则以判断根的性质。三者互为补充，遇到任何二次函数问题都能从容应对。建议配合历年真题反复练习判别式与参数范围类题目，巩固这三种方法的灵活切换能力。

Mastering this chapter rests on three points: being fluent in factorising for speed, knowing completing the square to read off the vertex directly, and memorising the discriminant rules to classify roots. The three methods complement one another, so you can approach any quadratic problem with confidence. We recommend practising discriminant and parameter-range questions repeatedly with past papers to consolidate your ability to switch flexibly between the three methods.

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