📚 Exercise 4C: Completing the Square and Quadratic Graphs | 练习4C：配方法与二次函数图像
In this AQA A-Level Mathematics revision article, we focus on Exercise 4C, which builds confidence in completing the square, sketching quadratic functions, and interpreting the discriminant. These techniques appear throughout AS and A2 papers, from pure algebra to applied modelling.
在这篇 AQA A-Level 数学复习文章中，我们聚焦练习 4C，重点训练配方法、二次函数草图绘制以及判别式的解读。这些技巧贯穿 AS 和 A2 试卷，从纯代数到应用建模都会出现。
1. Why Completing the Square Matters | 为什么配方法重要
Completing the square transforms a general quadratic of the form ax² + bx + c into an equivalent expression a(x + p)² + q. This new form immediately reveals the minimum or maximum value, the coordinates of the turning point, and the axis of symmetry.
配方法将 ax² + bx + c 这样的一般二次式转换为一个等价表达式 a(x + p)² + q。这个新形式可以立刻显示出最小值或最大值、顶点坐标以及对称轴。
The method is also essential when a quadratic does not factorise over the integers. In AQA exam papers, an instruction such as 'complete the square' often signals that the vertex, the range, or the exact roots of the function will be required next.
当二次式不能进行整数因式分解时，配方法也至关重要。在 AQA 试卷中，“complete the square”这样的指令通常提示接下来需要求出函数的顶点、值域或精确根。
2. The Core Identity of Completing the Square | 配方法的核心恒等式
For a monic quadratic in the form x² + bx, the key identity is to take half of the coefficient of x, square it, then add and subtract that square inside the expression.
对于形如 x² + bx 的首一二次式，核心恒等式是取 x 的系数的一半，将其平方，然后在表达式中加上并减去这个平方项。
x² + bx = (x + b/2)² − (b/2)²
For example, x² + 6x becomes (x + 3)² − 9. The constant term is adjusted carefully because (x + 3)² already produces x² + 6x + 9, so we must subtract 9 to keep equality.
例如，x² + 6x 可以写成 (x + 3)² − 9。这里必须小心调整常数项，因为 (x + 3)² 已经产生了 x² + 6x + 9，所以需要减去 9 才能保持相等。
This identity is the foundation for the more general case, where the coefficient of x² is not equal to 1. Mastery of this basic step prevents most sign and factor errors later.
这个恒等式是更一般情形的基础，即当 x² 的系数不等于 1 时的配方。掌握这个基本步骤可以避免后面出现的大部分符号和系数错误。
3. Finding the Vertex Form of a Quadratic | 求二次函数的顶点式
irst factor the leading coefficient a from the first two terms. Then complete the square inside the bracket and adjust the constant term at the end.
对于 y = ax² + bx + c 这样的二次函数，首先从前两项中提取首项系数 a。然后在括号内进行配方，并在最后调整常数项。
y = a(x + b/(2a))² + (c − b²/(4a))
pleting the square inside the bracket gives y = 2[(x + 2)² − 4] + 5, so y = 2(x + 2)² − 3.
例如，y = 2x² + 8x + 5 可以先写成 y = 2(x² + 4x) + 5。在括号内配方得到 y = 2[(x + 2)² − 4] + 5，因此 y = 2(x + 2)² − 3。
2, -3). Since a = 2 is positive, the graph opens upwards and the minimum value of y is -3.
顶点式显示顶点为 (-2, -3)。由于 a = 2 为正数，图像开口向上，y 的最小值为 -3。
4. Solving Quadratics by Completing the Square | 用配方法解二次方程
Completing the square is also a reliable way to solve quadratic equations, especially when factorisation is not obvious or when exact surd answers are required.
配方法也是解二次方程的可靠方法，尤其是在因式分解不明显或题目要求精确根式答案时。
tion as x² + 6x = 4. Completing the square gives x² + 6x + 9 = 13, which is (x + 3)² = 13.
对于 x² + 6x − 4 = 0，先移项得到 x² + 6x = 4。配方得到 x² + 6x + 9 = 13，也就是 (x + 3)² = 13。
Taking square roots on both sides gives x + 3 = ±√13, so x = −3 ± √13. Keep the exact surd form unless a decimal approximation is specifically requested.
两边开平方得到 x + 3 = ±√13，因此 x = −3 ± √13。除非题目特别要求小数近似值，否则应保留精确根式形式。
5. Sketching Quadratic Graphs Efficiently | 高效绘制二次函数图像
int (-p, q), the axis of symmetry x = -p, and the direction of opening from the sign of a.
顶点式可以让草图绘制快得多。从 a(x + p)² + q 中，我们可以直接读出顶点 (-p, q)、对称轴 x = -p，以及从 a 的符号判断图像开口方向。
by solving y = 0. This gives enough information to draw a clear, labelled quadratic curve.
要完成草图，还需要令 x = 0 求出 y 轴截距，并令 y = 0 求出所有 x 轴截距。这样就有足够信息画出一条清晰、带标记的二次曲线。
le, y = (x + 2)² − 3 has vertex (-2, -3), opens upwards, y-intercept 1, and roots x = −2 ± √3. Plot these points before drawing the smooth curve.
例如，y = (x + 2)² − 3 的顶点为 (-2, -3)，开口向上，y 轴截距为 1，根为 x = −2 ± √3。先标出这些点，再画出光滑曲线。
6. Reading Roots and Turning Points from Graphs | 从图像读取根与顶点
coordinate of the turning point is always the midpoint of the two roots, so x = (p + q)/2.
当二次函数以因式分解形式 y = (x − p)(x − q) 给出时，根可以直接读出为 p 和 q。顶点的 x 坐标始终是两个根的中点，即 x = (p + q)/2。
s at x = 4, and substituting back gives y = (4 − 2)(4 − 6) = −4, so the vertex is (4, -4).
例如，y = (x − 2)(x − 6) 的根为 x = 2 和 x = 6。顶点位于 x = 4，代回得到 y = (4 − 2)(4 − 6) = −4，因此顶点为 (4, -4)。
This midpoint idea is very useful for checking that a completed square form and a factorised form represent the same quadratic function.
这个中点性质非常有用，可以用来检验配方形式和因式分解形式是否表示同一个二次函数。
7. The Discriminant and Curve Position | 判别式与图像位置
= b² − 4ac. It controls how many times the quadratic graph crosses or touches the x-axis.
二次式 ax² + bx + c 的判别式定义为 Δ = b² − 4ac。它决定了二次函数图像与 x 轴相交或相切的次数。
Δ = b² − 4ac
ated real root and the curve touches the x-axis at the vertex. If Δ < 0, there are no real roots and the curve lies entirely above or below the x-axis depending on the sign of a.
如果 Δ > 0，方程有两个不同的实根，曲线与 x 轴相交两次。如果 Δ = 0，则有一个重根，曲线在顶点处与 x 轴相切。如果 Δ < 0，则没有实根，曲线完全位于 x 轴上方或下方，具体取决于 a 的符号。
8. Quadratic Inequalities from a Graph | 由图像解二次不等式
Quadratic inequalities are best solved by sketching the curve and deciding which regions satisfy the inequality. The roots split the x-axis into intervals, and the sign of the quadratic changes or stays the same according to the shape of the curve.
二次不等式最好通过绘制草图并判断哪些区域满足不等式来求解。根将 x 轴分成若干区间，二次式的符号根据图像形状发生变化或保持不变。
ince the coefficient of x² is positive, the curve opens upwards and is above the x-axis when x < 2 or x > 3.
对于 x² − 5x + 6 > 0，因式分解得到 (x − 2)(x − 3) > 0。根为 x = 2 和 x = 3。由于 x² 的系数为正，图像开口向上，当 x < 2 或 x > 3 时曲线位于 x 轴上方。
Therefore the solution is x < 2 or x > 3. If the inequality had been ≤
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