Solving Quadratic Equations and Graphing Quadratics | 解二次方程与二次函数图像

📚 Solving Quadratic Equations and Graphing Quadratics | 解二次方程与二次函数图像

o solve these equations and how to interpret their graphs is one of the most examinable skills in IGCSE Mathematics.

二次方程是任何可以写成 ax² + bx + c = 0（其中 a ≠ 0）形式的方程。掌握如何解这些方程以及如何理解其图像，是 IGCSE 数学中最常考的技能之一。

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1. The Standard Form and Key Terminology | 标准形式与关键术语

e coefficient of x², b is the coefficient of x, and c is the constant term. The value of a matters: if a > 0 the parabola opens upwards, and if a < 0 it opens downwards.

每个二次方程的最高次项都是平方项。标准形式为 ax² + bx + c = 0，其中 a 是 x² 的系数，b 是 x 的系数，c 是常数项。a 的值很重要：若 a > 0，抛物线开口向上；若 a < 0，抛物线开口向下。

• A quadratic in the form ax² + bx + c = 0 may have two, one, or zero real solutions.
• ax² + bx + c = 0 形式的二次方程可能有二个、一个或零个实数解。
• The constant term c gives the y-intercept of the graph, since setting x = 0 makes y = c.
• 常数项 c 给出图像的 y 轴截距，因为令 x = 0 时 y = c。
• If the equation has a common factor, always divide through by it first before solving.
• 若方程有公因数，先约去公因数再求解。

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2. Solving by Factorisation | 因式分解法

= 0. By the zero product property, either x + 2 = 0 or x + 3 = 0, giving x = -2 or x = -3.

因式分解法是方程有整数解时最快的解法。解 x² + 5x + 6 = 0 时，要找两个乘积为 6、和为 5 的数，即 2 和 3，因此方程可分解为 (x + 2)(x + 3) = 0。根据零乘积性质，要么 x + 2 = 0，要么 x + 3 = 0，所以 x = -2 或 x = -3。

(x + 2)(x + 3) = 0 → x = -2 或 x = -3

d factors of 4 that add to -5 (these are -1 and -4), split the middle term, then factor by grouping.

对于非首一二次方程，如 2x² - 5x + 2 = 0，将 a 与 c 相乘：2 × 2 = 4。找到和为 -5 的 4 的因数（即 -1 和 -4），拆开中间项，再分组分解。

2x² - 5x + 2 = 2x² - 4x - x + 2 = (2x - 1)(x - 2) = 0 → x = ½ 或 x = 2

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3. Solving by Completing the Square | 配方法

r nicely. For example, x² + 6x + 1 = 0 becomes (x + 3)² - 9 + 1 = 0, that is (x + 3)² = 8.

配方法将 x² + bx 改写为 (x + b/2)² - (b/2)²。此方法适用于任何二次方程，尤其当方程难以因式分解时。例如，x² + 6x + 1 = 0 可化为 (x + 3)² - 9 + 1 = 0，即 (x + 3)² = 8。

(x + 3)² = 8 → x + 3 = ±√8 = ±2√2 → x = -3 ± 2√2

d-square form y = (x - p)² + q, the vertex of the parabola is directly readable as (p, q).

记住将 x 的系数除以 2、平方，然后加一项减一项以保持等式平衡。在配方式 y = (x - p)² + q 中，抛物线的顶点可以直接读出为 (p, q)。

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4. Solving by the Quadratic Formula | 公式法

lutions are given by the quadratic formula. This method works for every equation, including those with irrational or no real solutions.

对于任何二次方程 ax² + bx + c = 0，其解都可使用求根公式得到。此方法适用于所有方程，包括含有无理数解或无实数解的方程。

x = (-b ± √b² - 4ac) / 2a

b and c directly. For example, in 3x² - 4x - 2 = 0, a = 3, b = -4 and c = -2, so x = (4 ± √16 + 24) / 6 = (4 ± √40) / 6. Simplify √40 to 2√10 and cancel the common factor of 2 to write (2 ± √10) / 3.

直接把系数 a、b、c 代入公式即可。例如，在 3x² - 4x - 2 = 0 中，a = 3，b = -4，c = -2，所以 x = (4 ± √16 + 24) / 6 = (4 ± √40) / 6。将 √40 化简为 2√10，再约去公因数 2，得到 (2 ± √10) / 3。

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5. The Discriminant | 判别式

The discriminant is the value b² - 4ac inside the square root of the quadratic formula. It tells us how many real roots a quadratic equation has without solving it, and it is also essential for curve-tangent problems.

判别式是求根公式中根号内的 b² - 4ac 的值。不用解方程，它就能告诉我们二次方程有几个实根，它在曲线与切线问题中也至关重要。

b² - 4ac > 0	Two distinct real roots | 两个不等实根
b² - 4ac = 0	One repeated real root | 一个重根
b² - 4ac < 0	No real roots | 无实根

A tangent touches a curve at exactly one point, so when you set the equation of a line equal to a quadratic, the discriminant must be zero.

切线与曲线只有一个交点，因此当把直线方程与二次方程联立时，判别式必须为零。

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6. Shape and Key Features of Quadratic Graphs | 二次函数图像：形状与关键特征

. Its key features are the vertex, the axis of symmetry, the y-intercept and the roots, which are also called x-intercepts or zeros.

y = ax² + bx + c 的图像是一条抛物线。其关键特征包括顶点、对称轴、y 轴截距以及根（也称为 x 轴截距或零点）。

the vertex. Since the parabola is symmetric, the two roots are equidistant from this line; the vertex is exactly midway between them along the x-axis.

对称轴是经过顶点的竖直线 x = -b/(2a)。由于抛物线具有对称性，两个根到这条线的距离相等；顶点的 x 坐标正好位于两个根的正中间。

对称轴：x = -b/(2a)

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7. Finding the Vertex | 顶点的求法

The vertex, or turning point, is the point where the graph changes direction. Its x-coordinate is -b/(2a); substitute this value back into the original equation to find the y-coordinate.

顶点（即转向点）是图像改变方向的点。其 x 坐标为 -b/(2a)，将此值代回原方程即可求得 y 坐标。

es y = 4 - 8 + 3 = -1, so the vertex is (2, -1). Since a = 1 > 0, this is a minimum point.

对于 y = x² - 4x + 3，对称轴为 x = 4/2 = 2。代入得 y = 4 - 8 + 3 = -1，所以顶点为 (2, -1)。因为 a = 1 > 0，它是最小值点。

)² + q as (p, q). This saves time in exam questions that give the equation in vertex form.

另一种方法是直接从配方式 y = (x - p)² + q 中读出顶点 (p, q)。当题目给出顶点式

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