📚 Mastering Quadratic Functions | 掌握二次函数
Quadratic functions are a central topic in IGCSE Mathematics. They appear in algebra, graphing, and modelling problems, and understanding them can significantly boost your exam score.
二次函数是 IGCSE 数学的核心内容,贯穿代数、图像和实际建模问题。掌握它们能大幅提升你的考试成绩。
1. Standard Form of a Quadratic Function | 二次函数的标准形式
A quadratic function is a polynomial of degree 2. Its general form is:
二次函数是次数为 2 的多项式,其一般形式为:
f(x) = ax² + bx + c
where a, b, and c are constants and a ≠ 0. The coefficient a is the leading coefficient, b is the linear coefficient, and c is the constant term.
其中 a、b、c 为常数,且 a ≠ 0。a 称为二次项系数,b 称为一次项系数,c 为常数项。
The value of c represents the y-intercept of the parabola, since f(0) = c. This is often the easiest point to plot when sketching a graph.
常数项 c 表示抛物线的 y 截距,因为 f(0) = c。这通常是绘图时最容易标出的点。
2. The Graph of a Quadratic Function | 二次函数的图像
When we graph y = ax² + bx + c, we obtain a symmetric curve called a parabola. Every parabola has a vertex, which is either the lowest point (when a > 0) or the highest point (when a < 0).
当我们画出 y = ax² + bx + c 的图像时,得到的是一条对称曲线,称为抛物线。每条抛物线都有一个顶点,它是最低点(当 a > 0)或最高点(当 a < 0)。
The parabola is symmetric about a vertical line called the axis of symmetry, which passes through the vertex.
抛物线关于一条竖直直线对称,这条直线称为对称轴,对称轴经过顶点。
For example, y = x² – 2x has a = 1, so it opens upward and its vertex is a minimum point.
例如,y = x² – 2x 中 a = 1,因此开口向上,顶点是最小值点。
3. Finding the Vertex | 求顶点
The x-coordinate of the vertex of y = ax² + bx + c is given by the formula:
二次函数 y = ax² + bx + c 的顶点横坐标由以下公式给出:
x = -b / (2a)
Then substitute this x value into the function to find the y-coordinate.
然后将这个 x 值代入函数,即可求出纵坐标。
For example, if y = 2x² – 4x + 1, then x = -(-4)/(2×2) = 4/4 = 1. Substituting gives y = 2(1)² – 4(1) + 1 = -1, so the vertex is (1, -1).
例如,对于 y = 2x² – 4x + 1,x = -(-4)/(2×2) = 4/4 = 1,代入得 y = 2(1)² – 4(1) + 1 = -1,因此顶点为 (1, -1)。
4. The Axis of Symmetry | 对称轴
Since the parabola is symmetric, its axis of symmetry is the vertical line passing through the vertex. Its equation is simply x = -b / (2a).
因为抛物线是对称的,其对称轴是经过顶点的竖直线,它的方程就是 x = -b / (2a)。
This line divides the parabola into two mirror images. It is also the line about which the two halves coincide.
这条线把抛物线分成两个镜面对称的部分,左右两半完全重合。
When asked to find the maximum or minimum value of a quadratic, you first find the axis of symmetry and then evaluate the function at that x value.
当需要求二次函数的最大值或最小值时,先求对称轴,然后计算该 x 值所对应的函数值。
5. Roots / x-Intercepts | 根或 x 截距
The roots of a quadratic equation are the values of x where f(x) = 0. Graphically, these are the points where the parabola crosses the x-axis.
二次方程的根是使 f(x) = 0 的 x 值。在图像上,这些点是抛物线与 x 轴的交点。
Depending on the discriminant, a quadratic can have two distinct roots, one repeated root, or no real roots.
根据判别式,二次方程可以有两个不同实根、一个重根,或无实根。
For example, y = x² – 4 has roots x = 2 and x = -2, so its graph crosses the x-axis at (-2, 0) and (2, 0).
例如,y = x² – 4 的根为 x = 2 和 x = -2,因此它的图像在 (-2, 0) 和 (2, 0) 处与 x 轴相交。
6. The Discriminant | 判别式
The discriminant is the expression inside the square root of the quadratic formula, usually denoted by Δ:
判别式是二次公式中根号内的表达式,通常用 Δ 表示:
Δ = b² – 4ac
If Δ > 0, there are two distinct real roots. If Δ = 0, there is exactly one repeated root. If Δ < 0, there are no real roots.
若 Δ > 0,有两个不同实根;若 Δ = 0,有一个重根;若 Δ < 0,没有实根。
For the equation 2x² + 3x + 4 = 0, Δ = 3² – 4×2×4 = 9 – 32 = -23, so it has no real roots.
对于方程 2x² + 3x + 4 = 0,Δ = 3² – 4×2×4 = 9 – 32 = -23,因此它没有实根。
7. Solving by Factorisation | 因式分解法
When a quadratic can be factorised, solving is straightforward. For example, x² – 5x + 6 = 0 can be written as (x – 2)(x – 3) = 0.
当二次方程可以因式分解时,求解就很简单。例如,x² – 5x + 6 = 0 可以写成 (x – 2)(x – 3) = 0。
Then by the zero product property, either x – 2 = 0 or x – 3 = 0, so x = 2 or x = 3.
根据零乘积性质,要么 x – 2 = 0,要么 x – 3 = 0,所以 x = 2 或 x = 3。
For quadratics with a ≠ 1, look for a product of brackets or use the method of “splitting the middle term”. For instance, 2x² + 5x + 2 = 0 becomes (2x + 1)(x + 2) = 0, giving x = -1/2 or x = -2.
对于 a ≠ 1 的二次方程,可以使用十字相乘或“拆项”的方法。例如,2x² + 5x + 2 = 0 可化为 (2x + 1)(x + 2) = 0,得到 x = -1/2 或 x = -2。
8. Solving by Completing the Square | 配方法
Completing the square rewrites a quadratic in the form a(x – h)² + k, where (h, k) is the vertex.
配方法将二次式
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