📚 Investigation 3 - Finding Quadratic Functions | 探究3 - 求二次函数
In this investigation, you will learn how to determine the equation of a quadratic function from a given set of conditions, such as points on the graph, the vertex, or the roots. Quadratic functions appear in many real-world contexts, and being able to find their algebraic form is a fundamental skill in IB Mathematics.
在本探究中，您将学习如何根据给定条件（如图像上的点、顶点或根）确定二次函数的解析式。二次函数出现在许多现实情境中，能够求出它们的代数形式是IB数学的一项基本技能。
1. Forms of Quadratic Functions | 二次函数的形式
A quadratic function can be written in three common forms: standard form, vertex form, and factored form. Each form provides different insights and is useful for different types of problems.
二次函数可以写成三种常见形式：标准式、顶点式和因式分解式。每种形式提供不同的信息，适用于不同类型的问题。
Standard form: f(x) = ax² + bx + c, where a ≠ 0. This form directly shows the y-intercept (0, c) and the coefficients for polynomial operations.
标准式：f(x) = ax² + bx + c，其中 a ≠ 0。这种形式直接显示y轴截距(0, c)和多项式运算的系数。
Vertex form: f(x) = a(x - h)² + k, where (h, k) is the vertex. This is ideal when the vertex is known, and the sign of a indicates the direction of opening.
顶点式：f(x) = a(x - h)² + k，其中 (h, k) 是顶点。当已知顶点时，这种形式最理想，a的符号指示开口方向。
Factored form: f(x) = a(x - p)(x - q), where p and q are the x-intercepts (roots). This is useful when the roots are given or easily found.
因式分解式：f(x) = a(x - p)(x - q)，其中 p 和 q 是x轴截距（根）。当已知根或容易求得根时，这种形式很有用。
2. Finding a Quadratic from Three Points | 由三点求二次函数
o obtain three linear equations in a, b, and c. Solving this system gives the unique quadratic function passing through all three points, provided the points are not collinear and represent a function.
如果已知抛物线上的三个不同点，可以将每一个点代入标准式 y = ax² + bx + c，得到关于 a、b 和 c 的三个线性方程。解这个方程组即可得到穿过这三个点的唯一二次函数，前提是这些点不共线且符合函数的定义。
Example: Find the quadratic through (1, 2), (2, 3), and (3, 6). Substitute:
示例：求经过 (1, 2)、(2, 3) 和 (3, 6) 的二次函数。代入：
For (1,2): a(1)² + b(1) + c = 2 → a + b + c = 2
对于 (1,2)：a(1)² + b(1) + c = 2 → a + b + c = 2
For (2,3): 4a + 2b + c = 3
对于 (2,3)：4a + 2b + c = 3
For (3,6): 9a + 3b + c = 6
对于 (3,6)：9a + 3b + c = 6
1, b = -2, c = 3. Hence f(x) = x² - 2x + 3. Always verify by substituting the points back.
解方程组得 a = 1, b = -2, c = 3。因此 f(x) = x² - 2x + 3。务必通过代回原坐标进行验证。
3. Using the Vertex and Another Point | 使用顶点和另一点
vertex coordinates to get y = a(x - h)² + k, then plug in the other point to solve for a.
当给定顶点 (h, k) 和另一个点 (x₁, y₁) 时，顶点式最为高效。代入顶点坐标得到 y = a(x - h)² + k，然后代入另一个点求解 a。
(5, 6). Write y = a(x - 2)² - 3. Then 6 = a(5 - 2)² - 3 → 6 = 9a - 3 → 9a = 9 → a = 1. The
Published by TutorHao | IB Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图