📚 Determining Quadratic Function Expressions from Graphs | 由函数图像确定二次函数表达式
In IB Mathematics, one common task is to determine the equation of a quadratic function from its graph. This skill combines algebraic manipulation with geometric interpretation, and appears in both Analysis and Approaches and Applications and Interpretation.
在IB数学中，一个常见任务是根据函数图像确定二次函数的表达式。这项技能将代数运算与几何解释相结合，在分析与方法（AA）以及应用与解释（AI）课程中都会出现。
1. Standard Form y = ax² + bx + c | 一般式 y = ax² + bx + c
The standard form of a quadratic function is y = ax² + bx + c, where a, b, and c are constants and a ≠ 0. The constant c represents the y-intercept, since setting x = 0 gives y = c.
二次函数的一般式为 y = ax² + bx + c，其中 a、b、c 为常数且 a ≠ 0。常数 c 表示 y 轴截距，因为令 x = 0 可得 y = c。
If the graph clearly shows the y-intercept and two other points with integer coordinates, you can substitute these values into the general form and solve a system of equations for a, b, and c.
如果图像清晰显示了 y 轴截距以及另外两个整数坐标点，可以将这些值代入一般式，通过解方程组求出 a、b、c。
This form is most useful when the graph does not clearly reveal the vertex or the roots.
当图像没有清晰显示顶点或根时，这种形式最为实用。
2. Vertex Form y = a(x – h)² + k | 顶点式 y = a(x – h)² + k
The vertex form is y = a(x – h)² + k, where (h, k) is the vertex of the parabola. The sign of a determines whether the parabola opens upward (a > 0) or downward (a < 0).
顶点式为 y = a(x – h)² + k，其中 (h, k) 是抛物线的顶点。a 的符号决定抛物线开口方向：a > 0 时开口向上，a < 0 时开口向下。
To use this form, first read the vertex directly from the graph. Then substitute any other known point on the parabola to solve for a.
使用这种形式时，先从图像上直接读取顶点，然后代入抛物线上另一个已知点求 a。
This is often the fastest method when the vertex is clearly identifiable.
当顶点清晰可辨时，这通常是最快捷的方法。
3. Factored Form y = a(x – p)(x – q) | 交点式 y = a(x – p)(x – q)
The factored form is y = a(x – p)(x – q), where p and q are the x-intercepts (roots) of the quadratic. This form is directly linked to the solutions of the equation y = 0.
交点式为 y = a(x – p)(x – q)，其中 p 和 q 是二次函数的 x 轴截距（根）。这种形式与方程 y = 0 的解直接相关。
If the parabola crosses the x-axis at two distinct points, read p and q from the graph, then substitute another point to find a.
如果抛物线与 x 轴有两个不同交点，从图像上读出 p 和 q，再代入另一点求 a。
This form is especially convenient when the roots are integers and clearly marked on the graph.
当根为整数且标注清晰时，这种形式特别方便。
4. Identifying the Vertex from the Graph | 从图像识别顶点
The vertex is the highest or lowest point of the parabola. On a graph, it is the point where the curve changes direction. If the vertex is not labeled, look for the axis of symmetry and find the midpoint between any two symmetric points.
顶点是抛物线的最高点或最低点。在图像上，它是曲线改变方向的点。如果顶点未标注，可以寻找对称轴，并求出任意两个对称点的中点。
For example, if the parabola passes through (0, 3) and (4, 3), the axis of symmetry is x = 2, so the vertex has x-coordinate 2.
例如，若抛物线经过 (0, 3) 和 (4, 3)，则对称轴为 x = 2，因此顶点的 x 坐标为 2。
Remember: the x-coordinate of the vertex is also given by the formula x = -b/(2a) in standard form.
记住：在一般式中，顶点的 x 坐标也可由公式 x = -b/(2a) 给出。
5. Identifying Roots and y-intercept | 识别根和 y 轴截距
Roots are the x-values where the graph intersects the x-axis (y = 0). The y-intercept is the point where the graph intersects the y-axis (x = 0).
根是图像与 x 轴交点的 x 值（此时 y = 0）。y 轴截距是图像与 y 轴交点的纵坐标（此时 x = 0）。
In standard form, the y-intercept is simply c. In factored form, roots are p and q. In vertex form, the y-intercept must be computed by setting x = 0.
在一般式中，y 轴截距就是 c。在交点式中，根就是 p 和 q。在顶点式中，需要令 x = 0 来计算 y 轴截距。
When the graph does not show exact integer intercepts, choose the form that best matches the visible information and use approximate values if necessary.
当图像未显示精确的整数截距时，选择与可见信息最匹配的形式，必要时使用近似值。
6. Determining the Sign of Leading Coefficient | 确定二次项系数的符号
The leading coefficient a controls the direction and “width” of the parabola. If the parabola opens upward, a > 0; if it opens downward, a < 0.
二次项系数 a 控制抛物线的开口方向和“宽窄”。若抛物线开口向上，则 a > 0；若开口向下，则 a < 0。
A larger absolute value of a makes the parabola narrower; a smaller absolute value makes it wider.
|a| 越大，抛物线越窄；|a| 越小，抛物线越宽。
When solving for a, check that your final expression produces the correct opening direction and shape.
在解出 a 时，检查最终表达式是否产生正确的开口方向和形状。
7. Step-by-Step Method: Using Vertex Form | 使用顶点式的逐步方法
Step 1: Locate the vertex (h, k) on the graph.
步骤1：在图像上找到顶点 (h, k)。
Step 2: Write y = a(x – h)² + k with the known h and k.
步骤2：写出 y = a(x – h)² + k，代入已知 h 和 k。
Step 3: Choose another point (x, y) on the parabola and substitute it into the equation.
步骤3：选择抛物线上的另一个点 (x, y) 并代入方程。
Step 4: Solve for a.
步骤4：解出 a。
Step 5: Write the complete expression and simplify if required.
步骤5：写出完整表达式，并根据需要化简。
Example: vertex (2, -1), point (0, 3) → 3 = a(0 – 2)² – 1 → 4 = 4a → a = 1 → y = (x – 2)² – 1
例：顶点 (2, -1)，点 (0, 3) → 3 = a(0 – 2)² – 1 → 4 = 4a → a = 1 → y = (x – 2)² – 1
8. Step-by-Step Method: Using Factored Form | 使用交点式的逐步方法
Step 1: Identify the x-intercepts p and q from the graph.
步骤1：从图像上识别 x 轴截距 p 和 q。
Step 2: Write y = a(x – p)(x – q).
步骤2：写出 y = a(x – p)(x – q)。
Step 3: Substitute another known point (x, y) into the equation.
步骤3：将另一个已知点 (x, y) 代入方程。
Step 4: Solve for a.
步骤4：解出 a。
Step 5: Expand the expression to standard form if needed.
步骤5：如有需要，将表达式展开为一般式。
Example: roots -1 and 3, point (0, -3) → -3 = a(0 + 1)(0 – 3) → -3 = -3a → a = 1 → y = (x + 1)(x – 3) = x² – 2x – 3
例：根 -1 和 3，点 (0, -3) → -3 = a(0 + 1)(0 – 3) → -3 = -3a → a = 1 → y = (x + 1)(x – 3) = x² – 2x – 3
9. Step-by-Step Method: Using Standard Form | 使用一般式的逐步方法
Step 1: Read the y-intercept c from the graph.
步骤1：从图像上读取 y 轴截距 c。
Step 2: Choose two other points on the parabola.
步骤2：选择抛物线上的另外两个点。
Step 3: Substitute each point into y = ax² + bx + c to create a system of two equations.
步骤3：将每个点代入 y = ax² + bx + c，建立包含两个方程的方程组。
Step 4: Solve the system for a and b.
步骤4：解方程组求出 a 和 b。
Step 5: Write the standard form expression.
步骤5：写出一般式表达式。
This method works even when the vertex and roots are not visible, as long as three points are clearly known.
即使顶点和根不可见，只要已知三个清晰点，这种方法也适用。
10. Worked Example | 综合例题
Consider the graph of a quadratic function whose vertex is (1, 4) and which passes through the point (3, 0). Determine its expression.
考虑一个二次函数的图像，其顶点为 (1, 4)，且经过点 (3, 0)。确定其表达式。
Since the vertex is known, use vertex form: y = a(x – 1)² + 4.
由于顶点已知，使用顶点式：y = a(x – 1)² + 4。
Substitute (3, 0): 0 = a(3 – 1)² + 4 → 0 = 4a + 4 → a = -1.
代入 (3, 0)：0 = a(3 – 1)² + 4 → 0 = 4a + 4 → a = -1。
Therefore, y = -(x – 1)² + 4. Expanding gives y = -x² + 2x + 3.
因此，y = -(x – 1)² + 4。展开得 y = -x² + 2x + 3。
Check: the parabola opens downward because a = -1, which matches a vertex at the maximum point.
检验：a = -1，抛物线开口向下，符合最大值顶点的情况。
11. Common Mistakes | 常见错误
Mistake 1: Confusing the sign of h in vertex form. Remember y = a(x – h)² + k, so if the vertex is (-3, 2), write (x + 3)², not (x – 3)².
错误1：混淆顶点式中 h 的符号。记住 y = a(x – h)² + k，因此若顶点为 (-3, 2)，应写 (x + 3)²，而不是 (x – 3)²。
Mistake 2: Using the wrong point or misreading coordinates from the graph.
错误2：用错点或从图像上读错坐标。
Mistake 3: Forgetting that a ≠ 0 and checking the opening direction.
错误3：忘记 a ≠ 0，且未检验开口方向。
Mistake 4: When using the factored form, the roots p and q must be used with the correct signs: y = a(x – p)(x – q), not y = a(x + p)(x + q) unless p and q are negative.
错误4：使用交点式时，根 p 和 q 的符号必须正确：应为 y = a(x – p)(x – q)，而不是 y = a(x + p)(x + q)，除非 p 和 q 为负数。
Mistake 5: Using only two points with unknown vertex; you need at least three points in standard form, or one point plus vertex/roots.
错误5：只使用两个点且未知顶点；一般式至少需要三个点，或者一个点加上顶点/根。
12. Practice Questions | 练习题
Question 1: A parabola has vertex (2, -5) and passes through (0, 3). Find its equation.
练习1：一条抛物线的顶点为 (2, -5)，且经过点 (0, 3)。求其方程。
Question 2: The x-intercepts of a quadratic are -2 and 4, and the y-intercept is 8. Find the expression in standard form.
练习2：一个二次函数的 x 轴截距为 -2 和 4，y 轴截距为 8。求其一般式表达式。
Question 3: A quadratic graph passes through (1, 2), (2, 1), and (3, 4). Determine the function.
练习3：一个二次函数图像经过 (1, 2)、(2, 1) 和 (3, 4)。确定该函数。
Question 4: The graph of y = ax² + bx + c has its vertex at (1, 1) and opens upward. If it passes through (0, 3), find a, b, and c.
练习4：函数 y = ax² + bx + c 的顶点为 (1, 1)，开口向上。若它经过 (0, 3)，求 a、b、c。
Answers: 1) y = 2(x – 2)² – 5 = 2x² – 8x + 3; 2) y = -(x + 2)(x – 4) = -x² + 2x + 8; 3) y = ax² + bx + c → solve a = 1, b = -2, c = 3, so y = x² – 2x + 3; 4) a = 2, b = -4, c = 3.
答案：1) y = 2(x – 2)² – 5 = 2x² – 8x + 3；2) y = -(x + 2)(x – 4) = -x² + 2x + 8；3) y = ax² + bx + c → 解得 a = 1, b = -2, c = 3，即 y = x² – 2x + 3；4) a = 2, b = -4, c = 3。
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