Determine the equation y = ax² + bx + c from conditions such as the vertex, a point on the graph, or the x-intercepts.
Example ①（Vertex and a Point）
Find the quadratic function with vertex (2, -3) that passes through (4, 5).
Answer: Substitute (4, 5) into y = a(x - 2)² - 3: 5 = 4a - 3 → a = 2. y = 2(x-2)² - 3
Find the quadratic function with vertex (2, -3) that passes through (4, 5).
Answer: Substitute (4, 5) into y = a(x - 2)² - 3: 5 = 4a - 3 → a = 2. y = 2(x-2)² - 3
Example ②（Passing Through Three Points）
Find the quadratic function passing through (0, 1), (1, 0), and (3, 4).
Answer: Substitute into y = ax² + bx + c: c = 1、a + b + 1 = 0、9a + 3b + 1 = 4。
Solving gives a = 1, b = -2, c = 1. y = x² - 2x + 1
Find the quadratic function passing through (0, 1), (1, 0), and (3, 4).
Answer: Substitute into y = ax² + bx + c: c = 1、a + b + 1 = 0、9a + 3b + 1 = 4。
Solving gives a = 1, b = -2, c = 1. y = x² - 2x + 1
Key Point
- If the vertex is known, find a using the form y = a(x-p)² + q
- If three points are known, use simultaneous equations with y = ax² + bx + c
- If two x-intercepts are known, use the form y = a(x - α)(x - β)
Practice Problems
- Find the quadratic function with vertex (-1, 4) that passes through (1, 0).
Answers and Explanations
- Answer: y = -(x+1)² + 4. Substitute (1,0) into y = a(x+1)² + 4: 0 = 4a + 4 → a = -1.