生成讲解动画视频---**Question Content:** In a planar Cartesian coordinate system, a quadratic function `y = x^2 + bx + c` is given. The coordinates of the vertex of its graph are (1, -4). (1) Find the expression of this quadratic function. (2) How can the graph of this quadratic function be obtained by translating the graph of `y = (x+1)^2`? **Chart/Diagram Description:** * **Type:** Cartesian coordinate system showing a parabolic curve. * **Coordinate Axes:** * X-axis: Labeled 'x', with tick marks and labels from -5 to 5. * Y-axis: Labeled 'y', with tick marks and labels from -5 to 5. * **Parabolic Curve:** * The curve represents the function `y = (x-1)^2 - 4`. * **"目标顶点" (Target Vertex):** Marked with a red circular dot at coordinates (1, -4). This is the vertex of the drawn parabola. * **"原顶点" (Original Vertex):** Marked with a green circular dot at coordinates (-1, 0). This point corresponds to the vertex of the function `y = (x+1)^2`, which is referenced in the question for translation. * **Grid:** A grid is present, helping to read coordinates. **Problem Solving Approach (Solution Hints/Steps):** **解题思路 (Problem Solving Approach):** **解 (Solution):** **(1) 顶点式形式 (Vertex form):** `y = a(x-1)^2 - 4` 顶点坐标 (Vertex coordinates): (1, -4) **提示 (Hint):** The coefficient `a` in the general form `y = ax^2 + bx + c` and the vertex form `y = a(x-h)^2 + k` are equal. ∵ `a = 1` (Since the given function is `y = x^2 + bx + c`, the coefficient of `x^2` is 1) 顶点式 (Vertex form): `y = (x-1)^2 - 4` 即表达式为 (The expression is): `y = x^2 - 2x - 3` **(2) 原函数 (Original function):** `y = (x+1)^2`, 顶点 (Vertex): (-1, 0) 目标函数 (Target function): `y = (x-1)^2 - 4`, 顶点 (Vertex): (1, -4) `y = (x+1)^2` 的图像先向右平移2个单位，再向下平移4个单位可得到 `y = (x-1)^2 - 4` (The graph of `y = (x+1)^2` first shifts 2 units to the right, then shifts 4 units downwards to obtain `y = (x-1)^2 - 4`)