Find the vertex form of the function. Then find each of the following. \( \begin{array}{llll}\text { (A) Intercepts } & \text { (B) Vertex } & \text { (C) Maximum or minimum } & \text { (D) Range } \\ s(x)=x^{2}-8 x+12\end{array} \) \( s(x)=\square \) (Type your answer in vertex form.)

To analyze the quadratic function \( s(x) = x^{2} - 8x + 12 \), we'll convert it to vertex form and then determine its intercepts, vertex, whether it has a maximum or minimum, and its range. ### **Vertex Form** The **vertex form** of a quadratic function is: \[ s(x) = a(x - h)^2 + k \] where \((h, k)\) is the vertex of the parabola. **Steps to Convert to Vertex Form:** 1. **Start with the standard form:** \[ s(x) = x^2 - 8x + 12 \] 2. **Complete the square:** - Take the coefficient of \( x \), which is \(-8\). - Divide by 2: \(-8 / 2 = -4\). - Square it: \((-4)^2 = 16\). 3. **Add and subtract this square inside the equation:** \[ s(x) = (x^2 - 8x + 16) - 16 + 12 \] 4. **Simplify the equation:** \[ s(x) = (x - 4)^2 - 4 \] **Vertex Form:** \[ s(x) = (x - 4)^2 - 4 \] --- ### **A. Intercepts** 1. **X-Intercepts (Points where \( s(x) = 0 \)):** Solve for \( x \) in \( x^2 - 8x + 12 = 0 \): \[ x = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 1 \cdot 12}}{2 \cdot 1} = \frac{8 \pm \sqrt{64 - 48}}{2} = \frac{8 \pm \sqrt{16}}{2} = \frac{8 \pm 4}{2} \] - \( x = \frac{8 + 4}{2} = 6 \) - \( x = \frac{8 - 4}{2} = 2 \) **X-Intercepts:** \( (2, 0) \) and \( (6, 0) \) 2. **Y-Intercept (Point where \( x = 0 \)):** \[ s(0) = 0^2 - 8(0) + 12 = 12 \] **Y-Intercept:** \( (0, 12) \) --- ### **B. Vertex** From the vertex form \( s(x) = (x - 4)^2 - 4 \), the vertex is: \[ \boxed{(4, -4)} \] --- ### **C. Maximum or Minimum** The coefficient of \( (x - 4)^2 \) is **positive** (\( a = 1 \)), which means the parabola **opens upwards**. Therefore, the function has a **minimum** point. - **Minimum Value:** \( -4 \) at \( x = 4 \) --- ### **D. Range** Since the parabola opens upwards and has a minimum value of \( -4 \), the range of \( s(x) \) is all real numbers **greater than or equal to** \( -4 \). \[ \text{Range}: \boxed{y \geq -4} \] --- ### **Summary** - **Vertex Form:** \( s(x) = (x - 4)^2 - 4 \) - **Intercepts:** - **X-Intercepts:** \( (2, 0) \) and \( (6, 0) \) - **Y-Intercept:** \( (0, 12) \) - **Vertex:** \( (4, -4) \) - **Minimum Value:** Yes, minimum at \( (4, -4) \) - **Range:** \( y \geq -4 \)