Find the vertex form of the function. Then find each of the following. \( \begin{array}{lll}\text { (A) Intercepts } & \text { (B) Verlex } & \text { (C) Maximum or minimum } \\ s(x)=x^{2}-12 x+32\end{array} \) \( s(x)=\square \) (Type yange \( s \) (Iy answer in vertex form.)

To convert the function \( s(x) = x^2 - 12x + 32 \) into vertex form, we can complete the square. 1. **Complete the square**: Start with \( s(x) = x^2 - 12x + 32 \). Take half of the coefficient of \( x \) (which is -12), square it, and add and subtract that square: \[ s(x) = (x^2 - 12x + 36 - 36) + 32 = (x - 6)^2 - 4. \] Thus, the vertex form is \( s(x) = (x - 6)^2 - 4 \). 2. **Vertex**: The vertex \((h, k)\) is \((6, -4)\). 3. **Intercepts**: To find the x-intercepts, set \( s(x) = 0 \): \[ (x - 6)^2 - 4 = 0 \implies (x - 6)^2 = 4 \implies x - 6 = \pm 2 \implies x = 4 \text{ or } x = 8. \] So, the x-intercepts are (4, 0) and (8, 0). To find the y-intercept, set \( x = 0 \): \[ s(0) = (0 - 6)^2 - 4 = 36 - 4 = 32. \] The y-intercept is (0, 32). 4. **Maximum or Minimum**: Since the parabola opens upwards (the coefficient of \( x^2 \) is positive), the vertex represents a minimum point. Therefore, the minimum value of \( s(x) \) is -4 at \( x = 6 \). So, summarizing: - Vertex form: \( s(x) = (x - 6)^2 - 4 \) - Intercepts: x-intercepts at (4, 0) and (8, 0); y-intercept at (0, 32). - Minimum: Minimum value is -4 at \( x = 6 \).