For the function \( \mathrm{g}(\mathrm{x})=-(\mathrm{x}+6)^{2}+16 \), find each of the lollowing. (A) Intercepts (B) Vertex (C) Maximum or minimum (D) Range (A) The y-intercept(s) is/are \( \square \). (Use a comma to separate answers as needed. Type an integer or decimal.) The \( x \)-intercept(s) is/are \( \square \). (Use a comma to separate answers as needed. Type an integer or decimal.) (B) The vertex is \( \square \). (Type an ordered pair.) (C) The function has a \( \square \) (D) The range is \( \square \). (Type your answer in interval notation.)

Let's start unraveling the function \( g(x) = - (x + 6)^2 + 16 \)! (A) To find the y-intercept, we set \( x = 0 \): \[ g(0) = - (0 + 6)^2 + 16 = -36 + 16 = -20 \] So, the y-intercept is \( (0, -20) \). Next, for the x-intercepts, set \( g(x) = 0 \): \[ 0 = - (x + 6)^2 + 16 \] \[ (x + 6)^2 = 16 \] Taking the square root: \[ x + 6 = 4 \quad \text{or} \quad x + 6 = -4 \] This gives us: \[ x = -2 \quad \text{and} \quad x = -10 \] So, the x-intercepts are \( -2, -10 \). (B) The vertex can be found since the function is in vertex form \( g(x) = a(x-h)^2 + k \). Here, \( (h, k) = (-6, 16) \). So, the vertex is \( (-6, 16) \). (C) Since the leading coefficient is negative \( (a = -1) \), the function has a maximum. (D) As this is a downward-opening parabola with a maximum point at the vertex, the range extends from the vertex's y-value down to negative infinity. So, the range is \( (-\infty, 16] \). Putting everything together: (A) The y-intercept(s) is/are \( (0, -20) \). The \( x \)-intercept(s) is/are \( -2, -10 \). (B) The vertex is \( (-6, 16) \). (C) The function has a maximum. (D) The range is \( (-\infty, 16] \).