Consider the given function \( g(x)=-(x-2)^{2}+4 \). (a) Determine whether the graph of the parabola opens upward or downward. (b) Identify the vertex. (c) Determine the \( x \)-intercept(s). (d) Determine the \( y \)-intercept(s). (e) Sketch the function. (f) Determine the axis of symmetry. (g) Determine the minimum or maximum value of the function. (h) Write the domain and range in interval notation. Write your answers in exact form.

To understand the behavior of the function \( g(x) = -(x - 2)^2 + 4 \), let's go through the parts step-by-step: (a) The graph of the parabola opens downward because the coefficient of \((x - 2)^2\) is negative. (b) The vertex of the parabola can be identified from the vertex form of a quadratic function, which is given by the point \((h, k)\). Here, this point is \((2, 4)\). (c) To find the \( x \)-intercept(s), set \( g(x) = 0 \): \[ -(x - 2)^2 + 4 = 0 \implies (x - 2)^2 = 4 \implies x - 2 = \pm 2. \] This gives \( x = 4 \) and \( x = 0 \). Thus, the \( x \)-intercepts are \((0, 0)\) and \((4, 0)\). (d) To find the \( y \)-intercept, set \( x = 0 \): \[ g(0) = -(0 - 2)^2 + 4 = -4 + 4 = 0. \] So, the \( y \)-intercept is \((0, 0)\). (e) For the sketch, you would plot the vertex at \((2, 4)\), the \( x \)-intercepts at \((0, 0)\) and \((4, 0)\) and recognize it opens downward, creating a 'U' shape. (f) The axis of symmetry for the parabola is the vertical line that passes through the vertex, which is \( x = 2 \). (g) The maximum value of the function occurs at the vertex, thus the maximum value is \( 4 \). (h) The domain of the function is all real numbers \((-∞, ∞)\) since it is a quadratic function. The range, however, is limited by the maximum value at the vertex, so the range is \( (-∞, 4] \) since it approaches negative infinity but does not exceed 4. Now go ahead and enjoy exploring parabolas and how they interact with their surroundings!