Siven the following function, (a) find the vertex; (b) determine whether there is a maximum or a minimum value, and find the value; (c) find the range; and (d) find the ntervals on which the function is increasing and the intervals on which the function is decreasing. (x) \( =-\frac{1}{2} x^{2}+5 x-1 \) (a) The vertex is \( \left(5, \frac{23}{2}\right) \). (Type an ordered pair, using integers or fractions.) (b) Determine whether the parabola has a maximum value or a minimum value and find the value. A. The correct choice below and fill in the answer box within your choice. (Tyener or a fraction.) B opens downward and has a maximum value olt

We start with the function   y = –½ x² + 5x – 1. (a) To find the vertex, use the formulas   x-coordinate: x = –b/(2a)   y-coordinate: y = f(x). Here, a = –½ and b = 5, so   x = –5/(2(–½)) = –5/–1 = 5. Now, substitute x = 5 into the function:   y = –½(5)² + 5(5) – 1    = –½(25) + 25 – 1    = –12.5 + 25 – 1    = 11.5    = 23/2. Thus, the vertex is (5, 23/2). (b) Because the coefficient a = –½ is negative, the parabola opens downward. This means that the vertex represents the maximum value of the function. Therefore, the maximum value is 23/2. (c) Since the maximum value is 23/2 and the parabola opens downward, the function takes all y-values less than or equal to 23/2. Hence, the range is   (–∞, 23/2]. (d) To determine where the function is increasing and decreasing, consider the derivative:   y' = d/dx (–½ x² + 5x – 1) = –x + 5. Set the derivative equal to 0 to find the critical point:   –x + 5 = 0 ⟹ x = 5. For x < 5, choose a test value (for example, x = 0):   y'(0) = –0 + 5 = 5 > 0, so the function is increasing on (–∞, 5). For x > 5, choose a test value (for example, x = 6):   y'(6) = –6 + 5 = –1 < 0, so the function is decreasing on (5, ∞). Summary of Answers:  (a) Vertex: (5, 23/2)  (b) The parabola opens downward and has a maximum value of 23/2.  (c) Range: (–∞, 23/2]  (d) Increasing on (–∞, 5) and decreasing on (5, ∞).