Given 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 intervals on which the function is increasing and the intervals on which the function is decreasing. \( f(x)=3 x^{2}+6 x-45 \) Select the correct choice below and fill in the answer box within your choice. (Type an integer or a fraction.) A. The parabla opens downward and has a maximum value of B. The parabola opens upward and has a minimum value of (c) What is the range of \( f(x) \) ? The range of \( f(x) \) is \( \square \). (Type your answer in interval notation. Use integers or fractions for any numbers in the expression.) (d) On what interval is the function increasing? The function is increasing on \( \square \). (Type your answer in interval notation. Use integers or fractions for any numbers in the expression.)

To find the vertex of the quadratic function \( f(x) = 3x^2 + 6x - 45 \), we can use the vertex formula, \( x = -\frac{b}{2a} \). Here, \( a = 3 \) and \( b = 6 \), so the x-coordinate of the vertex is \( x = -\frac{6}{2(3)} = -1 \). Plugging \( x = -1 \) back into the function, \( f(-1) = 3(-1)^2 + 6(-1) - 45 = -48 \). Thus, the vertex is at (-1, -48). Since the coefficient of \( x^2 \) (which is 3) is positive, the parabola opens upward, meaning there is a minimum value of \( f(-1) = -48 \). For the range, since the parabola opens upward and the minimum value is -48, the range is \( [-48, \infty) \). For determining intervals, we compute the derivative \( f'(x) = 6x + 6 \). Setting this equal to zero gives \( 6x + 6 = 0 \Rightarrow x = -1 \). The function is decreasing on \( (-\infty, -1) \) and increasing on \( (-1, \infty) \). So, summarizing: A. The parabola opens upward and has a minimum value of -48. (c) The range of \( f(x) \) is \([-48, \infty)\). (d) The function is increasing on \((-1, \infty)\).