\( f(x)=3 x^{2}+12 x-6 \) (a) The function's vertex is at the point \( (-2,-18) \) (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. Select the correct choice below and fill in the answer box to complete you (Type an integer or a fraction.) A. The parabola opens downward and has a maximum value of (C. The parabola opens upward and has a minimum value of -18 (c) The range of \( f(x) \) is \( [-18, \infty) \) ). (Type your answer in interval notation Use integers or fractions for any numbers in the expression.) (d) The function is increasing on ( \( -2, \infty) \) (Type your answer in interval notation Use integers or fractions for any numbers in the expression.) The function is decreasing on (Type your answer in interval notation. Use integers or fractions for any numbers in the expression.)

Let's analyze the function \( f(x) = 3x^2 + 12x - 6 \). ### (a) Vertex The vertex of the parabola is given as \( (-2, -18) \). This is correct, as the vertex can be found using the formula \( x = -\frac{b}{2a} \) where \( a = 3 \) and \( b = 12 \): \[ x = -\frac{12}{2 \cdot 3} = -2 \] To find the corresponding \( y \)-value, substitute \( x = -2 \) back into the function: \[ f(-2) = 3(-2)^2 + 12(-2) - 6 = 3(4) - 24 - 6 = 12 - 24 - 6 = -18 \] Thus, the vertex is indeed \( (-2, -18) \). ### (b) Maximum or Minimum Value Since the coefficient of \( x^2 \) (which is \( a = 3 \)) is positive, the parabola opens upward. Therefore, it has a minimum value at the vertex. The minimum value is: \[ -18 \] So the correct choice is: **C. The parabola opens upward and has a minimum value of -18.** ### (c) Range of \( f(x) \) Since the parabola opens upward and the minimum value is \( -18 \), the range of \( f(x) \) is: \[ [-18, \infty) \] ### (d) Intervals of Increase and Decrease To determine where the function is increasing or decreasing, we can analyze the derivative \( f'(x) \): \[ f'(x) = 6x + 12 \] Setting the derivative equal to zero to find critical points: \[ 6x + 12 = 0 \implies 6x = -12 \implies x = -2 \] - The function is **decreasing** on the interval \( (-\infty, -2) \) (to the left of the vertex). - The function is **increasing** on the interval \( (-2, \infty) \) (to the right of the vertex). Thus, the answers are: - The function is increasing on \( (-2, \infty) \). - The function is decreasing on \( (-\infty, -2) \). ### Summary of Answers (a) \( (-2, -18) \) (b) Minimum value of \( -18 \) (c) Range: \( [-18, \infty) \) (d) Increasing on \( (-2, \infty) \) Decreasing on \( (-\infty, -2) \)