a) Find the vertex. b) Determine whether there is a maximum or a minimum value and find that value. c) Find the range. d) Find the intervals on which the function is increasing and the intervals on which the function is decreasing. \( f(x)=3 x^{2}+12 x-6 \) (a) The function's vertex is at the point (Type an ordered pair, using integers or fractions.)

Find the second derivative with respect to \( x \) for \( 3x^2 + 12x - 6 \). Evaluate the derivative by following steps: - step0: Evaluate the derivative: \(\frac{d}{dx}\left(\frac{d}{dx}\left(3x^{2}+12x-6\right)\right)\) - step1: Calculate: \(\frac{d}{dx}\left(6x+12\right)\) - step2: Use differentiation rules: \(\frac{d}{dx}\left(6x\right)+\frac{d}{dx}\left(12\right)\) - step3: Calculate: \(6+0\) - step4: Remove 0: \(6\) The second derivative of the function \( f(x) = 3x^2 + 12x - 6 \) is 6. Since the second derivative is positive (6), the function has a minimum value. To find the vertex of the parabola represented by the function, we can use the formula for the x-coordinate of the vertex: \( x = -\frac{b}{2a} \), where the function is in the form \( f(x) = ax^2 + bx + c \). In this case, \( a = 3 \) and \( b = 12 \). Substituting these values into the formula, we get: \[ x = -\frac{12}{2(3)} = -2 \] Therefore, the vertex of the function is at the point (-2, f(-2)). To find the y-coordinate of the vertex, we substitute x = -2 into the function: \[ f(-2) = 3(-2)^2 + 12(-2) - 6 = 12 - 24 - 6 = -18 \] So, the vertex of the function is at the point (-2, -18).