what is the y - value of the vertex of the function f(x)=-(x - 3)(x + 11)?\n-8\n-4\n33\n49

Explanation:

Step1: Expand the function

First, expand (f(x)=-(x - 3)(x + 11)) using FOIL method. [ \begin{align*} f(x)&=-(x^{2}+11x-3x - 33)\ &=-(x^{2}+8x - 33)\ &=-x^{2}-8x + 33 \end{align*} ]

Step2: Find the x - coordinate of the vertex

For a quadratic function in the form (y = ax^{2}+bx + c) (here (a=-1), (b = - 8), (c = 33)), the x - coordinate of the vertex is given by (x=-\frac{b}{2a}). [x=-\frac{-8}{2\times(-1)}=-\frac{-8}{-2}=-4]

Step3: Find the y - coordinate of the vertex

Substitute (x = - 4) into the function (y=-x^{2}-8x + 33). [ \begin{align*} y&=-(-4)^{2}-8\times(-4)+33\ &=-16 + 32+33\ &=49 \end{align*} ]

Answer:

49