Identify where the minimum of the square root function \( y=2 \sqrt{4 x+16}+2 \) occurs. (1 point) \( (2,-4) \) \( (-16,2) \) \( (-4,2) \)

To find where the minimum of the function \( y = 2\sqrt{4x + 16} + 2 \) occurs, we start by simplifying the function inside the square root. The expression \( \sqrt{4x + 16} \) can be rewritten as \( \sqrt{4(x + 4)} \), which gives us: \[ y = 2\sqrt{4(x + 4)} + 2 = 2(2\sqrt{x + 4}) + 2 = 4\sqrt{x + 4} + 2 \] The function \( \sqrt{x + 4} \) has its minimum when \( x + 4 = 0 \), which is at \( x = -4 \). Plugging \( x = -4 \) back into the function to find the corresponding \( y \)-value: \[ y = 4\sqrt{-4 + 4} + 2 = 4\sqrt{0} + 2 = 2 \] Thus, the minimum occurs at the point \( (-4, 2) \). Answer: \( (-4, 2) \)