What is the range of (f(x)=\sqrt{x^2+8x+20})?

Correct answer: ([2,\infty))

The direct answer is option A: the range is [2,∞). Start with the expression inside the square root: x²+8x+20. Complete the square: x²+8x+20=(x+4)²+4. A square such as (x+4)² is always at least 0, so the expression inside the root is at least 4. Its smallest value is 4, reached at x=−4. Therefore the smallest function value is √4=2. As x moves far from −4, the square becomes larger without limit, so the function can take every value from 2 upward. Option A is correct. Option B, [20,∞), forgets that the square root changes the minimum 4 into 2. Option C, [0,∞), is too broad because values between 0 and 2 cannot occur. Option D, all real numbers, is wrong because a real square root cannot be negative. Memory cue: complete the square, find the minimum inside the root, then take its square root.