Find the range of function y = Log & nbsp; 12 (- x2-2x + 3)

Find the range of function y = Log & nbsp; 12 (- x2-2x + 3)

Let t = - x2-2x + 3, then t = - (x + 1) 2 + 4, so 0 < T ≤ 4, because the function y = Log & nbsp; 12t monotonically decreases, so y = log ⁡ 12t ≥ log ⁡ 124 = − 2, that is, the value range of the function is [- 2, + ∞)