Function 1 has a maximum at y = 1 Now we need to find the maximum of Function 2 by completing the square: -x^2 + 2x - 3 = -(x^2 - 2x) - 3 = -(x - 1)^2 +1 - 3 = -(x - 1)^2 - 2 Therefor the turning point is at (1, -2) and the maximum is at y = -2
-2 < 1, therefor Function 1 has the larger maximum
