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
