Complete the square for y = a^2 - 4a + 5.

• A. y = (a + 2)^2 + 1
• B. y = (a - 2)^2 - 1
• C. y = (a - 2)^2 + 1
• D. y = (a - 2)^2 + 5

Answer: (C)

Completing the square rewrites a quadratic into a perfect square plus a constant, revealing the parabola’s vertex form. For y = a^2 - 4a + 5, take half of -4, which is -2, square it to get 4, and adjust: y = (a^2 - 4a + 4) + (5 - 4) = (a - 2)^2 + 1. This exactly matches the form (a - h)^2 + k with h = 2 and k = 1. Expanding (a - 2)^2 + 1 gives a^2 - 4a + 4 + 1 = a^2 - 4a + 5, confirming the rewrite. The vertex is at a = 2, y = 1. Other forms would change the linear coefficient or the constant and wouldn’t reproduce the original polynomial.