What is the minimum value of the function f(x) = x² - 6x + 5?

Complete the square. Half of $-6$ is $-3$, and $(-3)^2 = 9$. Then $x^2 - 6x + 5 = (x^2 - 6x + 9) - 9 + 5 = (x - 3)^2 - 4$. Since a square is never negative, the minimum occurs when $x = 3$ and equals $-4$, choice C. Choice A (5) is the $y$-intercept, $f(0)$. Choice B (3) is the $x$-coordinate of the vertex, not the minimum value. Choice D ($-6$) is the $x$-coefficient. The question asks for the minimum value of the function, which is the $y$-coordinate of the vertex.