What is x² - 6x + 13 written in vertex form?

Complete the square. Take half of $-6$, which is $-3$, and square it to get $9$: $x^2-6x+13=(x^2-6x+9)+13-9=(x-3)^2+4$, choice A. The vertex is $(3,4)$, and the parabola opens upward. Choice C ($(x-3)^2-4$) has the right binomial but the wrong sign on the constant, since $13-9=+4$. Choice D ($(x+3)^2+4$) has the wrong sign inside the square. Choice B ($(x-6)^2+13$) uses the full coefficient instead of half. Expanding the answer is the best check: $(x-3)^2+4=x^2-6x+13$.