What is the x-coordinate of the vertex of the graph of f(x) = x² + 12x + 40?

For a quadratic $ax^2 + bx + c$, the vertex has $x = -\frac{b}{2a}$. Here $a = 1$ and $b = 12$, so $x = -\frac{12}{2} = -6$, choice A. Completing the square agrees: $x^2 + 12x + 40 = (x + 6)^2 + 4$, so the vertex is $(-6, 4)$. Choice B (6) has the wrong sign, forgetting the negative in the formula. Choice C (12) is the $x$-coefficient and choice D (40) is the constant, which is also the $y$-intercept. The vertex's $x$-coordinate is half of the $x$-coefficient with the opposite sign.