1. The problem asks for the vertex of the quadratic function $f(x) = x^2 + 4x + 6$. 2. The vertex of a parabola given by $f(x) = ax^2 + bx + c$ can be found using the formula for the $x$-coordinate of the vertex: $$x = -\frac{b}{2a}$$ 3. Here, $a = 1$ and $b = 4$. Substitute these values: $$x = -\frac{4}{2 \times 1} = -\frac{4}{2} = -2$$ 4. To find the $y$-coordinate of the vertex, substitute $x = -2$ back into the function: $$f(-2) = (-2)^2 + 4(-2) + 6 = 4 - 8 + 6$$ 5. Simplify the expression: $$4 - 8 + 6 = (4 - 8) + 6 = -4 + 6 = 2$$ 6. Therefore, the vertex is at the point $(-2, 2)$. 7. Among the given options, the correct answer is B. $(-2, 2)$.