What is the minimum possible value for $y$ in the equation $y = x^2 + 12x + 5$?
I notice that this is a quadratic equation, so I know that the graph of $y$ is a parabola that opens upward, since the coefficient of $x^2$ is positive. That means that the minimum value of $y$ occurs at the vertex of the parabola. I recall that the vertex of a parabola has an x-coordinate of $-\frac{b}{2a}$, where $a$ and $b$ are the coefficients of $x^2$ and $x$ respectively. In this case, $a = 1$ and $b = 12$, so the x-coordinate of the vertex is $-\frac{12}{2(1)} = -6$. To find the y-coordinate of the vertex, I plug in $x = -6$ into the equation and simplify: $y = (-6)^2 + 12(-6) + 5 = 36 - 72 + 5 = -31$. 