I recognize that this is a quadratic function, and I know that the graph of a quadratic function is a parabola. I also know that the coefficient of $x^2$ is positive, so the parabola opens upward and has a minimum point at its vertex. To find the vertex of a parabola, I can use the formula $x=-\frac{b}{2a}$, where $a$ and $b$ are the coefficients of $x^2$ and $x$ respectively. In this case, $a=3$ and $b=6$, so $x=-\frac{6}{2(3)}=-1$. To find the corresponding $y$-value, I can plug in $x=-1$ into the original function and simplify. I get $y=3(-1)^2+6(-1)+9=3-6+9=6$. Therefore, the minimum value of $y$ is 6. 
What is the minimum value of $y$ if $y=3x^2+6x+9?$