For the parabola y^2 + 6y + 9 = 12 - 12x, identify the vertex, focus, axis of symmetry, and directrix.

Step-by-step solution

1. 1

   Recognise the left side as a perfect square. $y^2+6y+9=(y+3)^2$, because $9=3^2$ and $6=2\cdot 3$. No completing the square is needed; the square is already assembled.

2. 2

   Factor the right side to expose the vertex shift. $12-12x=-12(x-1)$, so the equation becomes $(y+3)^2=-12(x-1)$.

3. 3

   Match to the horizontal template. Here $y$ is squared, so the form is $(y-k)^2=4p(x-h)$ with $h=1$, $k=-3$ and $4p=-12$, hence $p=-3$. A negative $p$ means the parabola opens to the left.

4. 4

   State the vertex and axis. The vertex is $(h,k)=(1,-3)$, and for a horizontal parabola the axis of symmetry is the horizontal line $y=k=-3$.

5. 5

   Locate the focus and directrix. The focus is $(h+p,k)=(1-3,-3)=(-2,-3)$, inside the curve. The directrix is the vertical line $x=h-p=1-(-3)=4$, on the opposite side of the vertex.

6. 6

   Check with the focus-directrix definition. The point $(-2,3)$ satisfies $(3+3)^2=36=-12(-2-1)$. Its distance to the focus $(-2,-3)$ is $6$, and its horizontal distance to the line $x=4$ is $|4-(-2)|=6$. The two distances match.