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

Step-by-step solution

1. 1

   Spot the perfect-square trinomial instead of completing the square. The left side $y^2 + 6y + 9$ is already $(y+3)^2$, because $9 = (6/2)^2$. Recognising this skips the completing-the-square step entirely:

   $(y+3)^2 = 12 - 12x$

2. 2

   Factor the right side so it looks like $4p(x-h)$. Pull out $-12$:

   $(y+3)^2 = -12(x-1)$

   The sign matters — writing $12(1-x)$ instead would hide the value of $h$.

3. 3

   Match to the horizontal template. Here $y$ is the squared variable, so the parabola opens sideways and the form is $(y-k)^2 = 4p(x-h)$. Writing $(y-(-3))^2 = -12(x-1)$ gives $h = 1$, $k = -3$ and $4p = -12$, so

   $p = -3$

   Because $p < 0$ the parabola opens left.

4. 4

   Locate the vertex and focus. The vertex is $(h,k) = (1,-3)$. For a horizontal parabola the focus is $(h+p,\; k)$:

   $(1 + (-3),\, -3) = (-2, -3)$

5. 5

   Give the axis and directrix. The axis of symmetry is the horizontal line through the vertex, $y = k = -3$. The directrix is the vertical line $x = h - p = 1 - (-3) = 4$, three units to the right of the vertex — on the opposite side from the focus, as it must be.

6. 6

   Check one point against the definition. At $y = 3$ the equation gives $(3+3)^2 = 36 = -12(x-1)$, so $x = -2$ and $(-2, 3)$ lies on the curve. Its distance to the focus $(-2,-3)$ is $6$, and its distance to the directrix $x = 4$ is $|-2 - 4| = 6$. The two distances agree.