Convert the equation to standard form by completing the square on $x$ or $y$. Then find the vertex, focus, and directrix of the parabola. Finally, graph the parabola.
\[
x^{2}+12 x-12 y+12=0
\]
Answer
Thus, the vertex is \(\boxed{(-6, -2)}\), the focus is \(\boxed{(-6, 1)}\), and the directrix is \(\boxed{y = -5}\).
Step 1 ：First, we rewrite the equation in the form of a parabola. We can rewrite the equation as \(x^2 + 12x = 12y - 12\).
Step 2 ：Then, we complete the square on the left side of the equation. We add \((12/2)^2 = 36\) to both sides of the equation to get \((x + 6)^2 = 12y - 12 + 36\).
Step 3 ：Simplify the equation to get \((x + 6)^2 = 12(y + 2)\).
Step 4 ：Now, we can see that the equation is in the standard form of a parabola \((x - h)^2 = 4a(y - k)\), where \((h, k)\) is the vertex of the parabola, and \(a\) is the distance from the vertex to the focus and the directrix.
Step 5 ：From the equation, we can see that the vertex of the parabola is \((-6, -2)\).
Step 6 ：Also, we can see that \(4a = 12\), so \(a = 3\).
Step 7 ：Since the parabola opens upwards, the focus of the parabola is \((h, k + a) = (-6, -2 + 3) = (-6, 1)\).
Step 8 ：The equation of the directrix is \(y = k - a = -2 - 3 = -5\).
Step 9 ：Finally, we can graph the parabola with the vertex at \((-6, -2)\), the focus at \((-6, 1)\), and the directrix at \(y = -5\).
Step 10 ：Thus, the vertex is \(\boxed{(-6, -2)}\), the focus is \(\boxed{(-6, 1)}\), and the directrix is \(\boxed{y = -5}\).
