### Instruction: Find the focus of the parabola $y = -3x^2 - 6x.$.### Response: To find the focus of a parabola, I need to write it in the standard form $y = a(x - h)^2 + k$, where $(h, k)$ is the vertex and $a$ is a constant that determines the width and direction of the parabola.
To do that, I need to complete the square for the expression $-3x^2 - 6x$.
I can factor out a $-3$ to get $-3(x^2 + 2x)$, and then I need to add and subtract the square of half of the coefficient of $x$, which is $1$, inside the parentheses.
That gives me $-3(x^2 + 2x + 1 - 1) = -3((x + 1)^2 - 1)$.
Expanding the $-3$, I get $y = -3(x + 1)^2 + 3$.
Comparing this with the standard form, I can see that $a = -3$, $h = -1$, and $k = 3$.
So the vertex of the parabola is $(-1, 3)$, and the parabola opens downward since $a < 0$.
The formula for the focus of a parabola is $(h, k + \frac{1}{4a})$, where $a$ is the same constant as in the standard form.
Plugging in the values I found, I get $(-1, 3 + \frac{1}{4(-3)}) = (-1, 3 - \frac{1}{12})$.
This is the focus of the parabola.