### Instruction: Find the focus of the parabola $y = -3x^2 - 6x.$.### Response: To find the focus of a parabola, I need to rewrite the equation 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 $x$-terms.
I can factor out a $-3$ from the $x$-terms to get $y = -3(x^2 + 2x)$.
Then, I need to add and subtract a term inside the parentheses that makes it a perfect square.
The term I need is half of the coefficient of $x$, squared, which is $(2/2)^2 = 1.$
So I get $y = -3(x^2 + 2x + 1 - 1) = -3((x + 1)^2 - 1).$
I can distribute the $-3$ and simplify to get $y = -3(x + 1)^2 + 3.$
Now I have the standard form, and I can identify the vertex as $(-1, 3)$ and $a = -3.$
The focus of the parabola is a point on the axis of symmetry that is $|1/(4a)|$ units away from the vertex.
The axis of symmetry is the vertical line $x = -1.$
Since $a < 0$, the parabola opens downward, so the focus is below the vertex.
The distance from the vertex to the focus is $|1/(4a)| = |1/(4(-3))| = 1/12.$
Therefore, the focus is $1/12$ units below the vertex, which means its $y$-coordinate is $3 - 1/12 = 35/12.$
Its $x$-coordinate is the same as the vertex, which is $-1.$