Question

match the equation with the focus, vertex, directrix, and axis of symmetry. use the blank graph on the next question if you want to sketch the graph to help you determine the attributes.

equation	vertex	focus	axis of symmetry	directrix

$(4,-1)$ $(5,3)$ $(0,4)$ $(3,5)$ $(1,3)$ $(-1,-4)$ $(-2,-1)$ $(-1,-3)$ $y=-1$ $x = 0$ $y = 0$ $x=-3$ $x=-1$ $y = 3$ $y=-3$

Explanation:

Step1: Recall the standard - form of a parabola

The equation of a parabola in vertex - form is $y=a(x - h)^2+k$, where $(h,k)$ is the vertex. For the given equation $y =-\frac{1}{12}(x + 1)^2-3$, we have $h=-1$ and $k = - 3$. So the vertex is $(-1,-3)$.

Step2: Recall the relationship between $a$ and the focus and directrix

For a parabola $y=a(x - h)^2+k$, the distance from the vertex to the focus and from the vertex to the directrix is $|p|$, where $a=\frac{1}{4p}$. Given $a=-\frac{1}{12}$, then $-\frac{1}{12}=\frac{1}{4p}$, solving for $p$ gives $p=-3$.
Since the parabola opens downwards (because $a<0$), the focus is $(h,k + p)$. Substituting $h=-1,k=-3,p = - 3$ gives the focus $(-1,-6)$. The axis of symmetry of a parabola $y=a(x - h)^2+k$ is $x = h$. So the axis of symmetry is $x=-1$.
The directrix is $y=k - p$. Substituting $k=-3,p=-3$ gives $y=0$.

Answer:

Vertex: $(-1,-3)$
Axis of Symmetry: $x=-1$