Question

1. determine the vertex and the direction in which the parabola $x^{2}=-8z$ opens.

vertex
direction

1. determine the vertex and the direction in which the parabola $y^{2}=12x$ opens.

vertex
direction

1. determine the vertex and the direction in which the parabola $(z - 3)^{2}=4(y + 1)$ opens.

vertex
direction

1. determine the vertex and the direction in which the parabola $(y + 2)^{2}=-6(z - 4)$ opens.

vertex
direction

Explanation:

Step1: Recall parabola vertex - form

The standard - form of a parabola is either $(y - k)^2=4p(x - h)$ (opens left or right) or $(x - h)^2 = 4p(y - k)$ (opens up or down), where $(h,k)$ is the vertex.

Step2: Analyze $x^{2}=-8x$

Rewrite it as $x^{2}+8x = 0$. Completing the square: $x^{2}+8x + 16=16$, i.e., $(x + 4)^{2}=16$. In the form $(x - h)^2=4p(y - k)$, here $y = 0$, $h=-4$, $k = 0$, $4p=-8$ so $p=-2$.
The vertex is $(0,0)$ and since $p=-2<0$, it opens to the left.

Step3: Analyze $y^{2}=12x$

In the form $(y - k)^2=4p(x - h)$, $h = 0$, $k = 0$, and $4p=12$, so $p = 3$.
The vertex is $(0,0)$ and since $p = 3>0$, it opens to the right.

Step4: Analyze $(x - 3)^{2}=4(y + 1)$

Comparing with $(x - h)^2=4p(y - k)$, we have $h = 3$, $k=-1$, $p = 1$.
The vertex is $(3,-1)$ and since $p = 1>0$, it opens up.

Step5: Analyze $(y + 2)^{2}=-6(x - 4)$

Comparing with $(y - k)^2=4p(x - h)$, we have $h = 4$, $k=-2$, $4p=-6$ so $p=-\frac{3}{2}$.
The vertex is $(4,-2)$ and since $p=-\frac{3}{2}<0$, it opens to the left.

Answer:

1. Vertex: $(0,0)$; Direction: Left
2. Vertex: $(0,0)$; Direction: Right
3. Vertex: $(3,-1)$; Direction: Up
4. Vertex: $(4,-2)$; Direction: Left