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: Identify the form of the parabola equation

The given equation $x=\frac{1}{8}(y + 1)^2-2$ is of the form $x=a(y - k)^2+h$, which represents a parabola that opens either to the left or right. Here $h=-2,k = - 1,a=\frac{1}{8}$.

Step2: Recall the properties of a parabola in this form

For a parabola $x=a(y - k)^2+h$, the vertex is $(h,k)$, the focus is $(h+\frac{1}{4a},k)$ and the directrix is $x=h-\frac{1}{4a}$, and the axis of symmetry is $y = k$.

Step3: Calculate the focus

We know $h=-2,k=-1,a=\frac{1}{8}$. First, calculate $\frac{1}{4a}=\frac{1}{4\times\frac{1}{8}} = 2$. Then the focus is $(h+\frac{1}{4a},k)=(-2 + 2,-1)=(0,-1)$.

Step4: Calculate the directrix

The directrix is $x=h-\frac{1}{4a}=-2-2=-4$.

Step5: Determine the axis of symmetry

Since the equation is of the form $x=a(y - k)^2+h$, the axis of symmetry is $y=k$. So the axis of symmetry is $y=-1$.

Answer:

Equation	Vertex	Focus	Axis of Symmetry	Directrix