Question

a. $y = \frac{1}{8}(x - 1)^2$
vertex:
$p - value$:
axis of symmetry:
focus:
directrix:

Explanation:

Step1: Identify the vertex form of parabola

The standard form of a parabola is \(y=a(x - h)^2+k\), where \((h,k)\) is the vertex. For \(y=\frac{1}{8}(x - 1)^2\), \(h = 1\), \(k=0\).

Step2: Find the \(p - value\)

We know that \(a=\frac{1}{4p}\). Given \(a=\frac{1}{8}\), then \(\frac{1}{8}=\frac{1}{4p}\), solving for \(p\) gives \(p = 2\).

Step3: Determine the axis of symmetry

For a parabola \(y=a(x - h)^2+k\), the axis of symmetry is \(x=h\). Here \(h = 1\), so the axis of symmetry is \(x = 1\).

Step4: Calculate the focus

The focus of a parabola \(y=a(x - h)^2+k\) is \((h,k + p)\). Substituting \(h = 1\), \(k = 0\), \(p=2\), we get \((1,0 + 2)=(1,2)\).

Step5: Find the directrix

The directrix of a parabola \(y=a(x - h)^2+k\) is \(y=k-p\). Substituting \(k = 0\), \(p = 2\), we get \(y=0-2=-2\).

Answer:

Vertex: \((1,0)\)
\(p - value\): \(2\)
Axis of symmetry: \(x = 1\)
Focus: \((1,2)\)
Directrix: \(y=-2\)