Question

1. identify the vertex of the function y=(x - 1)^2-16 and then use a table of values to identify two ordered pairs on either side of the vertex.

the vertex is (1, - 16). the ordered pairs on either side of the vertex are (-3,0),(-1, - 12),(3, - 12), and (5,0)
the vertex is (-1, - 16). the ordered pairs on either side of the vertex are (-3,0),(-1, - 12),(3, - 12), and (5,0)
the vertex is (-1,16). the ordered pairs on either side of the vertex are (-3,0),(-1, - 12),(3, - 12), and (5,0)
the vertex is (1, - 16). the ordered pairs are (-3, - 32),(-1, - 20),(3, - 12), and (5,0)

Explanation:

Step1: Recall vertex - form of a parabola

The vertex - form of a parabola is $y=a(x - h)^2+k$, where $(h,k)$ is the vertex of the parabola. For the function $y=(x - 1)^2-16$, comparing it with the vertex - form $y=a(x - h)^2+k$ (here $a = 1$, $h = 1$, $k=-16$), the vertex is $(1,-16)$.

Step2: Find ordered - pairs

Let's find the $x$ - intercepts by setting $y = 0$.
\[

$$\begin{align*} 0&=(x - 1)^2-16\\ (x - 1)^2&=16\\ x-1&=\pm4 \end{align*}$$

\]
When $x-1 = 4$, $x = 5$; when $x - 1=-4$, $x=-3$. So the $x$ - intercepts are $(-3,0)$ and $(5,0)$.
Let's find the $y$ - value when $x=-1$.
\[

$$\begin{align*} y&=(-1 - 1)^2-16\\ y&=4 - 16\\ y&=-12 \end{align*}$$

\]
So another point is $(-1,-12)$.

Answer:

The vertex is $(1,-16)$. The ordered pairs on either side of the vertex are $(-3,0),(-1,-12),(3,-12),(5,0)$