To determine the [tex]\(x\)[/tex]-intercepts of a continuous function given in the table, we need to identify the [tex]\(x\)[/tex]-values where [tex]\(f(x) = 0\)[/tex]. These points are where the graph of the function crosses the [tex]\(x\)[/tex]-axis.
Let's examine each pair [tex]\((x, f(x))\)[/tex] from the table:
[tex]\[
\begin{array}{|c|c|}
\hline
x & f(x) \\
\hline
-4 & 0 \\
\hline
-2 & 2 \\
\hline
0 & 8 \\
\hline
2 & 2 \\
\hline
4 & 0 \\
\hline
6 & -2 \\
\hline
\end{array}
\][/tex]
We need to find the [tex]\(x\)[/tex]-values where [tex]\(f(x) = 0\)[/tex]. Let's look at each row:
- For [tex]\(x = -4\)[/tex], [tex]\(f(x) = 0\)[/tex]. Thus, [tex]\((-4, 0)\)[/tex] is an [tex]\(x\)[/tex]-intercept.
- For [tex]\(x = -2\)[/tex], [tex]\(f(x) = 2\)[/tex]. Since [tex]\(f(x) \neq 0\)[/tex], this is not an [tex]\(x\)[/tex]-intercept.
- For [tex]\(x = 0\)[/tex], [tex]\(f(x) = 8\)[/tex]. Since [tex]\(f(x) \neq 0\)[/tex], this is not an [tex]\(x\)[/tex]-intercept.
- For [tex]\(x = 2\)[/tex], [tex]\(f(x) = 2\)[/tex]. Since [tex]\(f(x) \neq 0\)[/tex], this is not an [tex]\(x\)[/tex]-intercept.
- For [tex]\(x = 4\)[/tex], [tex]\(f(x) = 0\)[/tex]. Thus, [tex]\((4, 0)\)[/tex] is an [tex]\(x\)[/tex]-intercept.
- For [tex]\(x = 6\)[/tex], [tex]\(f(x) = -2\)[/tex]. Since [tex]\(f(x) \neq 0\)[/tex], this is not an [tex]\(x\)[/tex]-intercept.
From this analysis, the [tex]\(x\)[/tex]-intercepts are [tex]\((-4, 0)\)[/tex] and [tex]\((4, 0)\)[/tex].
Therefore, the correct answer is:
[tex]\[ (-4, 0), (4, 0) \][/tex]
