The x-intercept of a function is the point where the function crosses the x-axis. This means the value of [tex]\( f(x) \)[/tex] must be zero at that point.
Given the table:
[tex]\[
\begin{tabular}{|c|c|}
\hline
$x$ & $f(x)$ \\
\hline
-2 & 20 \\
\hline
-1 & 0 \\
\hline
0 & -6 \\
\hline
1 & -4 \\
\hline
2 & 0 \\
\hline
3 & 0 \\
\hline
\end{tabular}
\][/tex]
We need to look for the value of [tex]\( x \)[/tex] where [tex]\( f(x) = 0 \)[/tex].
Let's go through the table entries:
- At [tex]\( x = -2 \)[/tex], [tex]\( f(x) = 20 \)[/tex]
- At [tex]\( x = -1 \)[/tex], [tex]\( f(x) = 0 \)[/tex]
- At [tex]\( x = 0 \)[/tex], [tex]\( f(x) = -6 \)[/tex]
- At [tex]\( x = 1 \)[/tex], [tex]\( f(x) = -4 \)[/tex]
- At [tex]\( x = 2 \)[/tex], [tex]\( f(x) = 0 \)[/tex]
- At [tex]\( x = 3 \)[/tex], [tex]\( f(x) = 0 \)[/tex]
From this, we see that [tex]\( f(x) = 0 \)[/tex] at the following [tex]\( x \)[/tex]-values:
- [tex]\( x = -1 \)[/tex]
- [tex]\( x = 2 \)[/tex]
- [tex]\( x = 3 \)[/tex]
Therefore, the correct answer among the given options is the point where [tex]\( x = -1 \)[/tex] because all the provided points should have [tex]\( f(x) = 0 \)[/tex]. The given points are:
- [tex]\((-1, 0)\)[/tex]
- [tex]\((0, -6)\)[/tex]
- [tex]\((-6, 0)\)[/tex]
- [tex]\((0, -1)\)[/tex]
Thus, the [tex]\( x \)[/tex]-intercept of the continuous function is:
[tex]\[
(-1, 0)
\][/tex]
