To determine the [tex]$x$[/tex]-intercept of the continuous function shown in the table, we need to find the value of [tex]\( x \)[/tex] where the function [tex]\( f(x) \)[/tex] equals 0. The [tex]$x$[/tex]-intercept occurs where [tex]\( y = 0 \)[/tex] on the graph.
Let's examine the given table of values:
[tex]\[
\begin{array}{|c|c|}
\hline
x & f(x) \\
\hline
-1 & -9 \\
\hline
0 & -8 \\
\hline
1 & -7 \\
\hline
2 & 0 \\
\hline
3 & 19 \\
\hline
4 & 56 \\
\hline
\end{array}
\][/tex]
From the table, we observe that:
- When [tex]\( x = -1 \)[/tex], [tex]\( f(x) = -9 \)[/tex]
- When [tex]\( x = 0 \)[/tex], [tex]\( f(x) = -8 \)[/tex]
- When [tex]\( x = 1 \)[/tex], [tex]\( f(x) = -7 \)[/tex]
- When [tex]\( x = 2 \)[/tex], [tex]\( f(x) = 0 \)[/tex]
- When [tex]\( x = 3 \)[/tex], [tex]\( f(x) = 19 \)[/tex]
- When [tex]\( x = 4 \)[/tex], [tex]\( f(x) = 56 \)[/tex]
We see that [tex]\( f(x) = 0 \)[/tex] when [tex]\( x = 2 \)[/tex]. Therefore, the [tex]$x$[/tex]-intercept occurs at the point [tex]\((2, 0)\)[/tex].
Hence, the [tex]$x$[/tex]-intercept of the function is [tex]\((2,0)\)[/tex].
