Answer :
To determine the [tex]\(x\)[/tex]-intercept of the continuous function represented by the given table, we need to identify the point where the function [tex]\(f(x)\)[/tex] crosses the x-axis. The [tex]\(x\)[/tex]-intercept occurs when [tex]\(f(x) = 0\)[/tex].
Examining the table:
[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]
We see that at [tex]\(x = 2\)[/tex], [tex]\(f(x) = 0\)[/tex]. This means the function crosses the x-axis at the point [tex]\((2, 0)\)[/tex].
Therefore, the [tex]\(x\)[/tex]-intercept of the function is [tex]\((2, 0)\)[/tex].
Examining the table:
[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]
We see that at [tex]\(x = 2\)[/tex], [tex]\(f(x) = 0\)[/tex]. This means the function crosses the x-axis at the point [tex]\((2, 0)\)[/tex].
Therefore, the [tex]\(x\)[/tex]-intercept of the function is [tex]\((2, 0)\)[/tex].
