Let's analyze the given table of values to discuss the properties of the continuous function that generated them, particularly focusing on its [tex]\(x\)[/tex]-intercepts:
[tex]\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
-3 & 4 \\
\hline
0 & -3 \\
\hline
1 & 2 \\
\hline
5 & 0 \\
\hline
11 & 7 \\
\hline
\end{array}
\][/tex]
An [tex]\(x\)[/tex]-intercept of a function is a point where the function crosses the [tex]\(x\)[/tex]-axis, which occurs when [tex]\(y = 0\)[/tex]. From the table, we notice:
- For [tex]\( x = -3, y = 4 \)[/tex]
- For [tex]\( x = 0, y = -3 \)[/tex]
- For [tex]\( x = 1, y = 2 \)[/tex]
- For [tex]\( x = 5, y = 0 \)[/tex]
- For [tex]\( x = 11, y = 7 \)[/tex]
From this table, we see that [tex]\( y = 0 \)[/tex] when [tex]\( x = 5 \)[/tex].
This means that the function crosses the [tex]\(x\)[/tex]-axis exactly once, at [tex]\(x = 5\)[/tex]. Since we are given a table with discrete data points and can consider the function continuous based on this data:
- There is exactly one [tex]\(x\)[/tex]-intercept at [tex]\(x = 5\)[/tex], where [tex]\( y = 0 \)[/tex].
Given the information and based on the analysis of the tabulated values, we can conclude:
C. The function has exactly one [tex]\(x\)[/tex]-intercept
