To determine the nature of the function and its [tex]\( x \)[/tex]-intercepts, we need to examine the given table of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] values:
[tex]\[
\begin{tabular}{|c|c|}
\hline
x & y \\
\hline
-4 & -0.906 \\
\hline
-2 & -0.625 \\
\hline
0 & 0.5 \\
\hline
2 & 5 \\
\hline
4 & 23 \\
\hline
\end{tabular}
\][/tex]
### Step-by-step analysis:
1. Identify possible intervals where [tex]\( y = 0 \)[/tex]:
- An [tex]\( x \)[/tex]-intercept occurs where the function [tex]\( y = f(x) \)[/tex] crosses the [tex]\( x \)[/tex]-axis (i.e., where [tex]\( y = 0 \)[/tex]).
2. Check the values for [tex]\( y \)[/tex]:
- At [tex]\( x = -4 \)[/tex], [tex]\( y = -0.906 \)[/tex]
- At [tex]\( x = -2 \)[/tex], [tex]\( y = -0.625 \)[/tex]
- At [tex]\( x = 0 \)[/tex], [tex]\( y = 0.5 \)[/tex]
- At [tex]\( x = 2 \)[/tex], [tex]\( y = 5 \)[/tex]
- At [tex]\( x = 4 \)[/tex], [tex]\( y = 23 \)[/tex]
3. Evaluate intervals for sign changes (crossing the [tex]\( x \)[/tex]-axis):
- Between [tex]\( x = -4 \)[/tex] and [tex]\( x = -2 \)[/tex], [tex]\( y \)[/tex] goes from -0.906 to -0.625: no sign change (still negative).
- Between [tex]\( x = -2 \)[/tex] and [tex]\( x = 0 \)[/tex], [tex]\( y \)[/tex] goes from -0.625 to 0.5: sign change from negative to positive.
- Between [tex]\( x = 0 \)[/tex] and [tex]\( x = 2 \)[/tex], [tex]\( y \)[/tex] goes from 0.5 to 5: no sign change (remains positive).
- Between [tex]\( x = 2 \)[/tex] and [tex]\( x = 4 \)[/tex], [tex]\( y \)[/tex] goes from 5 to 23: no sign change (remains positive).
4. Conclusion:
- There is exactly one sign change in the interval [tex]\( -2 < x < 0 \)[/tex], indicating one [tex]\( x \)[/tex]-intercept.
Therefore, based on the given data and analysis, the function has exactly one [tex]\( x \)[/tex]-intercept.
The correct answer is:
[tex]\[ \text{B. the function has exactly one } x\text{-intercept} \][/tex]
