Let's analyze the table of values and the differences to determine the nature of the function and its growth rate.
[tex]\[
\begin{array}{|c|c|}
\hline
x & f(x) \\
\hline
-2 & -2 \\
\hline
2 & 8 \\
\hline
6 & 18 \\
\hline
10 & 28 \\
\hline
14 & 38 \\
\hline
\end{array}
\][/tex]
First, we need to check if the function [tex]\( f \)[/tex] is linear. A linear function grows by equal differences over equal intervals. Let's calculate the differences between successive [tex]\( f(x) \)[/tex] values:
[tex]\[
\begin{aligned}
&f(2) - f(-2) = 8 - (-2) = 10, \\
&f(6) - f(2) = 18 - 8 = 10, \\
&f(10) - f(6) = 28 - 18 = 10, \\
&f(14) - f(10) = 38 - 28 = 10.
\end{aligned}
\][/tex]
We see that the differences between successive [tex]\( f(x) \)[/tex] values are all equal to 10. This means [tex]\( f(x) \)[/tex] is linear and grows by equal differences over equal intervals.
Now, to determine the growth rate, we look at the change in [tex]\( f(x) \)[/tex] over a standard interval of [tex]\( x \)[/tex]:
The interval [tex]\( 2 - (-2) = 4 \)[/tex], and over this interval, the change in [tex]\( f(x) \)[/tex] is 10.
The growth rate per unit interval can be calculated as:
[tex]\[
\text{Growth rate} = \frac{\text{Change in } f(x)}{\text{Change in } x} = \frac{10}{4} = 2.5
\][/tex]
Thus, the function [tex]\( f(x) \)[/tex] is linear and grows at a rate of 2.5.
Based on this analysis, the correct statement is:
D. The function [tex]\( f \)[/tex] is linear and is growing by equal differences over equal intervals. The growth rate is 2.5.
