Let's investigate the rates of change for the given functions.
### Function [tex]\( A \)[/tex]
The equation for Function [tex]\( A \)[/tex] is [tex]\( y = 3x + 2 \)[/tex].
The rate of change for a linear function [tex]\( y = mx + b \)[/tex] is the coefficient of [tex]\( x \)[/tex], which in this case is [tex]\( 3 \)[/tex]. Therefore, the rate of change for Function [tex]\( A \)[/tex] is [tex]\( 3 \)[/tex].
### Function [tex]\( B \)[/tex]
The information for Function [tex]\( B \)[/tex] is given in a table:
[tex]\[
\begin{array}{|c|c|c|c|}
\hline
x & 0 & 3 & 5 \\
\hline
y & 3 & 9 & 13 \\
\hline
\end{array}
\][/tex]
We need to calculate the rate of change (slope) for Function [tex]\( B \)[/tex]. The slope between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is given by:
[tex]\[ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
#### Calculate the rate of change between [tex]\((0, 3)\)[/tex] and [tex]\((3, 9)\)[/tex]
[tex]\[ \text{slope}_{1} = \frac{9 - 3}{3 - 0} = \frac{6}{3} = 2 \][/tex]
#### Calculate the rate of change between [tex]\((3, 9)\)[/tex] and [tex]\((5, 13)\)[/tex]
[tex]\[ \text{slope}_{2} = \frac{13 - 9}{5 - 3} = \frac{4}{2} = 2 \][/tex]
Both intervals show a consistent rate of change, so the rate of change for Function [tex]\( B \)[/tex] is uniformly [tex]\( 2 \)[/tex].
### Comparing the Rates of Change
- For Function [tex]\( A \)[/tex], the rate of change is [tex]\( 3 \)[/tex].
- For Function [tex]\( B \)[/tex], the rate of change is [tex]\( 2 \)[/tex].
Since [tex]\( 3 \)[/tex] is greater than [tex]\( 2 \)[/tex], the rate of change for Function [tex]\( A \)[/tex] is greater than that for Function [tex]\( B \)[/tex].
### Conclusion
The correct statement is:
C. The rate of change is greater for Function [tex]\( A \)[/tex] than for Function [tex]\( B \)[/tex].
