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To determine whether a function is linear, we need to check if the relationship between [tex]\( x \)[/tex] and [tex]\( y \)[/tex] varies proportionally, leading to a constant rate of change (slope).
### Let's analyze both functions:
#### Function A
The data points for Function A are:
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline x & 3 & 6 & 9 & 12 & 15 \\ \hline y & 9 & 36 & 81 & 144 & 225 \\ \hline \end{array} \][/tex]
To check if [tex]\( y \)[/tex] is a linear function of [tex]\( x \)[/tex], we look at the differences in [tex]\( y \)[/tex] for equal changes in [tex]\( x \)[/tex]:
[tex]\[ \begin{aligned} \Delta y_{1} &= y_{2} - y_{1} = 36 - 9 = 27, \quad \Delta x_{1} = x_{2} - x_{1} = 6 - 3 = 3, \quad \text{slope} = \frac{\Delta y_{1}}{\Delta x_{1}} = \frac{27}{3} = 9\\ \Delta y_{2} &= y_{3} - y_{2} = 81 - 36 = 45, \quad \Delta x_{2} = x_{3} - x_{2} = 9 - 6 = 3, \quad \text{slope} = \frac{\Delta y_{2}}{\Delta x_{2}} = \frac{45}{3} = 15\\ \Delta y_{3} &= y_{4} - y_{3} = 144 - 81 = 63, \quad \Delta x_{3} = x_{4} - x_{3} = 12 - 9 = 3, \quad \text{slope} = \frac{\Delta y_{3}}{\Delta x_{3}} = \frac{63}{3} = 21\\ \Delta y_{4} &= y_{5} - y_{4} = 225 - 144 = 81, \quad \Delta x_{4} = x_{5} - x_{4} = 15 - 12 = 3, \quad \text{slope} = \frac{\Delta y_{4}}{\Delta x_{4}} = \frac{81}{3} = 27 \end{aligned} \][/tex]
Since the slopes are not constant, Function A is not linear.
#### Function B
The data points for Function B are:
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline x & 5 & 10 & 15 & 20 & 25 \\ \hline y & 8 & 16 & 24 & 32 & 40 \\ \hline \end{array} \][/tex]
To check if [tex]\( y \)[/tex] is a linear function of [tex]\( x \)[/tex], we look at the differences in [tex]\( y \)[/tex] for equal changes in [tex]\( x \)[/tex]:
[tex]\[ \begin{aligned} \Delta y_{1} &= y_{2} - y_{1} = 16 - 8 = 8, \quad \Delta x_{1} = x_{2} - x_{1} = 10 - 5 = 5, \quad \text{slope} = \frac{\Delta y_{1}}{\Delta x_{1}} = \frac{8}{5} = \frac{8}{5}\\ \Delta y_{2} &= y_{3} - y_{2} = 24 - 16 = 8, \quad \Delta x_{2} = x_{3} - x_{2} = 15 - 10 = 5, \quad \text{slope} = \frac{\Delta y_{2}}{\Delta x_{2}} = \frac{8}{5}\\ \Delta y_{3} &= y_{4} - y_{3} = 32 - 24 = 8, \quad \Delta x_{3} = x_{4} - x_{3} = 20 - 15 = 5, \quad \text{slope} = \frac{\Delta y_{3}}{\Delta x_{3}} = \frac{8}{5}\\ \Delta y_{4} &= y_{5} - y_{4} = 40 - 32 = 8, \quad \Delta x_{4} = x_{5} - x_{4} = 25 - 20 = 5, \quad \text{slope} = \frac{\Delta y_{4}}{\Delta x_{4}} = \frac{8}{5} \end{aligned} \][/tex]
Since the slopes are constant, Function B is linear.
### Conclusion
From our analysis, only Function B has a constant rate of change and is therefore linear. The answer is B) Function B.
### Let's analyze both functions:
#### Function A
The data points for Function A are:
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline x & 3 & 6 & 9 & 12 & 15 \\ \hline y & 9 & 36 & 81 & 144 & 225 \\ \hline \end{array} \][/tex]
To check if [tex]\( y \)[/tex] is a linear function of [tex]\( x \)[/tex], we look at the differences in [tex]\( y \)[/tex] for equal changes in [tex]\( x \)[/tex]:
[tex]\[ \begin{aligned} \Delta y_{1} &= y_{2} - y_{1} = 36 - 9 = 27, \quad \Delta x_{1} = x_{2} - x_{1} = 6 - 3 = 3, \quad \text{slope} = \frac{\Delta y_{1}}{\Delta x_{1}} = \frac{27}{3} = 9\\ \Delta y_{2} &= y_{3} - y_{2} = 81 - 36 = 45, \quad \Delta x_{2} = x_{3} - x_{2} = 9 - 6 = 3, \quad \text{slope} = \frac{\Delta y_{2}}{\Delta x_{2}} = \frac{45}{3} = 15\\ \Delta y_{3} &= y_{4} - y_{3} = 144 - 81 = 63, \quad \Delta x_{3} = x_{4} - x_{3} = 12 - 9 = 3, \quad \text{slope} = \frac{\Delta y_{3}}{\Delta x_{3}} = \frac{63}{3} = 21\\ \Delta y_{4} &= y_{5} - y_{4} = 225 - 144 = 81, \quad \Delta x_{4} = x_{5} - x_{4} = 15 - 12 = 3, \quad \text{slope} = \frac{\Delta y_{4}}{\Delta x_{4}} = \frac{81}{3} = 27 \end{aligned} \][/tex]
Since the slopes are not constant, Function A is not linear.
#### Function B
The data points for Function B are:
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline x & 5 & 10 & 15 & 20 & 25 \\ \hline y & 8 & 16 & 24 & 32 & 40 \\ \hline \end{array} \][/tex]
To check if [tex]\( y \)[/tex] is a linear function of [tex]\( x \)[/tex], we look at the differences in [tex]\( y \)[/tex] for equal changes in [tex]\( x \)[/tex]:
[tex]\[ \begin{aligned} \Delta y_{1} &= y_{2} - y_{1} = 16 - 8 = 8, \quad \Delta x_{1} = x_{2} - x_{1} = 10 - 5 = 5, \quad \text{slope} = \frac{\Delta y_{1}}{\Delta x_{1}} = \frac{8}{5} = \frac{8}{5}\\ \Delta y_{2} &= y_{3} - y_{2} = 24 - 16 = 8, \quad \Delta x_{2} = x_{3} - x_{2} = 15 - 10 = 5, \quad \text{slope} = \frac{\Delta y_{2}}{\Delta x_{2}} = \frac{8}{5}\\ \Delta y_{3} &= y_{4} - y_{3} = 32 - 24 = 8, \quad \Delta x_{3} = x_{4} - x_{3} = 20 - 15 = 5, \quad \text{slope} = \frac{\Delta y_{3}}{\Delta x_{3}} = \frac{8}{5}\\ \Delta y_{4} &= y_{5} - y_{4} = 40 - 32 = 8, \quad \Delta x_{4} = x_{5} - x_{4} = 25 - 20 = 5, \quad \text{slope} = \frac{\Delta y_{4}}{\Delta x_{4}} = \frac{8}{5} \end{aligned} \][/tex]
Since the slopes are constant, Function B is linear.
### Conclusion
From our analysis, only Function B has a constant rate of change and is therefore linear. The answer is B) Function B.
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