From simple questions to complex issues, IDNLearn.com has the answers you need. Discover detailed and accurate answers to your questions from our knowledgeable and dedicated community members.
Sagot :
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.
Thank you for contributing to our discussion. Don't forget to check back for new answers. Keep asking, answering, and sharing useful information. For clear and precise answers, choose IDNLearn.com. Thanks for stopping by, and come back soon for more valuable insights.
