Why will the output values for linear function [tex]$A$[/tex] always be different from the corresponding output values for linear function [tex]$B$[/tex]?

Linear Function [tex]$A$[/tex]
\begin{tabular}{|c|c|}
\hline
[tex]$x$[/tex] & [tex]$y$[/tex] \\
\hline
1 & 3 \\
\hline
3 & 7 \\
\hline
5 & 11 \\
\hline
7 & 15 \\
\hline
9 & 19 \\
\hline
\end{tabular}

Linear Function [tex]$B$[/tex]
\begin{tabular}{|c|c|}
\hline
[tex]$v$[/tex] & [tex]$w$[/tex] \\
\hline
1 & 4 \\
\hline
3 & 8 \\
\hline
5 & 12 \\
\hline
7 & 16 \\
\hline
9 & 20 \\
\hline
\end{tabular}

A. The initial values of the two functions are different, and the rates of change of the two functions are also different.

B. The initial values of the two functions are different, and the rates of change of the two functions are the same.

C. The initial values of the two functions are the same, and the rates of change of the two functions are different.

D. The initial values of the two functions are the same, and the rates of change of the two functions are also the same.



Answer :

To determine why the output values for linear function [tex]\( A \)[/tex] will always be different from the corresponding output values for linear function [tex]\( B \)[/tex], we need to analyze the given data and calculate the linear functions represented by [tex]\( A \)[/tex] and [tex]\( B \)[/tex].

### Step-by-Step Solution:

1. Identify the input-output pairs for each function:
- For [tex]\( A \)[/tex]:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & 3 \\ \hline 3 & 7 \\ \hline 5 & 11 \\ \hline 7 & 15 \\ \hline 9 & 19 \\ \hline \end{array} \][/tex]
- For [tex]\( B \)[/tex]:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & 4 \\ \hline 3 & 8 \\ \hline 5 & 12 \\ \hline 7 & 16 \\ \hline 9 & 20 \\ \hline \end{array} \][/tex]

2. Calculate the slopes (rate of change) of the functions:
The slope formula for a linear function given two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is:
[tex]\[ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]

- For [tex]\( A \)[/tex], using points [tex]\((1, 3)\)[/tex] and [tex]\((3, 7)\)[/tex]:
[tex]\[ \text{slope}_A = \frac{7 - 3}{3 - 1} = \frac{4}{2} = 2 \][/tex]
- For [tex]\( B \)[/tex], using points [tex]\((1, 4)\)[/tex] and [tex]\((3, 8)\)[/tex]:
[tex]\[ \text{slope}_B = \frac{8 - 4}{3 - 1} = \frac{4}{2} = 2 \][/tex]

Hence, both functions [tex]\( A \)[/tex] and [tex]\( B \)[/tex] have the same slope of [tex]\( 2 \)[/tex].

3. Calculate the y-intercepts of the functions:
The y-intercept ([tex]\( b \)[/tex]) of a linear function [tex]\( y = mx + b \)[/tex] can be found by substituting one of the points and the slope into the equation.

- For [tex]\( A \)[/tex], using point [tex]\((1, 3)\)[/tex]:
[tex]\[ 3 = 2 \cdot 1 + b \implies 3 = 2 + b \implies b_A = 1 \][/tex]
- For [tex]\( B \)[/tex], using point [tex]\((1, 4)\)[/tex]:
[tex]\[ 4 = 2 \cdot 1 + b \implies 4 = 2 + b \implies b_B = 2 \][/tex]

Hence, the y-intercept of function [tex]\( A \)[/tex] is [tex]\( 1 \)[/tex] and for function [tex]\( B \)[/tex] it is [tex]\( 2 \)[/tex].

4. Compare the slopes and y-intercepts:
- The slopes of both functions [tex]\( A \)[/tex] and [tex]\( B \)[/tex] are equal ([tex]\( 2 \)[/tex]).
- The y-intercepts of the functions are different: [tex]\( 1 \)[/tex] for [tex]\( A \)[/tex] and [tex]\( 2 \)[/tex] for [tex]\( B \)[/tex].

### Conclusion:
The output values for function [tex]\( A \)[/tex] will always be different from the corresponding output values for function [tex]\( B \)[/tex] because although they have the same rate of change (slope), their initial values (y-intercepts) are different.

Thus, the correct statement is:
- The initial values of the two functions are different, and the rates of change of the two functions are the same.

### Answer:
The initial values of the two functions are different, and the rates of change of the two functions are the same.