onnnit
Answered

This is the input-output table for the linear function [tex]y=3x[/tex].

\begin{tabular}{|r|r|}
\hline [tex]$x$[/tex] & [tex]$y$[/tex] \\
\hline -2 & -6 \\
\hline -1 & -3 \\
\hline 0 & 0 \\
\hline 1 & 3 \\
\hline 2 & 6 \\
\hline 3 & 9 \\
\hline
\end{tabular}

Which best describes how the [tex]y[/tex]-values are increasing over each interval?

A. Subtract 10 over each interval.
B. Multiply by 2 over each interval.
C. Add 3 over each interval.



Answer :

Sure, let's examine how the [tex]\( y \)[/tex]-values are increasing over each interval step-by-step.

Given the input-output table for the linear function [tex]\( y = 3x \)[/tex]:

[tex]\[ \begin{array}{|r|r|} \hline x & y \\ \hline -2 & -6 \\ \hline -1 & -3 \\ \hline 0 & 0 \\ \hline 1 & 3 \\ \hline 2 & 6 \\ \hline 3 & 9 \\ \hline \end{array} \][/tex]

We need to determine how the [tex]\( y \)[/tex]-values increase as [tex]\( x \)[/tex] changes.

### Step-by-Step Analysis:
1. From [tex]\( x = -2 \)[/tex] to [tex]\( x = -1 \)[/tex]:
[tex]\[ y\text{-value changes from } -6 \text{ to } -3. \][/tex]
The difference is:
[tex]\[ -3 - (-6) = 3 \][/tex]

2. From [tex]\( x = -1 \)[/tex] to [tex]\( x = 0 \)[/tex]:
[tex]\[ y\text{-value changes from } -3 \text{ to } 0. \][/tex]
The difference is:
[tex]\[ 0 - (-3) = 3 \][/tex]

3. From [tex]\( x = 0 \)[/tex] to [tex]\( x = 1 \)[/tex]:
[tex]\[ y\text{-value changes from } 0 \text{ to } 3. \][/tex]
The difference is:
[tex]\[ 3 - 0 = 3 \][/tex]

4. From [tex]\( x = 1 \)[/tex] to [tex]\( x = 2 \)[/tex]:
[tex]\[ y\text{-value changes from } 3 \text{ to } 6. \][/tex]
The difference is:
[tex]\[ 6 - 3 = 3 \][/tex]

5. From [tex]\( x = 2 \)[/tex] to [tex]\( x = 3 \)[/tex]:
[tex]\[ y\text{-value changes from } 6 \text{ to } 9. \][/tex]
The difference is:
[tex]\[ 9 - 6 = 3 \][/tex]

### Summary:
After calculating the differences for each interval, we observe that the [tex]\( y \)[/tex]-value increases by 3 in each interval.

Thus, the best description of how the [tex]\( y \)[/tex]-values are increasing over each interval is:
[tex]\[ \textbf{Add 3 over each interval.} \][/tex]