To determine which table can be created using the given equation [tex]\( y = -2 + 4x \)[/tex], we need to validate the points listed in each table against the equation. Here's a step-by-step verification process for each table:
### Equation
[tex]\[ y = -2 + 4x \][/tex]
### Table 1
[tex]\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
-5 & -22 \\
\hline
0 & -2 \\
\hline
3 & 10 \\
\hline
\end{array}
\][/tex]
1. For [tex]\( x = -5 \)[/tex]:
[tex]\[ y = -2 + 4(-5) = -2 - 20 = -22 \][/tex]
This matches the table entry.
2. For [tex]\( x = 0 \)[/tex]:
[tex]\[ y = -2 + 4(0) = -2 \][/tex]
This matches the table entry.
3. For [tex]\( x = 3 \)[/tex]:
[tex]\[ y = -2 + 4(3) = -2 + 12 = 10 \][/tex]
This matches the table entry.
Since all entries in Table 1 match the equation, this table is valid.
### Table 2
[tex]\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
-22 & -5 \\
\hline
-2 & 0 \\
\hline
10 & 3 \\
\hline
\end{array}
\][/tex]
1. For [tex]\( x = -22 \)[/tex]:
[tex]\[ y = -2 + 4(-22) = -2 - 88 = -90 \][/tex]
This does not match the table entry [tex]\( y = -5 \)[/tex].
2. Since the first entry does not match, we can conclude that Table 2 is not valid.
### Table 3
[tex]\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
-5 & -18 \\
\hline
0 & -2 \\
\hline
3 & 10 \\
\hline
\end{array}
\][/tex]
1. For [tex]\( x = -5 \)[/tex]:
[tex]\[ y = -2 + 4(-5) = -2 - 20 = -22 \][/tex]
This does not match the table entry [tex]\( y = -18 \)[/tex].
2. Since the first entry does not match, we can conclude that Table 3 is not valid.
### Conclusion
After verifying the entries in all three tables, only Table 1 matches the given equation [tex]\( y = -2 + 4x \)[/tex].
Therefore, the correct table is:
[tex]\[
\begin{tabular}{|c|c|}
\hline
x & y \\
\hline
-5 & -22 \\
\hline
0 & -2 \\
\hline
3 & 10 \\
\hline
\end{tabular}
\][/tex]
