To complete the tables for the original function \( y = f(x) \) and its translation \( y = f(x + 5) \), we start by looking at the original function values provided in the first table.
Original Function
\[ \begin{array}{|c|c|} \hline x & y = f(x) \ \hline -2 & 5 \ -1 & 2 \ 0 & 1 \ 1 & 2 \ 2 & 5 \ \hline \end{array} \]
Translated Function
To find the values of the translated function \( y = f(x + 5) \), we substitute \( x \) with \( x + 5 \) and use the original function's \( x \) values corresponding to those shifted \( x \) values.
So for each \( x \) in the translated function, we determine the original function value at \( x - 5 \):
- When \( x = 5 \), \( f(5 - 5) = f(0) = 1 \)
- When \( x = 2 \), \( f(2 - 5) = f(-3) \) (not provided originally)
- When \( x = 1 \), \( f(1 - 5) = f(-4) \) (not provided originally)
- When \( x = 2 \), \( f(2 - 5) = f(-3) \) (not provided originally)
- When \( x = 5 \), \( f(5 - 5) = f(0) = 1 \)
Since \( f(-3) \) and \( f(-4) \) are not specifically provided, we will fill in the known values for \( f(5) \) accordingly. Here’s how the completed table looks:
\[ \begin{array}{|c|c|} \hline x & y = f(x + 5) \ \hline 5 & 1 \ 2 & ? \ 1 & ? \ 2 & ? \ 5 & 1 \ \hline \end{array} \]
Though we cannot fill the unknown values for \( f(-3) \) and \( f(-4) \), we can highlight that:
The values for \( y \) in the translated function depend on the original function output, so two values remain unknown unless more original data points for \( f(x) \) are provided.
Final Translated Table
Given what we know: \[ \begin{array}{|c|c|} \hline x & y = f(x + 5) \ \hline 5 & 1 \ 2 & \text{(unknown, potentially needs data)} \ 1 & \text{(unknown, potentially needs data)} \ 2 & \text{(unknown, potentially needs data)} \ 5 & 1 \ \hline \end{array} \] The only calculated values are at \( x = 5 \) yielding \( y = 1 \). The \( y \) values corresponding to \( x = 2, 1 \) remain undefined without additional information.