To complete the table for the function \( y = f(x + 5) \), we will replace \( x \) in the original function \( y = f(x) \) with \( x + 5 \). This means that for each value of \( x \) in the original function, we will find the corresponding \( y \) value for \( x + 5 \).
Let's first list the original function values (based on the provided x-values):
Original Function Table \( y = f(x) \)
| x | y = f(x) | |-------|------------| | -2 | 5 | | -1 | 2 | | 1 | 0 | | 2 | 2 | | 5 | 2 |
Now, to create the new table for the translated function \( y = f(x + 5) \), we will use values \( x - 5 \) (since \( f(x + 5) \) shifts the graph to the left by 5 units).
Here's the process:
- Substitute each \( x \) in \( y = f(x) \) with \( x - 5 \) for the new function.
- Calculate the output for each substituted value.
Here are the corresponding \( x \) values and their translations:
Translated Function Table \( y = f(x + 5) \)
| x | y = f(x + 5) | |-------|---------------| | -7 | 5 | (since \( f(-2) = 5 \)) | | -6 | 2 | (since \( f(-1) = 2 \)) | | -4 | 0 | (since \( f(1) = 0 \)) | | -3 | 2 | (since \( f(2) = 2 \)) | | 0 | 2 | (since \( f(5) = 2 \)) |
Now the completed tables are:
Original Function Table \( y = f(x) \)
| x | y = f(x) | |-------|------------| | -2 | 5 | | -1 | 2 | | 1 | 0 | | 2 | 2 | | 5 | 2 |
Translated Function Table \( y = f(x + 5) \)
| x | y = f(x + 5) | |-------|---------------| | -7 | 5 | | -6 | 2 | | -4 | 0 | | -3 | 2 | | 0 | 2 |