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### Part (a): Completing the table for the graph of \( y = -x + 1 \)

To fill in the missing values for \( A \) and \( B \), we substitute the given values of \( x \) into the equation \( y = -x + 1 \).

1. **For \( x = -2 \):**
   \[
   y = -(-2) + 1 = 2 + 1 = 3
   \]
   So, \( A = 3 \).

2. **For \( x = 1 \):**
   \[
   y = -(1) + 1 = -1 + 1 = 0
   \]
   So, \( B = 0 \).

The completed table is:

\[
\begin{array}{|c|c|c|c|c|c|}
\hline
x & -2 & -1 & 0 & 1 & 2 \\
\hline
y & 3 & 2 & 1 & 0 & -1 \\
\hline
\end{array}
\]

### Part (b): Graph of \( y = -x + 1 \)

For graphing the line \( y = -x + 1 \), we can plot the points from the completed table:

- \( (-2, 3) \)
- \( (-1, 2) \)
- \( (0, 1) \)
- \( (1, 0) \)
- \( (2, -1) \)

Plot these points on the given axes and connect them to form a straight line.

Would you like to ask more questions about this or need any clarification? Here are five questions you can explore further:

1. What is the slope of the line \( y = -x + 1 \)?
2. How does changing the equation to \( y = -x + 2 \) affect the graph?
3. Can you find the x-intercept and y-intercept from the equation \( y = -x + 1 \)?
4. How would the graph look if the equation was \( y = x + 1 \) instead?
5. What is the domain and range of the function \( y = -x + 1 \)?

**Tip:** The slope in the equation \( y = -x + 1 \) is the coefficient of \( x \), which tells you how the graph behaves (in this case, it has a negative slope).

Complete the table and graph the equation y = -x + 1.

Learn how to complete a table for the linear equation y = -x + 1 and graph the line. This example covers graphing and understanding slope-intercept form, suitable for students in Grades 7-10.

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