Checkpoint Sketch the graph of \( (x+2)^{2}+(y-2)^{2}<16 \)

To sketch the graph of the inequality \( (x+2)^{2}+(y-2)^{2}<16 \), we can follow these steps: 1. **Identify the type of graph**: The expression \( (x+2)^{2}+(y-2)^{2}=16 \) represents a circle. The inequality \( < 16 \) indicates that we are interested in the area inside this circle. 2. **Determine the center and radius**: - The center of the circle is at the point \( (-2, 2) \). - The radius \( r \) can be found by taking the square root of 16, which gives \( r = 4 \). 3. **Draw the circle**: - Plot the center at \( (-2, 2) \). - From the center, measure 4 units in all directions (up, down, left, right) to find the points on the circle: - Up: \( (-2, 6) \) - Down: \( (-2, -2) \) - Left: \( (-6, 2) \) - Right: \( (2, 2) \) 4. **Sketch the circle**: - Draw a circle with the center at \( (-2, 2) \) and passing through the points identified above. 5. **Shade the interior**: - Since the inequality is \( < 16 \), shade the area inside the circle to indicate that all points within this region satisfy the inequality. 6. **Boundary**: - The boundary of the circle itself (where \( (x+2)^{2}+(y-2)^{2}=16 \)) is not included in the solution set, so you can represent this with a dashed line. This process will give you a clear sketch of the graph for the inequality \( (x+2)^{2}+(y-2)^{2}<16 \).