https://file.aiquickdraw.com/mathbot/file/1725984669241qpw5pium.jpg

To solve the problem, we need to translate the parallelogram ABCD by **5 units to the left** and **2 units down**. This means we will subtract 5 from the x-coordinates and 2 from the y-coordinates of each point.

The original coordinates of the vertices are:
- \( A(0, 4) \)
- \( B(4, 4) \)
- \( C(6, 0) \)
- \( D(2, 0) \)

Let's apply the translation to each vertex.

### Translation Rule: 
- \( (x', y') = (x - 5, y - 2) \)

### Translated Points:
1. \( A(0, 4) \rightarrow A'(0 - 5, 4 - 2) = A'(-5, 2) \)
2. \( B(4, 4) \rightarrow B'(4 - 5, 4 - 2) = B'(-1, 2) \)
3. \( C(6, 0) \rightarrow C'(6 - 5, 0 - 2) = C'(1, -2) \)
4. \( D(2, 0) \rightarrow D'(2 - 5, 0 - 2) = D'(-3, -2) \)

The new vertices after the translation are:
- \( A'(-5, 2) \)
- \( B'(-1, 2) \)
- \( C'(1, -2) \)
- \( D'(-3, -2) \)

Now, you can plot these points on the graph to get the translated parallelogram.

Would you like more details or have any questions?

Here are some related questions you can explore:
1. How do you find the translation matrix for a 2D shape?
2. What other transformations could be applied to a parallelogram besides translation?
3. How does translation differ from rotation and reflection in geometry?
4. Can you derive a general formula for translating any polygon?
5. How do translations affect the area and perimeter of shapes?

**Tip:** When translating shapes, always double-check the direction and magnitude of movement for both the x- and y-coordinates to ensure accuracy.

The figure shows parallelogram ABCD. Graph the image of the parallelogram after a translation of 5 units to the left and 2 units down.

Learn how to translate parallelogram ABCD on a coordinate plane by 5 units to the left and 2 units down. This problem involves key concepts of geometry and transformations, suitable for students in Grades 6-8.

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