As the position of an object (u) reflected from a concave mirror is varied, the position of the image (v) also varies. By latting the u changes from 0 to `+ oo` the graph between v versus u will be

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Answer

To solve the problem of finding the relationship between the object distance (u) and the image distance (v) for a concave mirror, we will use the mirror formula and analyze the behavior of the graph as u varies from 0 to +∞.

### Step-by-Step Solution:

1. **Understand the Mirror Formula**:
The mirror formula for a concave mirror is given by:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Rearranging this gives:
\[ \frac{1}{v} = \frac{1}{f} - \frac{1}{u} \]
or equivalently,
\[ v = \frac{uf}{u - f} \]

2. **Analyze the Behavior as u Approaches 0**:
- When \( u = 0 \):
\[ v = \frac{0 \cdot f}{0 - f} = 0 \]
- Thus, when the object is at the pole of the mirror, the image is also at the pole.

3. **Analyze the Behavior as u Approaches +∞**:
- When \( u \) approaches \( +\infty \):
\[ v = \frac{u \cdot f}{u - f} \approx f \quad \text{(since } u \text{ is much larger than } f\text{)} \]
- Therefore, as the object moves far away, the image distance approaches the focal length \( f \).

4. **Identify Key Points**:
- At \( u = 0 \), \( v = 0 \).
- As \( u \to +\infty \), \( v \to f \).

5. **Graphing the Relationship**:
- The graph starts at the origin (0,0) and approaches the line \( v = f \) as \( u \) increases.
- The curve will be hyperbolic in nature, reflecting that as \( u \) increases, \( v \) approaches a constant value (the focal length).

6. **Conclusion**:
- The graph of \( v \) versus \( u \) will start from the origin and will asymptotically approach the line \( v = f \) as \( u \) increases.