To determine whether the function \( y = 3\sqrt{x-1} \) is increasing or decreasing, we can analyze the behavior of the function as \( x \) changes.
1. **Domain**: The function is defined for \( x \geq 1 \) (because of the square root).
2. **Derivative**: To find out if the function is increasing or decreasing, we can look at the derivative of the function:
\[ y = 3\sqrt{x-1} \]
Using the chain rule to differentiate:
\[ y' = 3 \cdot \frac{1}{2\sqrt{x-1}} \cdot (1) = \frac{3}{2\sqrt{x-1}} \]
3. **Sign of the Derivative**: The derivative \( y' = \frac{3}{2\sqrt{x-1}} \) is positive for all \( x > 1 \). Since the derivative is positive in the domain \( x \geq 1 \), this means that the function is increasing in this interval.
**Conclusion**: The graph of the function \( y = 3\sqrt{x-1} \) shows that the function is **increasing**.
So the correct response is:
**increasing**.