The maximum distance of a point on the graph of function `y =sqrt( 3) sin x + cos x ` from x -axis is `"…………."`

Answer

To find the maximum distance of the point on the graph of the function \( y = \sqrt{3} \sin x + \cos x \) from the x-axis, we need to determine the maximum value of \( y \).

### Step-by-Step Solution:

1. **Identify the function**:
The given function is:
\[ y = \sqrt{3} \sin x + \cos x \]

2. **Rewrite the function**:
We can rewrite this function in a more manageable form using the sine addition formula. We know that:
\[ a \sin x + b \cos x = R \sin(x + \phi) \]
where \( R = \sqrt{a^2 + b^2} \) and \( \tan \phi = \frac{b}{a} \).

Here, \( a = \sqrt{3} \) and \( b = 1 \).

3. **Calculate \( R \)**:
\[ R = \sqrt{(\sqrt{3})^2 + (1)^2} = \sqrt{3 + 1} = \sqrt{4} = 2 \]

4. **Determine \( \phi \)**:
\[ \tan \phi = \frac{1}{\sqrt{3}} \implies \phi = \frac{\pi}{6} \]

5. **Rewrite the function**:
Thus, we can rewrite \( y \) as:
\[ y = 2 \sin\left(x + \frac{\pi}{6}\right) \]

6. **Find the maximum value of \( y \)**:
The maximum value of \( \sin \) function is 1. Therefore, the maximum value of \( y \) is:
\[ y_{\text{max}} = 2 \cdot 1 = 2 \]

7. **Conclusion**:
The maximum distance of a point on the graph from the x-axis is:
\[ \boxed{2} \]