To find the equation of a line parallel to \( y = 3x \) that passes through the point \((-2, -1)\), we need to recognize that parallel lines have the same slope.
The given line \( y = 3x \) has a slope of \( 3 \). Therefore, the line we want to find will also have a slope of \( 3 \).
Now, we can use the point-slope form of the equation of a line, which is:
\[ y - y_1 = m(x - x_1) \]
where \( m \) is the slope and \((x_1, y_1)\) is a point on the line. We have:
- \( m = 3 \)
- \( (x_1, y_1) = (-2, -1) \)
Substituting these values in, we get:
\[ y - (-1) = 3(x - (-2)) \]
This simplifies to:
\[ y + 1 = 3(x + 2) \]
Now, distributing the \( 3 \):
\[ y + 1 = 3x + 6 \]
Next, we subtract \( 1 \) from both sides to isolate \( y \):
\[ y = 3x + 5 \]
So, the equation of the line that is parallel to \( y = 3x \) and passes through the point \((-2, -1)\) is:
\[ \boxed{y = 3x + 5} \]