1. The problem is to graph the linear inequality $y \geq 3x + 3$ on the coordinate plane. 2. First, graph the boundary line given by the equation $y = 3x + 3$. This line has a slope of 3 and a y-intercept at (0, 3). 3. Since the inequality is $y \geq 3x + 3$, the boundary line should be solid to indicate that points on the line satisfy the inequality. 4. To graph the line, plot the y-intercept at (0, 3). From there, use the slope 3 (which is $\frac{3}{1}$) to plot another point by going up 3 units and right 1 unit, so points like (1, 6) are on the line. 5. Shade the region above the line because $y$ is greater than or equal to $3x + 3$. This means for any point above or on the line, the inequality holds true. 6. In terms of graph features, the line crosses the y-axis at (0, 3) and crosses the x-axis where $y=0$: solving $0 = 3x + 3$ gives $x = -1$. 7. The shaded region is all points $(x,y)$ such that $y$ is above or on the line $y = 3x + 3$. Final answer: The graph is the solid line $y = 3x + 3$ with shading above it showing all points where $y \geq 3x + 3$.