1. The problem states we need to graph a line that passes through the point $(-4,0)$ and has a slope of $\frac{3}{5}$. 2. Recall the slope-intercept form of a line is given by $$y = mx + b$$ where $m$ is the slope and $b$ is the y-intercept. 3. We know $m = \frac{3}{5}$ and the line passes through $(-4, 0)$, so we substitute to find $b$: $$0 = \frac{3}{5} \times (-4) + b$$ $$0 = -\frac{12}{5} + b$$ $$b = \frac{12}{5}$$ 4. Therefore, the equation of the line is: $$y = \frac{3}{5}x + \frac{12}{5}$$ 5. This line will cross the y-axis at $y = \frac{12}{5} = 2.4$ and rise 3 units for every 5 units it moves right. 6. To plot the line, mark the point $(-4, 0)$, then move right 5 units to $x = 1$ and up 3 units to $y = 3.6$. 7. Connect these points to form the line on the graph. Final answer: $$y = \frac{3}{5}x + \frac{12}{5}$$