Given the equation:

To be able to graph the equation, let's identify at least two points that pass through the graph of the equation. We can get this by getting them at two conditions: when x = 0 and y = 0.
At x = 0:



Therefore, Point A: (0,1) = x1,y1.
At y = 0:




Therefore, Point B: (3/2,0) = x2,y2.
Let's now plot the graph of the equation.