1. The problem is to plot the graph of the line given by the equation $$y = 4x + 3$$ where the gradient (slope) is positive. 2. The equation of a straight line in slope-intercept form is $$y = mx + c$$ where $$m$$ is the gradient and $$c$$ is the y-intercept. 3. Here, $$m = 4$$ (positive gradient) and $$c = 3$$ (y-intercept). 4. To plot the line, find two points: - When $$x = 0$$, $$y = 4(0) + 3 = 3$$, so point (0, 3). - When $$x = 2$$, $$y = 4(2) + 3 = 8 + 3 = 11$$, so point (2, 11). 5. Draw a straight line through these points to represent the graph of $$y = 4x + 3$$. Final answer: The graph is a straight line with positive slope 4, crossing the y-axis at 3 and passing through (2, 11).