1. The problem is to graph the function $$y = 4 + \sqrt{x}$$. 2. Important note: The square root function $$\sqrt{x}$$ is only defined for $$x \geq 0$$, so the domain of this function is $$[0, \infty)$$. 3. The function shifts the basic square root graph $$y = \sqrt{x}$$ upward by 4 units. 4. The endpoint of the graph is at $$x=0$$, where $$y = 4 + \sqrt{0} = 4$$, so the point is $$(0,4)$$. 5. To plot points, choose values of $$x$$ starting from 0 and calculate $$y$$: - For $$x=0$$: $$y=4+\sqrt{0}=4$$ - For $$x=1$$: $$y=4+\sqrt{1}=4+1=5$$ - For $$x=4$$: $$y=4+\sqrt{4}=4+2=6$$ - For $$x=9$$: $$y=4+\sqrt{9}=4+3=7$$ 6. The graph starts at $$(0,4)$$ and curves upward to the right, increasing slowly as $$x$$ increases. 7. The function has no intercepts with the x-axis because $$y$$ is always greater than or equal to 4. 8. The graph has no extrema (no maximum or minimum) because the square root function is increasing. Final answer: The graph of $$y = 4 + \sqrt{x}$$ starts at $$(0,4)$$ and increases slowly to the right as $$x$$ increases.