1. The problem is to identify the graph of the function $$f(x) = \sqrt{x} + 1 + 5$$. 2. First, simplify the function: $$f(x) = \sqrt{x} + 6$$ 3. The square root function $$\sqrt{x}$$ is defined only for $$x \geq 0$$. 4. The graph of $$f(x)$$ will start at $$x=0$$, where: $$f(0) = \sqrt{0} + 6 = 6$$ 5. As $$x$$ increases, $$\sqrt{x}$$ increases slowly, so the graph will start at point $$(0,6)$$ and increase slowly upward to the right. 6. The graph will not exist for $$x < 0$$ because the square root of a negative number is not real. 7. Comparing with the descriptions: - Bottom-left and bottom-center graphs start near $$y=5$$ at $$x=-1$$, which is outside the domain. - Bottom-right graph starts near $$y=2$$ at $$x=-3$$, also outside the domain. 8. None of the graphs start at $$x=0$$ with $$y=6$$, but the closest description is the bottom-center graph which covers $$x$$ from $$-2$$ to $$8$$ and starts near $$y=5$$ at $$x=-1$$. 9. Since the function is only defined for $$x \geq 0$$ and starts at $$y=6$$, the correct graph should start at $$(0,6)$$ and increase slowly to the right. 10. Therefore, none of the given graphs perfectly match the function $$f(x) = \sqrt{x} + 6$$, but the bottom-center graph is the closest if we consider domain restrictions. Final answer: The function $$f(x) = \sqrt{x} + 6$$ starts at $$(0,6)$$ and increases slowly to the right, defined only for $$x \geq 0$$.