1. **State the problem:** We are given the function $$y = \sqrt[3]{x} + 2$$ and want to understand its graph and behavior. 2. **Formula and rules:** The cube root function is defined as $$y = \sqrt[3]{x} = x^{\frac{1}{3}}$$. It is defined for all real numbers and is an odd function. Adding 2 shifts the graph vertically upward by 2 units. 3. **Intermediate work:** The original cube root function passes through points like $(-8, -2)$ because $$\sqrt[3]{-8} = -2$$. Adding 2 shifts this to $$-2 + 2 = 0$$, so the point becomes $(-8, 0)$ on the shifted graph. 4. **Evaluate key points:** - At $x = -8$: $$y = \sqrt[3]{-8} + 2 = -2 + 2 = 0$$ - At $x = -1$: $$y = \sqrt[3]{-1} + 2 = -1 + 2 = 1$$ - At $x = 0$: $$y = \sqrt[3]{0} + 2 = 0 + 2 = 2$$ - At $x = 8$: $$y = \sqrt[3]{8} + 2 = 2 + 2 = 4$$ 5. **Explanation:** The graph of $$y = \sqrt[3]{x} + 2$$ looks like the cube root curve shifted up by 2 units. It passes through points $(-8,0)$, $(-1,1)$, $(0,2)$, and $(8,4)$. The shape is an S-curve passing through these points smoothly. Final answer: The function $$y = \sqrt[3]{x} + 2$$ is a cube root curve shifted vertically up by 2 units, passing through the points calculated above.