1. The problem is to describe the function $y = x^3$. 2. This is a cubic function where each input $x$ is raised to the third power. 3. It is an odd function, meaning $f(-x) = -f(x)$, so it is symmetric about the origin. 4. The graph passes through the origin $(0,0)$. 5. For positive $x$, $y = x^3$ grows positively and for negative $x$, it decreases negatively. 6. The function is continuous and smooth everywhere. 7. The derivative is $y' = 3x^2$, which is zero at $x=0$, indicating a stationary point there. 8. Since $y'' = 6x$, the function is concave down for $x<0$ and concave up for $x>0$, making $x=0$ an inflection point. 9. To summarize, $y=x^3$ is a cubic function with a shape like an S curve passing through the origin. The final answer is the function itself: $$y = x^3$$