1. The problem is to understand and analyze the function $y=\frac{1}{3}x^2$. 2. This is a quadratic function in the form $y=ax^2$, where $a=\frac{1}{3}$. 3. The graph of this function is a parabola opening upwards because $a>0$. 4. The vertex of the parabola is at the origin $(0,0)$ since there is no linear or constant term. 5. The formula for the vertex of $y=ax^2+bx+c$ is $\left(-\frac{b}{2a}, c-\frac{b^2}{4a}\right)$, but here $b=0$ and $c=0$, so vertex is $(0,0)$. 6. The y-intercept is found by evaluating $y$ at $x=0$: $y=\frac{1}{3}\times 0^2=0$. 7. The x-intercepts are found by solving $\frac{1}{3}x^2=0$, which gives $x=0$. 8. The parabola is wider than $y=x^2$ because the coefficient $\frac{1}{3}$ is less than 1, which compresses the graph vertically. 9. Summary: The parabola opens upwards, vertex at $(0,0)$, intercepts at $(0,0)$, and is wider than the standard parabola $y=x^2$.