1. The problem asks if the function $y=\frac{1}{3}x^2 - 3$ is the same in standard form and vertex form. 2. The standard form of a quadratic function is generally written as $y = ax^2 + bx + c$. 3. The vertex form of a quadratic function is $y = a(x-h)^2 + k$, where $(h,k)$ is the vertex of the parabola. 4. For the given function, $y=\frac{1}{3}x^2 - 3$, it is already in standard form with $a=\frac{1}{3}$, $b=0$, and $c=-3$. 5. To convert to vertex form, we complete the square: $$y = \frac{1}{3}x^2 - 3 = \frac{1}{3}(x^2) - 3$$ Since $b=0$, the vertex is at $x=0$. 6. Thus, the vertex form is: $$y = \frac{1}{3}(x-0)^2 - 3 = \frac{1}{3}x^2 - 3$$ 7. Therefore, the function is the same in both standard and vertex forms because the vertex is at the origin shifted down by 3. Final answer: Yes, $y=\frac{1}{3}x^2 - 3$ is the same in standard form and vertex form.