The given function is $f(x) = -3x^2 + 1$. This is a quadratic function, and it represents a parabola that opens downwards because the coefficient of $x^2$ is negative.

Key Features of the Function:

1. Vertex: The vertex form of a quadratic function is $f(x) = a(x - h)^2 + k$, where $(h, k)$ is the vertex. In this case, the vertex occurs at $x = 0$, and the function simplifies to $f(0) = -3(0)^2 + 1 = 1$. Thus, the vertex is $(0, 1)$.
2. Axis of Symmetry: The axis of symmetry is the vertical line that passes through the vertex, $x = 0$.
3. Direction: Since the coefficient of $x^2$ is negative, the parabola opens downward.
4. Y-intercept: The y-intercept occurs when $x = 0$, so $f(0) = 1$. Therefore, the y-intercept is $(0, 1)$.
5. X-intercepts: To find the x-intercepts, set $f(x) = 0$ and solve for $x$: $-3x^2 + 1 = 0 \quad \Rightarrow \quad x^2 = \frac{1}{3} \quad \Rightarrow \quad x = \pm \sqrt{\frac{1}{3}} = \pm \frac{\sqrt{3}}{3}$ So, the x-intercepts are $\left( \frac{\sqrt{3}}{3}, 0 \right)$ and $\left( -\frac{\sqrt{3}}{3}, 0 \right)$.

Would you like more details on any of these features or have any questions?

Here are 5 related questions to expand on this:

1. How do you convert the quadratic function to vertex form?
2. Can you explain how to calculate the vertex of a parabola algebraically?
3. How would this graph change if the coefficient of $x^2$ was positive?
4. What is the range of the function $f(x) = -3x^2 + 1$?
5. How do you calculate the domain of any quadratic function?

Tip: For any quadratic function, the domain is always all real numbers $(-\infty, \infty)$, but the range depends on whether the parabola opens upwards or downwards.