The two equations in the images you've provided are:

1. First equation: $f(0) = -(0)^2 - 1$ Simplifying this: $f(0) = -0 - 1 = -1$

2. Second equation: $f(x) = -(x)^2 + 1$ This represents a quadratic function: $f(x) = -x^2 + 1$ which is a downward-facing parabola with its vertex at $(0, 1)$.

Let me know if you need more details or have any further questions.

Here are some related questions to consider:

1. What is the domain and range of the function $f(x) = -x^2 + 1$?
2. What are the x-intercepts and y-intercepts of $f(x) = -x^2 + 1$?
3. How does the graph of $f(x) = -x^2 + 1$ compare to $f(x) = x^2$?
4. What is the vertex of the parabola $f(x) = -x^2 + 1$?
5. How would you solve for $f(x) = 0$?

Tip: When working with parabolas, the sign of the coefficient of $x^2$ determines whether the parabola opens upwards (positive) or downwards (negative).