The function given is:

$f(x) = -x^2 - 1$

Analysis:

1. Type of Function: This is a quadratic function in the form $f(x) = ax^2 + bx + c$, where $a = -1$, $b = 0$, and $c = -1$.

2. Shape of the Graph: The graph will be a parabola. Since the coefficient $a = -1$ is negative, the parabola opens downwards.

3. Vertex: The vertex of the parabola can be calculated using the formula for the $x$-coordinate of the vertex:
   $x = -\frac{b}{2a}$ Substituting $b = 0$ and $a = -1$: $x = -\frac{0}{2(-1)} = 0$ Now, substituting $x = 0$ into $f(x)$: $f(0) = -(0)^2 - 1 = -1$ So, the vertex is at $(0, -1)$.

4. Axis of Symmetry: The axis of symmetry is the vertical line through the vertex: $x = 0$

5. Y-intercept: The y-intercept occurs when $x = 0$: $f(0) = -1$ So, the y-intercept is $(0, -1)$.

6. X-intercepts: To find the x-intercepts, set $f(x) = 0$: $-x^2 - 1 = 0 \Rightarrow -x^2 = 1 \Rightarrow x^2 = -1$ Since $x^2 = -1$ has no real solutions, there are no x-intercepts.

Summary:

• The vertex is at $(0, -1)$.
• The parabola opens downwards.
• The axis of symmetry is $x = 0$.
• There is a y-intercept at $(0, -1)$.
• No x-intercepts.

Would you like to see a graph or have more details on this?

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Related Questions:

1. What would be the vertex if the function was $f(x) = -x^2 + 2x - 1$?
2. How does the sign of the coefficient $a$ affect the direction in which the parabola opens?
3. What transformations would shift the vertex of this function to $(2, -1)$?
4. How do you find the maximum or minimum value of a quadratic function?
5. How would you solve for $x$ if the function were $f(x) = -x^2 + 3x - 4$?

Tip:

For any quadratic function of the form $ax^2 + bx + c$, the vertex is always given by $x = -\frac{b}{2a}$. This helps quickly locate the highest or lowest point on the parabola.