Let's analyze the graph and the questions provided.

Observations from the graph:

• The graph is a parabola that opens upwards with a vertex at the point $(0, -4)$.
• The graph crosses the x-axis at $(-2, 0)$ and $(2, 0)$, which are the x-intercepts.
• The y-intercept occurs at $(0, -4)$.

Answering the questions:

1. The domain (part a): Since the parabola extends indefinitely in both the left and right directions, the domain is all real numbers.

   • Answer: $(-\infty, \infty)$.

2. The range (part b): The lowest point on the graph (the vertex) is $-4$, and the parabola opens upwards, so the range starts from $-4$ and goes to infinity.

   • Answer: $[-4, \infty)$.

3. The x-intercepts (part c): The graph crosses the x-axis at the points $(-2, 0)$ and $(2, 0)$.

   • Answer: $-2, 2$.

4. The y-intercept (part d): The graph crosses the y-axis at $(0, -4)$.

   • Answer: $-4$.

5. Missing function values (f(-2) and f(2)) (last part):

   • $f(-2) = 0$ and $f(2) = 0$, as both points are the x-intercepts, meaning the function equals 0 at these points.

Let me know if you'd like more details on any part of this or have additional questions.

Here are 5 related questions:

1. How do we determine the vertex of a parabola from its equation?
2. What are the general forms of a quadratic function?
3. How do the domain and range change if the parabola opens downwards?
4. What is the symmetry of the graph of a parabola, and how does it help in solving problems?
5. How do you find the roots of a quadratic equation algebraically?

Tip:

For quadratic functions, the vertex can be found using the formula $x = \frac{-b}{2a}$ when the function is in standard form $ax^2 + bx + c$.