Let's analyze and solve each part of the question based on the graph.

1. Solutions:

a) Determine the equation of the axis of symmetry for the parabola.

The axis of symmetry of a parabola is a vertical line that passes through its vertex. From the graph, the vertex appears to be at $x = 4$. Therefore, the equation of the axis of symmetry is: $x = 4$

b) Determine the coordinates of the vertex of the parabola.

The vertex is the lowest point on the parabola in this case (since it opens upwards). From the graph, the vertex is located at $(4, -16)$.

c) State the domain and range of the function.

For any quadratic function (parabola), the domain is all real numbers because the parabola extends infinitely in both directions along the x-axis. Therefore: $\text{Domain: } x \in (-\infty, \infty)$

Since the parabola opens upwards and the lowest point is the vertex, the range is all y-values from the y-coordinate of the vertex upward. Thus: $\text{Range: } y \geq -16$

Summary of Answers:

1. Axis of Symmetry: $x = 4$
2. Vertex: $(4, -16)$
3. Domain: $x \in (-\infty, \infty)$
4. Range: $y \geq -16$

Would you like further details on any part of this solution?

Related Questions:

1. How do you determine the direction a parabola opens?
2. What is the general form of a quadratic function and how does it relate to the vertex?
3. How can you find the vertex if you are given a quadratic equation in standard form?
4. What changes to a quadratic function affect the position of its axis of symmetry?
5. How does the domain of a quadratic function compare to other types of functions?

Tip:

The axis of symmetry for a parabola $y = ax^2 + bx + c$ is always $x = -\frac{b}{2a}$. This formula is helpful when you don't have the graph but know the equation of the parabola.