Let's analyze the function given in the image and extract the necessary information:

The function is: $f(x) = \frac{1}{2}(x - 2)^2 - 3$

This is a quadratic function in vertex form, $f(x) = a(x - h)^2 + k$, where:

• $h$ and $k$ represent the vertex of the parabola.
• $a$ controls the direction and width of the parabola (if $a > 0$, the parabola opens upwards; if $a < 0$, it opens downwards).

1. Vertex

Since the function is in vertex form, we can identify the vertex directly from $(h, k)$:

• Here, $h = 2$ and $k = -3$.
• Therefore, the vertex is $(2, -3)$.

2. Axis of Symmetry

The axis of symmetry of a parabola in vertex form $f(x) = a(x - h)^2 + k$ is the line $x = h$.

• For this function, the axis of symmetry is $x = 2$.

3. Domain

Since this is a quadratic function, the domain is all real numbers.

• In interval notation, the domain is $(-\infty, \infty)$.

4. Range

Since $a = \frac{1}{2} > 0$, the parabola opens upwards, and the minimum value of $f(x)$ is at the vertex, $y = -3$.

• Therefore, the range is $[-3, \infty)$.

Summary of Answers

• Vertex: $(2, -3)$
• Axis of Symmetry: $x = 2$
• Domain: $(-\infty, \infty)$
• Range: $[-3, \infty)$

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

Here are some additional questions to consider:

1. What would change in the vertex if we modified the value of $h$ or $k$?
2. How does the coefficient $a$ affect the width of the parabola?
3. What if $a$ was negative? How would that affect the direction and range?
4. Can the axis of symmetry ever be something other than $x = h$ in vertex form?
5. How do you find the vertex if the equation is in standard form instead of vertex form?

Tip: When graphing parabolas, always identify the vertex and axis of symmetry first; they serve as central guides for sketching the shape accurately.