From the image, we are analyzing a quadratic function given as:

$f(x) = 2(x - 3)^2 - 1$

Key Features:

1. Vertex Form: The equation $f(x) = 2(x - 3)^2 - 1$ is already in vertex form, $f(x) = a(x - h)^2 + k$, where:

   • $h = 3$ (x-coordinate of the vertex),
   • $k = -1$ (y-coordinate of the vertex).
   • Thus, the vertex is at $(3, -1)$.

2. Axis of Symmetry: The axis of symmetry is the vertical line that passes through the vertex. Therefore: $x = 3$

3. Domain: A quadratic function is defined for all real numbers. Therefore, the domain is: $(-\infty, \infty)$

4. Range: Since $a = 2 > 0$, the parabola opens upwards. The lowest point of the graph is the vertex's y-coordinate, $-1$. Thus, the range is: $[-1, \infty)$

Instructions for Graphing:

• Plot the vertex at $(3, -1)$.

• Use the y-intercept: Setting $x = 0$: $f(0) = 2(0 - 3)^2 - 1 = 2(9) - 1 = 18 - 1 = 17$ The y-intercept is at $(0, 17)$.

• Reflect points symmetrically across the axis of symmetry, $x = 3$.

Final Answers:

1. Axis of symmetry: $x = 3$.
2. Domain: $(-\infty, \infty)$.

Would you like a more detailed explanation or visualization? Here are five related questions to deepen your understanding:

1. What is the significance of the vertex in a parabola?
2. How does the coefficient $a$ in $f(x) = a(x-h)^2 + k$ affect the graph's shape?
3. What steps can you take to find the intercepts of a quadratic function?
4. How would the graph change if $a$ were negative?
5. Can the vertex form of a quadratic function be converted to standard form? If so, how?

Tip: Always check the sign of $a$ to determine whether a parabola opens upward or downward!