The given function is $f(x) = 2(x - 6)^2 - 7$.

This is a quadratic function in vertex form, which is given by the equation:

$f(x) = a(x - h)^2 + k$

In this form, the vertex is $(h, k)$, where:

• $h$ is the x-coordinate of the vertex.
• $k$ is the y-coordinate of the vertex.

By comparing $f(x) = 2(x - 6)^2 - 7$ with the vertex form, we can see that:

• $h = 6$
• $k = -7$

Thus, the vertex of the function is $(6, -7)$.

Would you like further details or have any other questions?

Here are 5 related questions:

1. How do you find the vertex of a quadratic function in standard form?
2. What is the axis of symmetry for this function?
3. How does the value of $a$ (in this case, 2) affect the shape of the parabola?
4. What are the x- and y-intercepts of this quadratic function?
5. How would you graph this quadratic function step-by-step?

Tip: The sign of $a$ in the vertex form determines if the parabola opens upward (positive $a$) or downward (negative $a$).