The given function is:

$f(x) = 3(x - 4)^2$

This is a quadratic function, and it is in vertex form, which can be written as:

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

where:

• $a = 3$
• $h = 4$
• $k = 0$

From this, we can infer that:

• The vertex of the parabola is at $(h, k) = (4, 0)$.
• The parabola opens upwards because $a = 3 > 0$, and the value of $a$ also indicates that the parabola is narrower than the standard parabola $y = x^2$.

Key features:

1. Vertex: $(4, 0)$
2. Axis of symmetry: $x = 4$
3. Direction: Opens upwards
4. Stretch factor: The factor of 3 makes the parabola narrower than the standard parabola $y = x^2$.

Would you like to see details on how to graph it or explore its properties further?

Related Questions:

1. How do we find the x- and y-intercepts for this function?
2. How can we rewrite the function in standard form?
3. What is the domain and range of this quadratic function?
4. How does changing the value of $a$ affect the graph of the function?
5. How do we solve $f(x) = 0$?

Tip: The vertex form of a quadratic function is helpful for quickly identifying the vertex and the direction of the parabola.