The quadratic function provided is:

$f(x) = 3(x - 1)^2 - 4$

Let's analyze the key features of this function:

1. Axis of Symmetry

The given quadratic is in vertex form:

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

where $h$ represents the x-coordinate of the vertex and the axis of symmetry is $x = h$.

• For $f(x) = 3(x - 1)^2 - 4$, the vertex is at $(1, -4)$, so the axis of symmetry is:

$x = 1$

2. Domain

The domain of any quadratic function is all real numbers, as the function is defined for all values of $x$.

• So, the domain is:

$(-\infty, \infty)$

3. Range

The quadratic has a positive coefficient for the squared term (i.e., $a = 3$), meaning the parabola opens upwards. The vertex represents the lowest point of the parabola.

• The minimum value of $f(x)$ occurs at $x = 1$, where $f(1) = -4$.
• Since the parabola opens upwards, the range starts at $-4$ and goes to $\infty$.

Thus, the range is:

$[-4, \infty)$

Summary:

• The axis of symmetry is $x = 1$.
• The domain is $(-\infty, \infty)$.
• The range is $[-4, \infty)$.

Would you like further details or explanations?

Here are 5 related questions for further exploration:

1. How do you determine the vertex of a quadratic function in standard form?
2. What is the effect of changing the value of $a$ in a quadratic function?
3. How would the graph change if the function had a negative leading coefficient?
4. Can a quadratic function have a finite domain?
5. How does the range of a quadratic function depend on its vertex?

Tip: Always look at the sign of the leading coefficient to determine whether the parabola opens upwards or downwards!