The function $f(x) = -3(x - 2)^2 - 4$ is a quadratic function in vertex form, $f(x) = a(x - h)^2 + k$, where the vertex is given by $(h, k)$ and the value of $a$ determines whether the parabola opens upwards or downwards.

For the given function:

• The vertex is $(2, -4)$.
• Since $a = -3$, the parabola opens downwards.

Thus, the graph of the function has a maximum value of $-4$ at $x = 2$.

The completed statement is:

"The graph of the function $f(x) = -3(x - 2)^2 - 4$ has a maximum value of -4."

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Would you like more details or have any questions?

Here are some related questions:

1. What is the axis of symmetry for the quadratic function?
2. How do you determine if a parabola opens upwards or downwards?
3. How does the value of $a$ in the vertex form affect the shape of the parabola?
4. What is the domain and range of the function $f(x) = -3(x - 2)^2 - 4$?
5. How would the graph change if the function were $f(x) = 3(x - 2)^2 - 4$?

Tip: The vertex form of a quadratic function is a quick way to identify the vertex and the direction in which the parabola opens.