The given quadratic function is:

$y = -2(x - 1)^2 + 3$

This is in vertex form $y = a(x - h)^2 + k$, where the vertex is $(h, k)$.

• The coefficient $a = -2$, which indicates that the parabola opens downwards (since $a < 0$).
• The vertex $(h, k)$ is $(1, 3)$.

Since the parabola opens downwards, the vertex represents the maximum point of the function.

Thus, the maximum value of the function is the $y$-coordinate of the vertex, which is $3$.

Therefore, the correct answer is 3.

Would you like any additional details or have any questions?

Here are 5 related questions:

1. How do you determine if a quadratic function has a maximum or minimum value?
2. What is the axis of symmetry for the function $y = -2(x - 1)^2 + 3$?
3. How do you find the range of a quadratic function that opens downward?
4. What is the effect of changing the $k$ value in a quadratic function in vertex form?
5. How do you determine the x-intercepts of a downward-opening quadratic function?

Tip: For a quadratic function $y = a(x - h)^2 + k$, if $a < 0$, the vertex gives the maximum value. If $a > 0$, the vertex gives the minimum value.