The question asks for the coordinates of the x-intercepts for the function $g(x) = -x^2 + 4$.

To find the x-intercepts, we need to set $g(x) = 0$ because x-intercepts occur when the output of the function (y-value) is zero.

Step 1: Set the equation to 0

$0 = -x^2 + 4$

Step 2: Solve for $x$

$x^2 = 4$

Now, take the square root of both sides: $x = \pm 2$

Step 3: Find the coordinates

Thus, the x-intercepts are at $x = -2$ and $x = 2$. The corresponding coordinates are $(-2, 0)$ and $(2, 0)$.

Conclusion:

The correct answer is Option 2, which gives the x-intercepts $(-2, 0)$ and $(2, 0)$.

Do you want more details or have any questions?

Here are 5 related questions to expand on this:

1. What are the steps to find the y-intercept of a quadratic function?
2. How do the signs of the coefficients in a quadratic function affect its graph?
3. How would you determine the vertex of $g(x) = -x^2 + 4$?
4. What is the axis of symmetry for the function $g(x) = -x^2 + 4$?
5. How does the graph of $g(x) = -x^2 + 4$ differ from $g(x) = x^2 - 4$?

Tip: When finding x-intercepts of quadratic functions, always set $g(x) = 0$ and solve for $x$.