In the image, you are asked to find $h(0)$, where $h(x) = -2x^2 - 2x - 2$.

To solve for $h(0)$:

$h(0) = -2(0)^2 - 2(0) - 2$ $h(0) = 0 - 0 - 2$ $h(0) = -2$

So, $h(0) = -2$.

Would you like more details or have any other questions?

Here are five related questions:

1. How can you find the roots of the quadratic function $h(x) = -2x^2 - 2x - 2$?
2. What does the graph of $h(x) = -2x^2 - 2x - 2$ look like?
3. How do you determine the vertex of a quadratic function?
4. How can you find the y-intercept of $h(x)$ directly from its equation?
5. How would you find $h(1)$ using the same method?

Tip: For any quadratic function $ax^2 + bx + c$, the y-intercept is always the constant $c$.