https://file.aiquickdraw.com/mathbot/file/1726597776677nxcmqcnd.jpg

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 \( c \).

Find h(0) for the function h(x) = -2x^2 - 2x - 2.

Learn how to find h(0) for the quadratic function h(x) = -2x^2 - 2x - 2. This step-by-step solution involves using substitution in the quadratic equation. Suitable for high school students learning algebra.

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