To find $h(0)$ for the function $h(x) = -3x^2 - 11x + 4$:

1. Substitute $x = 0$ into the function: $h(0) = -3(0)^2 - 11(0) + 4$

2. Simplify each term: $h(0) = 0 + 0 + 4 = 4$

So, $h(0) = 4$.

Would you like further details, or do you have any questions?

1. What would $h(x)$ be for other values of $x$, such as $h(1)$ or $h(-1)$?
2. How do we find the vertex of the quadratic function $h(x) = -3x^2 - 11x + 4$?
3. What is the axis of symmetry for this function?
4. How do we determine if this parabola opens upwards or downwards?
5. How does changing the coefficient of $x^2$ affect the graph of a quadratic function?

Tip: Substituting $x = 0$ into any function gives the y-intercept, showing where the function crosses the y-axis.