To evaluate $h(n+1)$ for the given function $h(x) = 3x^2 + 2x - 1$, we substitute $n+1$ in place of $x$ in the function.

Step-by-step substitution:

$h(n+1) = 3(n+1)^2 + 2(n+1) - 1$

Now, let's expand and simplify the expression:

1. Expand $(n+1)^2$: $(n+1)^2 = n^2 + 2n + 1$

2. Substitute into the equation: $h(n+1) = 3(n^2 + 2n + 1) + 2(n+1) - 1$

3. Distribute the constants: $= 3n^2 + 6n + 3 + 2n + 2 - 1$

4. Combine like terms: $h(n+1) = 3n^2 + 8n + 4$

Thus, the expression for $h(n+1)$ is:

$h(n+1) = 3n^2 + 8n + 4$

Would you like further details on any part of the process?

Related questions:

1. How do you find the derivative of $h(x) = 3x^2 + 2x - 1$?
2. What happens when $h(x)$ is evaluated at a negative number?
3. Can we solve for $n$ when $h(n+1) = 0$?
4. How do quadratic functions like $h(x)$ behave graphically?
5. How does the general form $ax^2 + bx + c$ apply to this function?

Tip: When substituting variables into a function, always simplify terms carefully to avoid small errors.