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.

Learn how to substitute values into a quadratic function with the example h(x) = 3x^2 + 2x - 1. This step-by-step solution simplifies the expression for h(n+1), expanding and combining like terms. Suitable for Grades 8-10 students studying quadratic functions and algebra.

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