write first 4 terms      f(n) = 2n to the p owner of 2 + 4

Let's break down the function \( f(n) = 2n^2 + 4 \) and find its first four terms.

### Function:
\[ f(n) = 2n^2 + 4 \]

To calculate the first four terms, we substitute \( n = 1, 2, 3, 4 \) into the equation.

1. **When \( n = 1 \)**:
   \[
   f(1) = 2(1)^2 + 4 = 2(1) + 4 = 2 + 4 = 6
   \]

2. **When \( n = 2 \)**:
   \[
   f(2) = 2(2)^2 + 4 = 2(4) + 4 = 8 + 4 = 12
   \]

3. **When \( n = 3 \)**:
   \[
   f(3) = 2(3)^2 + 4 = 2(9) + 4 = 18 + 4 = 22
   \]

4. **When \( n = 4 \)**:
   \[
   f(4) = 2(4)^2 + 4 = 2(16) + 4 = 32 + 4 = 36
   \]

### Conclusion:
The first four terms of the function \( f(n) = 2n^2 + 4 \) are:
\[
6, 12, 22, 36
\]

Would you like further details on this solution or how to generalize this kind of pattern?

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Here are five follow-up questions:
1. What would the 5th term of the sequence be?
2. How does the function change if the constant is increased?
3. What happens if you replace \( n^2 \) with \( n^3 \)?
4. How would the graph of this quadratic function look?
5. How do we find the sum of the first four terms?

**Tip**: When evaluating functions, always square or cube the variable before applying multiplication to avoid errors.

write first 4 terms f(n) = 2n to the p owner of 2 + 4

Explore how to find the first four terms of the quadratic function f(n) = 2n^2 + 4. This problem introduces key concepts in algebra, specifically evaluating quadratic expressions for different values of n. Suitable for Grades 7-10.

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