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We are given the functions \( g(n) = n^2 + 4 + 2n \) and \( h(n) = -3n + 2 \), and we need to find \( (g \cdot h)(1) \), which means the value of the product of \( g(n) \) and \( h(n) \) when \( n = 1 \).

### Step-by-step solution:
1. First, evaluate \( g(1) \):
   \[
   g(1) = (1)^2 + 4 + 2(1) = 1 + 4 + 2 = 7
   \]

2. Next, evaluate \( h(1) \):
   \[
   h(1) = -3(1) + 2 = -3 + 2 = -1
   \]

3. Now, find the product \( (g \cdot h)(1) \):
   \[
   (g \cdot h)(1) = g(1) \cdot h(1) = 7 \cdot (-1) = -7
   \]

So, \( (g \cdot h)(1) = -7 \).

Would you like more details or have any further questions?

Here are some related questions to explore:
1. How do you find the product of two functions in general?
2. What is the domain of the product of two functions?
3. Can the product of two functions change the shape of their individual graphs?
4. How do you find the derivative of the product of two functions?
5. What happens when you divide two functions instead of multiplying them?

**Tip:** Always check whether the operations on the functions (addition, multiplication, etc.) can change their domain or range.

Find (g ⋅ h)(1) for the functions g(n) = n^2 + 4 + 2n and h(n) = -3n + 2.

Learn how to find the value of the product of two functions g(n) = n^2 + 4 + 2n and h(n) = -3n + 2 when n equals 1. This problem involves key algebraic concepts such as function evaluation and multiplication. Suitable for students in Grades 9-12.

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