We are given the function \( h(x) = 2x + 1 \) and we need to find \( h(a+1) \cdot h(a) \).

### Step 1: Calculate \( h(a+1) \)
Using the function definition \( h(x) = 2x + 1 \), substitute \( x = a + 1 \):
\[
h(a+1) = 2(a+1) + 1 = 2a + 2 + 1 = 2a + 3
\]

### Step 2: Calculate \( h(a) \)
Similarly, substitute \( x = a \) into the function:
\[
h(a) = 2a + 1
\]

### Step 3: Multiply \( h(a+1) \) and \( h(a) \)
Now multiply \( h(a+1) \) and \( h(a) \):
\[
h(a+1) \cdot h(a) = (2a + 3) \cdot (2a + 1)
\]

### Step 4: Expand the product
Using the distributive property:
\[
(2a + 3)(2a + 1) = 2a(2a + 1) + 3(2a + 1)
\]
\[
= 4a^2 + 2a + 6a + 3 = 4a^2 + 8a + 3
\]

Thus, \( h(a+1) \cdot h(a) = 4a^2 + 8a + 3 \).

Let me know if you need more details or have any questions! Here are 5 related questions for practice:

1. What is \( h(2a) \) for the same function \( h(x) = 2x + 1 \)?
2. How would you calculate \( h(a-1) \cdot h(a+2) \)?
3. What is the derivative of \( h(x) = 2x + 1 \)?
4. Can you express \( h(x) \cdot h(y) \) for two different values \( x \) and \( y \)?
5. What would \( h(h(x)) \) equal for \( h(x) = 2x + 1 \)?

**Tip**: When multiplying polynomials, always expand carefully using the distributive property (also called FOIL for binomials).

Explore how to solve for h(a+1) * h(a) using the linear function h(x) = 2x + 1. This solution walks through polynomial multiplication using the distributive property and covers algebraic concepts suitable for Grades 9-10.

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