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Let's go through each part of the problem and provide a solution.

### Problem Restatement:
- Paul sells **22 burger combos a day** at a price of **$4 each**.
- For every **$1 price increase**, Paul sells **2 fewer combos**.

### Part a) 
We are asked to determine a function that models the revenue of Paul's burger shop, based on the number of $1 price increases, in factored form.

Let:
- $ p(x) $ = price of a burger combo (in dollars)
- $ x $ = number of $1 price increases
- Initial price of the burger combo = $4, so the price after $ x $ increases will be $ 4 + x $.

The number of combos sold decreases by 2 for every $1 price increase. Initially, Paul sells 22 combos, and after $ x $ price increases, he will sell $ 22 - 2x $ combos.

The revenue function $ R(x) $ is the product of the price and the number of combos sold:
\[
R(x) = (4 + x)(22 - 2x)
\]
This is the function in factored form.

### Part b) 
We need to determine graphically the price that maximizes Paul's revenue and provide a fully labeled sketch.

The revenue function from part (a) is:
\[
R(x) = (4 + x)(22 - 2x)
\]
We can expand this to find the vertex, which will give us the maximum revenue:
\[
R(x) = 88 - 8x + 22x - 2x^2 = -2x^2 + 14x + 88
\]
This is a quadratic function in the form $ R(x) = ax^2 + bx + c $, where $ a = -2 $, $ b = 14 $, and $ c = 88 $. The maximum of this parabola occurs at the vertex, which is given by the formula:
\[
x = -\frac{b}{2a} = -\frac{14}{2(-2)} = 3.5
\]
Thus, the number of $1 price increases that maximizes the revenue is $ x = 3.5 $. The corresponding price is:
\[
p = 4 + 3.5 = 7.5 \text{ dollars}
\]

You can sketch the parabola of the function $ R(x) = -2x^2 + 14x + 88 $, with the vertex at $ x = 3.5 $, and label the axes as "Price increases" on the x-axis and "Revenue" on the y-axis.

### Part c) 
The maximum revenue occurs when $ x = 3.5 $. To find the maximum revenue, substitute $ x = 3.5 $ into the revenue function:
\[
R(3.5) = (4 + 3.5)(22 - 2(3.5)) = (7.5)(15) = 112.5 \text{ dollars}
\]

Thus, the maximum revenue Paul can achieve is **$112.50**.

### Part d) 
We need to state an applicable domain and range for this situation.

#### Domain:
The domain is based on the number of price increases Paul can realistically implement. Since he starts with 22 combos and sells 2 fewer for each $1 increase, the maximum $ x $ occurs when he sells 0 combos:
\[
22 - 2x = 0 \implies x = 11
\]
Thus, $ x $ can range from 0 to 11 (since $ x $ represents the number of price increases).

So the domain is:
\[
x \in [0, 11]
\]

#### Range:
The range corresponds to the revenue values. The minimum revenue occurs when $ x = 11 $, which gives:
\[
R(11) = (4 + 11)(22 - 2(11)) = 15 \times 0 = 0
\]
The maximum revenue occurs at $ x = 3.5 $, which we already found to be $112.50.

So the range is:
\[
R(x) \in [0, 112.5]
\]

---

### Do you need more details or have any questions?

Here are five related questions you might find useful:
1. How would the revenue change if Paul sold 3 fewer combos for every $1 price increase?
2. Can we use calculus to verify the maximum point of the revenue function?
3. What happens if the initial price is increased to $5 instead of $4?
4. How would the domain change if Paul can only sell combos for a certain number of days?
5. What is the significance of the vertex form of a quadratic function in revenue models?

### Tip:
When maximizing or minimizing a quadratic function, the vertex formula $ x = -\frac{b}{2a} $ is extremely useful for quickly finding the point of maximum or minimum.

At Paul’s burger shop, he can sell 22 burger combos a day at $4 each. Research has shown that for every $1 price increase, he sells 2 fewer burger combos. Determine a function that models the revenue of the burger shop, find the price that maximizes revenue, state the maximum revenue, and identify the applicable domain and range.

This problem involves modeling the revenue of a burger shop using a quadratic function. Explore how price increases affect sales, determine the maximum revenue, and understand the domain and range of the situation. Suitable for students in Grades 9-11.

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