5. explain why the function a(x) = 36 - x² represents the shaded area in the figure above.
6. draw a rough sketch of the graph.
7. what does the x-axis represent?
8. what does the y-axis represent?
9. if the square has sides of 3 ft, what is the area of the shaded region?
10. if the area of the shaded region is 10, what are the lengths of the sides of the square?
11. what are the x-intercepts?
12. what do the x-intercepts represent in this problem?
13. are the x-intercepts possible solutions for this problem? why/why not?
14. fill in the table. describe what happens to a(x) as x becomes infinitely small.
table with x values: 2, 1, 0.5, 0.25, 0.1 and a(x) column

Question

Accepted Answer

### Question 5: Explain why the function $ A(x) = 36 - x^2 $ represents the shaded area in the figure.
# Explanation:
## Step 1: Analyze the figure (rectangle with a square cut out)
The large rectangle has dimensions $ 6 \, \text{ft} \times 6 \, \text{ft} $ (since $ 6 \times 6 = 36 $, which is the constant term in $ A(x) $). The area of the large rectangle is $ 6 \times 6 = 36 \, \text{sq ft} $.

## Step 2: Analyze the square cut out
The square has a side length of $ x \, \text{ft} $, so its area is $ x \times x = x^2 \, \text{sq ft} $.

## Step 3: Determine the shaded area
The shaded area is the area of the large rectangle minus the area of the square cut out. So, Shaded Area $ = $ Area of Rectangle $ - $ Area of Square $ = 36 - x^2 $, which matches the function $ A(x) = 36 - x^2 $.
# Answer:
The large rectangle has an area of $ 6 \times 6 = 36 $ square feet. The square cut out has an area of $ x^2 $ square feet (since its side length is $ x $). The shaded area is the area of the rectangle minus the area of the square, so $ A(x) = 36 - x^2 $.

### Question 9: If the square has sides of $ 3 \, \text{ft} $, what is the area of the shaded region?
# Explanation:
## Step 1: Identify the function for shaded area
From question 5, the shaded area function is $ A(x) = 36 - x^2 $, where $ x $ is the side length of the square.

## Step 2: Substitute $ x = 3 $ into the function
Substitute $ x = 3 $ into $ A(x) $:
$ A(3) = 36 - (3)^2 $

## Step 3: Calculate the value
First, calculate $ (3)^2 = 9 $. Then, $ 36 - 9 = 27 $.
# Answer:
The area of the shaded region is $ 27 $ square feet.

### Question 10: If the area of the shaded region is $ 10 $, what are the lengths of the sides of the square?
# Explanation:
## Step 1: Set up the equation
Using the shaded area function $ A(x) = 36 - x^2 $, set $ A(x) = 10 $:
$ 36 - x^2 = 10 $

## Step 2: Solve for $ x^2 $
Subtract $ 36 $ from both sides:
$ -x^2 = 10 - 36 $
$ -x^2 = -26 $

Multiply both sides by $ -1 $:
$ x^2 = 26 $

## Step 3: Solve for $ x $
Take the square root of both sides (since side length is positive, we use the positive root):
$ x = \sqrt{26} \approx 5.1 \, \text{ft} $ (or leave it as $ \sqrt{26} $ for exact form).
# Answer:
The side length of the square is $ \sqrt{26} \approx 5.1 \, \text{ft} $ (or $ \sqrt{26} $ feet).

### Question 14: Fill in the table. Describe what happens to $ A(x) $ as $ x $ becomes infinitely small.

#### Filling the table:
For each $ x $, calculate $ A(x) = 36 - x^2 $:

- When $ x = 2 $:
$ A(2) = 36 - (2)^2 = 36 - 4 = 32 $

- When $ x = 1 $:
$ A(1) = 36 - (1)^2 = 36 - 1 = 35 $

- When $ x = 0.5 $:
$ A(0.5) = 36 - (0.5)^2 = 36 - 0.25 = 35.75 $ (Wait, the table has $ 0.25 $ for $ A(x) $ when $ x = 0.5 $? Wait, no—wait, maybe I misread. Wait, the table has $ x $ values: $ 2, 1, 0.5, 0.25, 0.1 $, and $ A(x) $ column. Wait, original table (from the image) shows:

$ x $: $ 2, 1, 0.5, 0.25, 0.1 $
$ A(x) $: (blank), (blank), (blank), (blank), (blank) — but the user’s image might have a typo? Wait, no, maybe I misread. Wait, let's recheck:

Wait, the table in the image:

$ x $: $ 2, 1, 0.5, 0.25, 0.1 $
$ A(x) $: (empty cells)

So we calculate $ A(x) = 36 - x^2 $ for each $ x $:

- $ x = 2 $: $ A(2) = 36 - 2^2 = 36 - 4 = 32 $
- $ x = 1 $: $ A(1) = 36 - 1^2 = 36 - 1 = 35 $
- $ x = 0.5 $: $ A(0.5) = 36 - (0.5)^2 = 36 - 0.25 = 35.75 $
- $ x = 0.25 $: $ A(0.25) = 36 - (0.25)^2 = 36 - 0.0625 = 35.9375 $
- $ x = 0.1 $: $ A(0.1) = 36 - (0.1)^2 = 36 - 0.01 = 35.99 $

#### Describing what happens to $ A(x) $ as $ x $ becomes infinitely small:
As $ x $ approaches $ 0 $ (infinitely small), $ x^2 $ approaches $ 0 $. So $ A(x) = 36 - x^2 $ approaches $ 36 - 0 = 36 $. Thus, $ A(x) $ approaches the area of the large rectangle (since the square’s area becomes negligible).

# Explanation (for filling the table):
## Step 1: For $ x = 2 $
Calculate $ A(2) = 36 - 2^2 = 36 - 4 = 32 $.

## Step 2: For $ x = 1 $
Calculate $ A(1) = 36 - 1^2 = 36 - 1 = 35 $.

## Step 3: For $ x = 0.5 $
Calculate $ A(0.5) = 36 - (0.5)^2 = 36 - 0.25 = 35.75 $.

## Step 4: For $ x = 0.25 $
Calculate $ A(0.25) = 36 - (0.25)^2 = 36 - 0.0625 = 35.9375 $.

## Step 5: For $ x = 0.1 $
Calculate $ A(0.1) = 36 - (0.1)^2 = 36 - 0.01 = 35.99 $.

# Explanation (for the behavior of $ A(x) $):
As $ x $ becomes infinitely small (approaches $ 0 $), $ x^2 $ approaches $ 0 $. Thus, $ A(x) = 36 - x^2 $ approaches $ 36 $ (the area of the large rectangle, since the square’s area becomes negligible).

# Answer (table):
| $ x $ | $ A(x) $ |
|--------|-----------|
| 2      | 32        |
| 1      | 35        |
| 0.5    | 35.75     |
| 0.25   | 35.9375   |
| 0.1    | 35.99     |

# Answer (behavior):
As $ x $ becomes infinitely small, $ A(x) $ approaches $ 36 $ (the area of the large rectangle, since the square’s area becomes negligible).

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QUESTION IMAGE

Question

Explanation:

Question 5: Explain why the function \( A(x) = 36 - x^2 \) represents the shaded area in the figure.

Step 1: Analyze the figure (rectangle with a square cut out)

Step 2: Analyze the square cut out

Step 3: Determine the shaded area

Step 1: Identify the function for shaded area

Step 2: Substitute \( x = 3 \) into the function

Step 3: Calculate the value

Step 1: Set up the equation

Step 2: Solve for \( x^2 \)

Step 3: Solve for \( x \)

Answer:

Question 9: If the square has sides of \( 3 \, \text{ft} \), what is the area of the shaded region?