Question

The function A = s^2 giving the area of a square as a function of its side length

Linear or Nonlinear:


Domain:


Range:


Increasing or Decreasing:


End behavior:

Answer 1

Let's analyze the function \( A = s^2 \) where \( A \) is the area of the square and \( s \) is the side length.

### Linear or Nonlinear:
- **Nonlinear**: The function \( A = s^2 \) is a quadratic function, which is nonlinear because it does not form a straight line when graphed.

### Domain:
- **Domain**: Since \( s \) represents the side length of a square, it must be non-negative. Therefore, the domain is \( [0, \infty) \).

### Range:
- **Range**: The area \( A \) is also non-negative, as the area cannot be negative. Thus, the range is \( [0, \infty) \).

### Increasing or Decreasing:
- **Increasing**: The function \( A = s^2 \) is increasing for \( s \geq 0 \) because as \( s \) increases, \( A \) increases. The derivative \( \frac{dA}{ds} = 2s \) is positive for \( s > 0 \).

### End behavior:
- **End behavior**: As \( s \) approaches infinity (\( s \to \infty \)), \( A \) also approaches infinity (\( A \to \infty \)). As \( s \) approaches 0 (\( s \to 0 \)), \( A \) approaches 0 (\( A \to 0 \)). Therefore, the end behavior can be summarized as:
- As \( s \to 0 \): \( A \to 0 \)
- As \( s \to \infty \): \( A \to \infty \)

In summary:
- Linear or Nonlinear: **Nonlinear**
- Domain: **[0, ∞)**
- Range: **[0, ∞)**
- Increasing or Decreasing: **Increasing**
- End behavior:
- \( A \to 0 \) as \( s \to 0 \)
- \( A \to \infty \) as \( s \to \infty \)