Certainly! Let's examine the depreciation model for Sharlene's truck based on the given exponential function:
[tex]\[ y = 44,445 \cdot (0.78)^x \][/tex]
Here, [tex]\( y \)[/tex] represents the value of the truck after [tex]\( x \)[/tex] months.
### Step-by-Step Explanation:
1. Identifying the Components of the Function:
- The initial value of the truck when [tex]\( x = 0 \)[/tex] is represented by the coefficient [tex]\( 44,445 \)[/tex].
- The base of the exponential function is [tex]\( 0.78 \)[/tex], which is less than 1, indicating depreciation (a decrease in value) over time.
2. Behavior of the Function:
- As [tex]\( x \)[/tex] increases, the exponent [tex]\( x \)[/tex] in [tex]\( (0.78)^x \)[/tex] makes [tex]\( (0.78)^x \)[/tex] decrease more because [tex]\( 0.78 \)[/tex] being raised to higher powers gets smaller.
- When [tex]\( x = 0 \)[/tex], the value is [tex]\( 44,445 \cdot (0.78)^0 = 44,445 \cdot 1 = 44,445 \)[/tex].
- As [tex]\( x \)[/tex] approaches infinity, [tex]\( (0.78)^x \)[/tex] approaches 0, meaning the truck's value [tex]\( y \)[/tex] asymptotically approaches zero but never actually reaches zero.
3. Range Determination:
- The highest value [tex]\( y \)[/tex] can achieve is [tex]\( 44,445 \)[/tex] when [tex]\( x = 0 \)[/tex].
- The lowest value [tex]\( y \)[/tex] can theoretically approach is 0 as [tex]\( x \)[/tex] grows larger, but [tex]\( y \)[/tex] will never actually become 0 due to the nature of exponential decay.
4. Considering the Context:
- Since the truck cannot have a negative value, and it cannot be zero exactly given the nature of the function and the context of the problem, we discard the possibility of [tex]\( y \leq 0 \)[/tex].
### Conclusion:
Given the discussion above, the range of the function [tex]\( y = 44,445 \cdot (0.78)^x \)[/tex] based on its equation and the context is:
[tex]\[ 0 < y \leq 44,445 \][/tex]
Therefore, the correct answer is:
[tex]\[ 0 < y \leq 44,445 \][/tex]
