Let's analyze the situation step by step.
1. Initial Condition:
- The initial amount of water in the bathtub is 40 gallons.
2. Draining Rate:
- Water is draining at a rate of 1.5 gallons per minute.
3. Function of Water Remaining:
- As the water drains, the amount of water remaining in the bathtub can be calculated. For every minute [tex]\( x \)[/tex], the water decreases by [tex]\( 1.5 \times x \)[/tex].
- Therefore, the amount of water [tex]\( y \)[/tex] remaining in the bathtub after [tex]\( x \)[/tex] minutes can be expressed as:
[tex]\[
y = 40 - 1.5x
\][/tex]
4. Range of Values for [tex]\( y \)[/tex]:
- To find the range of [tex]\( y \)[/tex], let's determine the possible values that [tex]\( y \)[/tex] can take.
- Initially, when [tex]\( x = 0 \)[/tex]:
[tex]\[
y = 40 - 1.5 \times 0 = 40 \text{ gallons}
\][/tex]
- As time progresses, the water keeps draining until the bathtub is empty. The bathtub is empty when all 40 gallons have been drained:
[tex]\[
40 - 1.5x = 0
\][/tex]
- Solving for [tex]\( x \)[/tex]:
[tex]\[
1.5x = 40 \implies x = \frac{40}{1.5} = 26.\overline{6} \text{ minutes}
\][/tex]
- After [tex]\( 26.\overline{6} \)[/tex] minutes, the water amount [tex]\( y \)[/tex] reaches 0:
[tex]\[
y = 0 \text{ gallons}
\][/tex]
5. Identifying the Range:
- The value of [tex]\( y \)[/tex] starts from 40 gallons and decreases to 0 gallons. Therefore, the possible values for [tex]\( y \)[/tex] lie between 0 and 40 gallons inclusive.
Thus, the range of this function is:
[tex]\[
\boxed{\text{all real numbers such that } 0 \leq y \leq 40}
\][/tex]
