Since any t-value is reasonable, we can draw a straight line through the points. Therefore, the graph is continuous. Also, the graph is restricted to Quadrant I because the water level also cannot be negative.
Alternative Solution
Use the Maximum and Minimum Points
The water level can only decrease as the pump empties the room and can never become a negative number. Consequently, only l values between 0 and 37 are reasonable.
0≤ l ≤ 37
Now let's use the maximum and minimum values of the level of water to determine their corresponding times.
l = 37-1.5t
Operation
Minimum t-Value
Maximum t-Value
Substitution
37= 37-1.5t
0= 37-1.5t
Calculation
t= 0
t= 24 23
Point
( 0, 37)
( 24 23, 0)
Based on our table, we see that the maximum time of the emptying the basement is 24 23hours. The pump time can only increase to this value and can never become negative, so it only makes sense to include values from 0 to 24 23 in our graph.
0≤ t ≤ 24 23
Let's plot our points and connect them with a line.
Since the pump runs continuously between 0 and 24 23 hours, water level between 0 and 37 is also applicable in this situation. Therefore, the function is continuous.
