+845 
Find the values of t for which the height of the water is decreasing.
I tried typing in the graph on desmos expecting a cubic graph but it just gave me a straight line
+775 That is where the graph caught you off guard! The graph is actually not a straight line. Zoom out, and you will see that the graph curves like an "N" at around y = 58!
- PM
+775 So there you have it... the height of the water decreases around \(t = 2.25\).
Hope this helps,
- PM
+129852 Take the derivative
h ' = 12t^2 - 59t + 72
Set this to 0 and set to 0
12t^2 - 59t + 72 = 0
(4t -9) (3t - 8) = 0
Setting both factors to 0 and solving for x gives
t = 9/4 t = 8/3
Take the second derivative
24t - 59
Plugging 9/4 into this gives - 4....so....we have a max at t = 9/4 sec = 2,25 sec
Plugging 8/3 into this gives 5....so....we have a min at t = 8/3 sec = 2.66 sec
So...the water height is decreasing from 2.25 sec to 2.66 sec
Heres the graph : https://www.desmos.com/calculator/ne2dnm3tks
YEEEEEET.......This does look like a straight line if you don't "zoom in"....very subtle....!!!!
