Explanation:
so, if I understand you correctly and nothing was left out :
f(t) = t² + 8t + 9
d(t) = t² + 11t + 4
just by looking at the 2 functions, it is clear that d(t) creates larger functional values for higher values of t.
because t² = t², but 11t > 8t. and the constants in the end are irrelevant for larger numbers.
so the pool will drain.
a.
w(t) = f(t) - d(t) = t² + 8t + 9 - t² - 11t - 4 = -3t + 5
b.
yes, it will completely drain, as -3t will drive it to 0, and if it was physically possible into the negative ask the way to -infinity.
c.
-3t + 5 = 0
5 = 3t
t = 5/3
it will be completely drained after
1 2/3 hours = 1 hour 40 minutes.
d.
to intersect they need to create the same result for a value of t :
t² + 8t + 9 = t² + 11t + 4
8t + 9 = 11t + 4
-3t + 5 = 0
t = 5/3 or 1 hour 40 minutes.
that means that the filling and the draining are only equal, when there is no water in the pool.
that also means that f(t) is actually calculating how much water there is in the pool, if there would not be no drain, with a starting value of 9.
e.
the domain is the interval or set of all valid input values (here for t).
negative t (hours) don't make any sense for any of the 3 functions.
but f(t) can go on theoretically forever. but usually it's limited by the size of the pool (once it reaches the top, there is no more filling). we just don't know anything about that.
so, the domain for f(t) is
0 <= t < +infinity
d(t) could go on and on as well, as long as there is water in the pool.
so, the domain for d(t) is also
0 <= t < +infinity
but w(t) based on its meaning and definition makes only sense between the start (t = 0) and until the water is fully drained by the combination of f(t) and d(t).
so, the domain of w(t) is
0 <= t <= 5/3
we could define w(t) as segmented function
w(t) = -3t + 5 0 <= t <= 5/3
= 0 t > 5/3
THEN the domain would be again
0 <= t < +infinity