An professor finds that the number of students enrolling 
in his class is approximated by s(t) = -t² + 20t + 80 
where t is the number of tests given in one semester. 

Here is the graph. t, the number of tests is on the 
x-axis and the number of students is on the y-axis.




(a) How many tests should the professor give so that the 
number of students enrolling in the class is a maximum? 

That will be the highest point, the vertex, of the graph.

The vertex is the point (h,k) where h = ,
and k = s(h)

s(t) = -t² + 20t + 80

a = -1 (the coefficient of t²
b = 20 (the coefficient of t)
c = 80 (the constant)

h =  =  = 10

k = s(h) = -10² + 20(10) + 80 = -100 + 200 + 80 = 180



The vertex (10,180) at the top of the graph is 
interpreted as "when the professor gives 10 tests,
there will be 180 students.

(b) If the professor does not want students to enroll, 
how many tests should he give?

We substitute 0 students for s(t) and solve for t:

            s(t) = -t² + 20t + 80 
               0 = -t² + 20t + 80
   t² - 20t - 80 = 0 
  
That doesn't factor so we have to use the quadratic
formula:

                  ______ 
            -b ± Öb²-4ac
        t = —————————————
                2a 

where a = 1; b = -20; c = -80

                       ________________
             -(-20) ± Ö(-20)²-4(1)(-80)
        x = —————————————————————————————
                       2(1) 
                   _______ 
             20 ± Ö400+320
        x = ————————————————
                   2

                   ___ 
             20 ± Ö720
        x = ————————————
                 2 

                   _____ 
             20 ± Ö144·5
        x = —————————————
                  2 

                     _ 
             20 ± 12Ö5
        x = ———————————
                 2 

                       _
             20     12Ö7
        x = ———— ± ——————
              2       2
                   _
        x = 10 ± 6Ö7 
                        _
Using the +, x = 10 + 6Ö5, which
is one solution and equals about 23.41640786
                        _ 
Using the -, x = 10 - 6Ö5, which
is the other solution and equals about -3.416407865. This is
meaningless for this problem, since it's negative, so we
disregard it.

Therefore the point where the graph crosses the horizontal
axis is (23.4, 0) which means that if the professor gives 
more than 23 tests, he will have no students.

Edwin