In a quadratic sequence we'll get a linear first difference and a constant second difference.  Let's verify that.
 n          1   2   3   4
f(n)       19  15  9   1
 1st diff   -4  -6  -8
2nd diff     2   2
We see that we got a constant second difference.  We could just extend that and work back up to get more values.
 n          1   2   3   4       5    6      7
f(n)       19  15  9   1      -9   -21   -35
 1st diff   -4  -6  -8   -10  -12   -14
2nd diff     2   2    2     2    2
That's just an aside; we're after the general formula.  We have
f(1)=19, f(2)=15, f(3)=9
In general we can assume
f(n) = an²  + bn + c
We get three equations in three unknowns,
19 = a(1²)+b(1)+c = a+b+c
15 = a(2²) + b(2) + c = 4a + 2b + c
9 = a(3²) + b(3) + c =  9a + 3b + c
That's a 3x3 linear system; it's easy to solve directly.  Subtracting pairs,
4 = -3a - b
6 = -5a - b
Subtracting those,
-2 = 2a
a = -1
b = -3a -4 = -1
c = 19-a-b = 21
Answer: f(n) = -n² - n + 21
Check:
f(1) = -1 - 1 + 21 = 19, good
f(2) = -4 - 2 + 21 = 15, good
f(3) = -9 - 3 + 21 = 9, good
f(4) = -16 - 4 + 21 = 1, good
Let's check our extended table, how about
f(7)= -49 - 7 + 21 = -35, good