D) f(n) = 3 + 4(n-1)
3 = 1st term
4 = common difference among the terms
n = term number you are looking for.
To check: 3, 7, 11, 15, ...
f(1) = 3 + 4(1-1) = 3 + 4(0) = 3 + 0 = 3
f(2) = 3 + 4(2-1) = 3 + 4(1) = 3 + 4 = 7
f(3) = 3 + 4(3-1) = 3 + 4(2) = 3 + 8 = 11
f(4) = 3 + 4(4-1) = 3 + 4(3) = 3 + 12 = 15
Answer:
no this is not a quadratic function
You only put one equation in
