Given the table:
x f(x)
1 -4
3 -1
4 3
For f(x) = 4(3)^(x - 1) + 4:
f(1) = 4(3)^(1 - 1) + 4 = 4(3)^0 + 4 = 4(1) + 4 = 4 + 4 = 8.
Thus, f(x) = 4(3)^(x - 1) + 4 does not satisfy the data in the table.
For f(x) = 4(3)^(x - 1) - 4:
f(1) = 4(3)^(1 - 1) - 4 = 4 - 4 = 0
Thus, f(x) = 4(3)^(x - 1) - 4 does not satisfy the data in the table.
For f(x) = 1/4 (3)^(x - 1) + 4:
f(1) = 1/4 (3)^(1 - 1) + 4 = 1/4 + 4 = 4.25
Thus, f(x) = 1/4 (3)^(x - 1) + 4 does not satisfy the data in the table.
For f(x) = 1/4 (3)^(x - 1) - 4:
f(1) = 1/4 (3)^(1 - 1) - 4 = 1/4 - 4 = -3.75 which is approximately -4.
f(3) = 1/4 (3)^(3 - 1) - 4 = 1/4 (3)^2 - 4 = 1/4 (9) - 4 = 2.25 - 4 = -1.75 which is close to -1.
f(4) = 1/4 (3)^(4 - 1) - 4 = 1/4 (3)^3 - 4 = 1/4 (27) - 4 = 6.75 - 4 = 2.75 which is close to 3.
Therefore, f(x) = 1/4 (3)^(x - 1) - 4 satisfy the data in the table.
