What exponential function is the best fit for the data in the table? a. xf(x). 1-4. 3-1. 43. b. f(x) = 4(3)x - 1 + 4. f(x) = 4(3)x - 1 - 4. c. f(x) = one fourth(3)x - 1 + 4. d. f(x) = one fourth(3)x - 1 - 4

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.

The answer is $9.90 with rounding to the nearest tenth