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NeTakaya
3 years ago
5

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
Mathematics
1 answer:
AnnyKZ [126]3 years ago
6 0
<span>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.</span>
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<u>the answer is</u>

x-\frac{5}{2}y= \frac{25}{2}



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