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Which set of ordered pairs could be generated by an exponential function?

A) (1, 0), (0, 0), (1, -2), (2, 1)
B) (-1, -1), (0, 0), (1, 1), (2, 8)
C) (-1, 2, 10), (0, 1), (1, 2), (2, 4)
D) (-1, 1), (0, 0), (1, 1), (2, 4)

User Harriv
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1 Answer

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Final answer:

The correct set of ordered pairs that could be generated by an exponential function is D) (-1, 1), (0, 0), (1, 1), (2, 4), which can be represented by the function f(x) = 2^x.

Step-by-step explanation:

The question asks which set of ordered pairs could be generated by an exponential function. The key characteristic of an exponential function is that it involves a constant base raised to a power that includes the variable; the function often takes the form f(x) = ab^x where a is the initial value, b is the base, and x is the exponent.

Looking at the options provided:

  • Option A: This set cannot represent an exponential function as it includes (0, 0), which should not occur if the function is of the form f(x) = ab^x since f(0) would equal a, and a should not equal zero in this case.
  • Option B: This set could represent an exponential function if there were a base that is raised to the powers of -1, 0, 1, and 2 to get the respective y-values of -1, 0, 1, and 8. However, this is not possible as the y-values should be positive for all x if a is positive, and should not cross the x-axis (0,0).
  • Option C: This set contains an error, as the first pair should have two elements, not three.
  • Option D: This set can represent an exponential function where a = 1 and b = 2, which would result in the function f(x) = 2^x. The pairs (-1, 1), (0, 1), and (1, 1) correspond to 2^-1, 2^0, and 2^1, respectively. Then (2, 4) corresponds to 2^2. All these satisfy the criteria for an exponential function.

Therefore, the correct answer is D) (-1, 1), (0, 0), (1, 1), (2, 4).

User Dutoitns
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