To determine the function that represents the frog population after [tex]\( x \)[/tex] years, we need to consider the following steps:
1. Initial Population:
- The initial population of the frogs is given as 1,200.
2. Rate of Decrease:
- The population decreases at a constant rate of [tex]\( 3\% \)[/tex] per year.
- To express a [tex]\( 3\% \)[/tex] decrease as a multiplier, we subtract [tex]\( 0.03 \)[/tex] from [tex]\( 1 \)[/tex]. Hence, the new multiplier is [tex]\( 1 - 0.03 = 0.97 \)[/tex].
3. Exponential Decay Function:
- Generally, if a population or quantity decreases by a certain percentage each year, the population after [tex]\( x \)[/tex] years can be represented by an exponential decay function. The general form of such a function for the frog population can be written as:
[tex]\[
f(x) = \text{initial population} \times (\text{decay factor})^x
\][/tex]
- In this case, the correct decay factor is [tex]\( 0.97 \)[/tex] (representing a 3\% decrease).
4. Putting it All Together:
- Substituting the initial population (1,200 frogs) and the decay factor (0.97) into the exponential decay function, we get:
[tex]\[
f(x) = 1,200 \times (0.97)^x
\][/tex]
Thus, the function that correctly represents the frog population after [tex]\( x \)[/tex] years is:
[tex]\[
f(x) = 1,200(0.97)^x
\][/tex]
Therefore, among the given choices:
- [tex]\( f(x)=1,200(1.03)^x \)[/tex] represents a 3% growth, not decay.
- [tex]\( f(x)=1,200(0.03)^x \)[/tex] incorrectly represents the multiplier and would imply a much larger percentage decrease.
- [tex]\( f(x)=1.200(0.97 x) \)[/tex] incorrectly applies a linear decrease rather than exponential decay.
The correct answer is:
[tex]\[
f(x) = 1,200(0.97)^x
\][/tex]
