To determine the end behavior of the exponential function [tex]\( f(x) = 2^{x-3} \)[/tex], we need to analyze how the function behaves as [tex]\( x \)[/tex] approaches very high values (i.e., [tex]\( x \to \infty \)[/tex]) and very low values (i.e., [tex]\( x \to -\infty \)[/tex]).
### 1. End Behavior for Very High [tex]\( x \)[/tex]-Values
When [tex]\( x \)[/tex] becomes very large (i.e., [tex]\( x \to \infty \)[/tex]):
- The expression [tex]\( x - 3 \)[/tex] will also become very large because subtracting 3 from a very large number still leaves a very large number.
- Therefore, [tex]\( 2^{x-3} \)[/tex] will be [tex]\( 2 \)[/tex] raised to a very large power.
Since [tex]\( 2 \)[/tex] is a number greater than 1, raising it to an increasingly large power results in an exponentially increasing value. Hence,
[tex]\[ \lim_{{x \to \infty}} 2^{x-3} = \infty. \][/tex]
Thus, for very high values of [tex]\( x \)[/tex], [tex]\( f(x) \)[/tex] moves toward positive infinity.
### 2. End Behavior for Very Low [tex]\( x \)[/tex]-Values
When [tex]\( x \)[/tex] becomes very small (i.e., [tex]\( x \to -\infty \)[/tex]):
- The expression [tex]\( x - 3 \)[/tex] will also become very small (i.e., very large negative).
- Therefore, [tex]\( 2^{x-3} \)[/tex] will be [tex]\( 2 \)[/tex] raised to a very large negative power.
Raising 2 to a large negative power results in a very small positive number (closer to zero but never negative or zero itself). Hence,
[tex]\[ \lim_{{x \to -\infty}} 2^{x-3} = 0. \][/tex]
Thus, for very low values of [tex]\( x \)[/tex], [tex]\( f(x) \)[/tex] moves toward the horizontal asymptote, which in this case is the x-axis or [tex]\( y = 0 \)[/tex].
### Conclusion
Given the above analysis, the correct statement describing the end behavior of the exponential function [tex]\( f(x) = 2^{x-3} \)[/tex] for very high [tex]\( x \)[/tex]-values is:
A. For very high [tex]\( x \)[/tex]-values, [tex]\( f(x) \)[/tex] moves toward positive infinity.
