To describe the end behavior of the exponential function [tex]\( f(x) = 2^{x-3} \)[/tex], let's analyze how the function behaves for both very high and very low values of [tex]\( x \)[/tex].
1. For very high [tex]\( x \)[/tex]-values:
- When [tex]\( x \)[/tex] becomes very large (i.e., [tex]\( x \rightarrow \infty \)[/tex]), the exponent [tex]\( x-3 \)[/tex] will also become very large.
- Since the base of the exponential function is 2 (which is greater than 1), raising it to a very large power will result in a very large value.
- Therefore, as [tex]\( x \)[/tex] approaches infinity, [tex]\( 2^{x-3} \)[/tex] will also grow without bound and move toward positive infinity.
So, for very high [tex]\( x \)[/tex]-values, [tex]\( f(x) \)[/tex] moves toward positive infinity.
2. For very low [tex]\( x \)[/tex]-values:
- When [tex]\( x \)[/tex] becomes very small (i.e., [tex]\( x \rightarrow -\infty \)[/tex]), the exponent [tex]\( x-3 \)[/tex] will become a very large negative number.
- Since we're raising 2 to a very large negative power, [tex]\( 2^{x-3} \)[/tex] will approach 0 (but not become negative as exponential functions do not yield negative results).
- Thus, for very low [tex]\( x \)[/tex], [tex]\( f(x) \)[/tex] approaches 0.
Given the options:
A. For very high [tex]\( x \)[/tex]-values, [tex]\( f(x) \)[/tex] moves toward positive infinity.
B. For very high [tex]\( x \)[/tex]-values, [tex]\( f(x) \)[/tex] moves toward negative infinity.
C. For very high [tex]\( x \)[/tex]-values, [tex]\( f(x) \)[/tex] moves toward the horizontal asymptote.
D. For very low [tex]\( x \)[/tex]-values, [tex]\( f(x) \)[/tex] moves toward negative infinity.
The correct statement is option A: For very high [tex]\( x \)[/tex]-values, [tex]\( f(x) \)[/tex] moves toward positive infinity.
