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What is the horizontal asymptote of the function [tex]f(x) = 2^x - 5[/tex]?

Enter your answer as a number, like this: 42


Sagot :

To determine the horizontal asymptote of the function [tex]\( f(x) = 2^x - 5 \)[/tex], we need to analyze the behavior of the function as [tex]\( x \)[/tex] approaches positive and negative infinity.

1. As [tex]\( x \)[/tex] approaches positive infinity ([tex]\( x \to \infty \)[/tex]):
- The term [tex]\( 2^x \)[/tex] grows exponentially. This means that [tex]\( 2^x \)[/tex] will get very large.
- Therefore, when [tex]\( x \)[/tex] is very large, [tex]\( f(x) \)[/tex] will also get very large because [tex]\( f(x) = 2^x - 5 \)[/tex] and subtracting 5 from a very large number still results in a very large number.
- In this case, there is no horizontal asymptote as [tex]\( x \to \infty \)[/tex] because the function increases without bound.

2. As [tex]\( x \)[/tex] approaches negative infinity ([tex]\( x \to -\infty \)[/tex]):
- The term [tex]\( 2^x \)[/tex] approaches zero because any number raised to a very large negative power tends towards zero.
- Hence, as [tex]\( x \)[/tex] gets more negative, [tex]\( 2^x \)[/tex] gets closer and closer to 0.
- Therefore, [tex]\( f(x) = 2^x - 5 \)[/tex] will approach [tex]\( 0 - 5 \)[/tex], which simplifies to [tex]\( -5 \)[/tex].
- As [tex]\( x \to -\infty \)[/tex], [tex]\( f(x) \)[/tex] approaches [tex]\( -5 \)[/tex].

So, the horizontal asymptote of the function [tex]\( f(x) = 2^x - 5 \)[/tex] is the value that [tex]\( f(x) \)[/tex] approaches as [tex]\( x \to -\infty \)[/tex].

Thus, the horizontal asymptote is:
[tex]\[ \boxed{-5} \][/tex]