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What are the vertical and horizontal asymptotes of [tex]$f(x) = \frac{x-1}{x}$[/tex]?

A. Horizontal asymptote at [tex]$y = 0$[/tex], vertical asymptote at [tex][tex]$x = 1$[/tex][/tex]
B. Horizontal asymptote at [tex]$y = 2$[/tex], vertical asymptote at [tex]$x = 1$[/tex]
C. Horizontal asymptote at [tex][tex]$y = 1$[/tex][/tex], vertical asymptote at [tex]$x = 0$[/tex]
D. Horizontal asymptote at [tex]$y = 1$[/tex], vertical asymptote at [tex][tex]$x = 2$[/tex][/tex]


Sagot :

To determine the vertical and horizontal asymptotes of the function [tex]\( f(x) = x - 1 \)[/tex], we need to analyze its behavior.

### Vertical Asymptotes:
Vertical asymptotes are found by identifying values of [tex]\( x \)[/tex] that make the denominator of a rational function equal to zero. Since [tex]\( f(x) = x - 1 \)[/tex] is a linear function and not a rational function (it has no denominator that could be zero), it does not have any vertical asymptotes.

### Horizontal Asymptotes:
Horizontal asymptotes are determined by analyzing the end behavior of the function as [tex]\( x \)[/tex] approaches infinity or negative infinity.

- As [tex]\( x \to \infty \)[/tex], [tex]\( f(x) = x - 1 \to \infty \)[/tex].
- As [tex]\( x \to -\infty \)[/tex], [tex]\( f(x) = x - 1 \to -\infty \)[/tex].

Since the function grows without bound (both positively and negatively) as [tex]\( x \)[/tex] moves away from zero, there are no horizontal asymptotes.

### Conclusion:
Based on the above analysis:

- Vertical asymptotes: None
- Horizontal asymptotes: None

None of the provided options match our findings. Therefore, the correct conclusion is that there are no vertical or horizontal asymptotes for the function [tex]\( f(x) = x - 1 \)[/tex].