The function [tex]$f(x)=\frac{1}{x+3}$[/tex] has a vertical asymptote at:

A. [tex]$x=3$[/tex]
B. [tex][tex]$x=1$[/tex][/tex]
C. [tex]$x=-3$[/tex]
D. [tex]$x=0$[/tex]



Answer :

To determine the vertical asymptote of the function [tex]\( f(x) = \frac{1}{x+3} \)[/tex], we need to identify where the function becomes undefined due to the denominator being zero. Here's the process step-by-step:

1. Identify the Denominator:
The function is given as [tex]\( f(x) = \frac{1}{x+3} \)[/tex]. The denominator of this function is [tex]\( x + 3 \)[/tex].

2. Set the Denominator Equal to Zero:
To find where the function is undefined, we solve for [tex]\( x \)[/tex] in the equation [tex]\( x + 3 = 0 \)[/tex].

3. Solve for [tex]\( x \)[/tex]:
Solving the equation [tex]\( x + 3 = 0 \)[/tex]:
[tex]\[ x + 3 = 0 \][/tex]
Subtract 3 from both sides:
[tex]\[ x = -3 \][/tex]

4. Conclusion:
The vertical asymptote occurs at the value of [tex]\( x \)[/tex] that makes the denominator zero, which in our case is [tex]\( x = -3 \)[/tex].

Therefore, the function [tex]\( f(x) = \frac{1}{x+3} \)[/tex] has a vertical asymptote at [tex]\( x = -3 \)[/tex].

The correct answer is:
C. [tex]\( x = -3 \)[/tex]