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Find the vertical​ asymptotes, if​ any, and the value of x corresponding to​ holes, if​ any, of the graph of the following rational function.

f left parenthesis x right parenthesis equals StartFraction x minus 10 Over x squared minus 100 EndFraction
Question content area bottom
Part 1
Select the correct choice below​ and, if​ necessary, fill in the answer box to complete your choice. ​(Type an integer or a fraction. Use a comma to separate answers as​ needed.)
A.
There are no vertical asymptotes but there is​ (are) hole(s) corresponding to xequals
  
enter your response here.
B.
Vertical​ asymptote(s) at xequals
  
enter your response here. There are no holes.
C.
Vertical​ asymptote(s) at xequals
  
enter your response here and​ hole(s) corresponding to xequals
  
enter your response here.
D.
There are no discontinuities.Find the vertical​ asymptotes, if​ any, and the value of x corresponding to​ holes, if​ any, of the graph of the following rational function.
f left parenthesis x right parenthesis equals StartFraction x minus 10 Over x squared minus 100 EndFraction
Question content area bottom
Part 1
Select the correct choice below​ and, if​ necessary, fill in the answer box to complete your choice. ​(Type an integer or a fraction. Use a comma to separate answers as​ needed.)
A.
There are no vertical asymptotes but there is​ (are) hole(s) corresponding to xequals
  
enter your response here.
B.
Vertical​ asymptote(s) at xequals
  
enter your response here. There are no holes.
C.
Vertical​ asymptote(s) at xequals
  
enter your response here and​ hole(s) corresponding to xequals
  
enter your response here.
D.
There are no discontinuities.

1 answer

To analyze the rational function \( f(x) = \frac{x - 10}{x^2 - 100} \), we will identify any vertical asymptotes and any holes in the function.

  1. Finding the vertical asymptotes: Vertical asymptotes occur where the denominator is zero but the numerator is not zero at the same point.

    First, let's factor the denominator: \[ x^2 - 100 = (x - 10)(x + 10). \] Setting the denominator equal to zero gives: \[ (x - 10)(x + 10) = 0. \] So, the solutions are: \[ x - 10 = 0 \quad \Rightarrow \quad x = 10, \] \[ x + 10 = 0 \quad \Rightarrow \quad x = -10. \]

    Next, we check the numerator: \[ x - 10 = 0 \quad \Rightarrow \quad x = 10. \] At \( x = 10 \), both the numerator and denominator are zero, which indicates a hole at this point.

    Since the denominator is zero at \( x = -10 \) and the numerator is not (since \( x - 10 \neq 0 \) at \( x = -10 \)), we have a vertical asymptote at \( x = -10 \).

  2. Conclusion: There is one vertical asymptote at \( x = -10 \) and a hole at \( x = 10 \).

Thus, the correct choice is: C. Vertical asymptote(s) at \( x = -10 \) and hole(s) corresponding to \( x = 10 \).