Question

Which of the following functions has a vertical asymptote at \(\mathrm{x}=-1\). (A) \(y=\frac{\left|x^{2}-1\right|}{x+1}\) (B) \(y=\frac{x^{2}-6 x-7}{x+1}\) (C) \(y=\frac{x^{2}+1}{x+1}\) (D) \(y=\frac{\sin (x+2)}{x+1}\)

Short answer

Answer: (B), (C), and (D)

Step-by-step solution

Function A

Evaluate the limit as x approaches -1 for the first function: \[y = \frac{|x^2-1|}{x+1}\] We need to check if the denominator approaches zero and the numerator does not. When x = -1, \[y = \frac{|(-1)^2-1|}{-1+1}\] \[y=\frac{0}{0}\] Since both the numerator and the denominator approach zero, we do not have a vertical asymptote at x=-1 for this function.

Function B

Evaluate the limit as x approaches -1 for the second function: \[y = \frac{x^2-6x-7}{x+1}\] When x = -1, \[y = \frac{(-1)^2-6(-1)-7}{-1+1}\] \[y = \frac{6}{0}\] Since the denominator approaches zero and the numerator does not, we have a vertical asymptote at x=-1 for this function.

Function C

Evaluate the limit as x approaches -1 for the third function: \[y = \frac{x^2+1}{x+1}\] When x = -1, \[y = \frac{(-1)^2+1}{-1+1}\] \[y = \frac{2}{0}\] Again, since the denominator approaches zero and the numerator does not, we have a vertical asymptote at x=-1 for this function.

Function D

Evaluate the limit as x approaches -1 for the fourth function: \[y = \frac{\sin(x+2)}{x+1}\] When x = -1, \[y = \frac{\sin(-1+2)}{-1+1}\] \[y = \frac{\sin(1)}{0}\] Again, since the denominator approaches zero and the numerator does not, we have a vertical asymptote at x=-1 for this function.

Conclusion

Both functions B, C, and D have a vertical asymptote at x=-1. So, the correct choices are (B), (C), and (D).