Answer:
Option B:
Function A has a vertical asymptote at x = 1
Function B has a vertical asymptote at x = -3
Step-by-step explanation:
A function f(x) has a vertical asymptote if:
[tex]\lim_{x \to\\k^+}f(x) = \±\infty\\\\ \lim_{x \to\\k^-}f(x) = \±\infty[/tex]
This means that if there is a value k for which f(x) has infinity or a -infinity then x = k is a vertical asymptote of f(x). Therefore, the closer x to k approaches, the closer the function becomes to infinity.
We can calculate the asymptote for function A.
[tex]\lim_{x \to \\1^+}(\frac{1}{x-1})\\\\ \lim_{x \to \\1^+}(\frac{1}{1^-1})\\\\ \lim_{x \to \\1^+}(\frac{1}{0}) = \infty\\\\and\\ \lim_{x \to \\1^-}(\frac{1}{x-1})\\\\\lim_{x \to \\1^-}(\frac{1}{0}) = -\infty[/tex]
Then, function A has a vertical asymptote at x = 1.
The asymptote of function B can be easily observed in the graph. Note that the function b is not defined for x = -3 and when x is closest to -3, f(x) approaches infinity.
Therefore x = -3 is asintota of function B.
Therefore the correct answer is option B.
