Final answer:
The correct answer is option B. The function has a vertical asymptote at x = -3 and a horizontal asymptote at y = 2.
Step-by-step explanation:
The correct answer is option B.
To determine the asymptotes of the function, we need to look at its behavior as x approaches infinity and as x approaches negative infinity.
For the vertical asymptote, we need to find the value of x for which the function approaches infinity or negative infinity. In this case, as x approaches negative infinity, the function y = 1/x approaches negative infinity. Therefore, there is a vertical asymptote at x = -3.
For the horizontal asymptote, we need to find the value of y for which the function approaches a constant value as x approaches infinity or negative infinity. In this case, as x approaches infinity, the function y = 1/x approaches 0. Therefore, there is a horizontal asymptote at y = 2.
The correct answer is option D. When determining asymptotes of a function, we look at the behavior of the function as the input (usually x) approaches certain critical values. For a function like f(x) = 1/(x-2) +3, which seems similar to the one described by the question, the vertical asymptote occurs where the function is undefined, which is at x = 2. As x approaches 2 from either direction, the value of f(x) becomes infinitely large or infinitely negative.
The horizontal asymptote is determined by the behavior of the function as x goes to infinity. Since the value of 1/(x-2) approaches 0 as x increases without bound, the y-value approaches the constant term, which is y = -3. So, the function has a horizontal asymptote at y = -3 and a vertical asymptote at x = 2.