The statement is false.
Let's break it down step-by-step.
A vertical asymptote occurs at the values of x for which the denominator of a rational function is zero (assuming the numerator is not also zero at that point, as that could be a hole rather than an asymptote).
The statement that "the function is always negative between two asymptotes" suggests that for any rational function, if we have two vertical asymptotes, the function must be negative for all x-values that are in between these two asymptotes. However, this is not necessarily true.
To show that the statement is incorrect, we simply need a counterexample—a rational function that has two vertical asymptotes but is not always negative between them.
Consider the rational function:
\[ f(x) = \frac{x^2 - 1}{x^2 - 4} \]
This function has vertical asymptotes at the points where the denominator is zero and the numerator is not zero. Solving the equation \(x^2 - 4 = 0\), we find that the vertical asymptotes are at x = 2 and x = -2.
Now we need to analyze the behavior of the function between these two asymptotes. We can rewrite the numerator as \( (x + 1)(x - 1) \), which means:
\[ f(x) = \frac{(x + 1)(x - 1)}{(x - 2)(x + 2)} \]
Between the vertical asymptotes, when x is in the interval (-2, 2), both the numerator and denominator will have the same sign because \( (x - 1) \) and \( (x - 2) \) will both be negative and \( (x + 1) \) and \( (x + 2) \) will both be positive for \( -1 < x < 1 \). This means their quotient, \( f(x) \), will be positive in the interval (-1, 1).
Hence, we have found a counterexample to the original assertion. Since \( f(x) \) is positive between the asymptotes at x = -2 and x = 2 (specifically in the interval (-1, 1)), the statement "the function is always negative between two asymptotes" is false.
Choice B (False) is the correct answer.
