Which of the following can be said about the vertical asymptotes of y = cot(x) and y = -tan(x - π/2)?
Understand the Problem
The question is asking about the vertical asymptotes of the two given functions, y = cot(x) and y = -tan(x - π/2), and how they relate to each other. The options suggest different possible outcomes regarding the vertical asymptotes' locations and relationship.
Answer
The vertical asymptotes of the graphs of the two functions coincide.
Answer for screen readers
The vertical asymptotes of the graphs of the two functions coincide.
Steps to Solve
- Identify vertical asymptotes of $y = \cot(x)$
The function $y = \cot(x)$ has vertical asymptotes where the function is undefined, which occurs when $\sin(x) = 0$; this happens at:
$$ x = n\pi, \text{ where } n \in \mathbb{Z} $$
- Identify vertical asymptotes of $y = -\tan\left(x - \frac{\pi}{2}\right)$
The function $y = -\tan\left(x - \frac{\pi}{2}\right)$ can be rewritten using the property of tangent:
$$ -\tan\left(x - \frac{\pi}{2}\right) = \cot(x) $$
The vertical asymptotes for $y = -\tan\left(x - \frac{\pi}{2}\right)$ are identical to those of $y = \cot(x)$, occurring at:
$$ x = n\pi, \text{ where } n \in \mathbb{Z} $$
- Compare the asymptotes
Since both functions have vertical asymptotes at the same locations, we conclude:
The vertical asymptotes of both functions coincide.
The vertical asymptotes of the graphs of the two functions coincide.
More Information
The functions $y=\cot(x)$ and $y=-\tan\left(x - \frac{\pi}{2}\right)$ exhibit similar behavior due to their relationship; both have vertical asymptotes at integer multiples of $\pi$, making their asymptotes identical.
Tips
- A common mistake is misinterpreting the transformation of tangent when shifted, leading to incorrect identification of asymptotes.
- Another mistake could involve assuming that the sine and cosine functions have different zeroes without recognizing their periodic nature.
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