which of the following represents $y = \cot x$ ?

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# Explanation: ## Step1: Recall properties of $ y = \cot x $ The cotangent function $ y=\cot x=\frac{\cos x}{\sin x} $ has vertical asymptotes where $ \sin x = 0 $, i.e., at $ x = n\pi $ for integer $ n $. The period of $ \cot x $ is $ \pi $, and in the interval $ (0, \pi) $, $ \cot x $ is negative (since $ \cos x $ is positive and $ \sin x $ is positive in $ (0, \pi/2) $, but $ \cos x $ is negative and $ \sin x $ is positive in $ (\pi/2, \pi) $? Wait, no: $ \cos x $ in $ (0, \pi) $: $ \cos x $ is positive in $ (0, \pi/2) $, negative in $ (\pi/2, \pi) $; $ \sin x $ is positive in $ (0, \pi) $. So $ \cot x=\frac{\cos x}{\sin x} $ is positive in $ (0, \pi/2) $ (positive/positive) and negative in $ (\pi/2, \pi) $ (negative/positive). Wait, actually, let's check the sign in different quadrants. In the first quadrant ($ 0 < x < \pi/2 $): $ \cos x>0 $, $ \sin x>0 $, so $ \cot x>0 $. In the second quadrant ($ \pi/2 < x < \pi $): $ \cos x<0 $, $ \sin x>0 $, so $ \cot x<0 $. In the third quadrant ($ \pi < x < 3\pi/2 $): $ \cos x<0 $, $ \sin x<0 $, so $ \cot x>0 $ (negative/negative). In the fourth quadrant ($ 3\pi/2 < x < 2\pi $): $ \cos x>0 $, $ \sin x<0 $, so $ \cot x<0 $. Also, the graph of $ \cot x $ passes through $ ( \pi/4, 1) $, $ ( 3\pi/4, -1) $, etc. ## Step2: Analyze the asymptotes and sign The vertical asymptotes are at $ x = -\pi, 0, \pi $ (for the interval shown). Now, let's check the sign in intervals: - Between $ -\pi $ and $ 0 $: Let's take $ x = -\pi/2 $. $ \cot(-\pi/2)=\frac{\cos(-\pi/2)}{\sin(-\pi/2)}=\frac{0}{-1}=0 $? Wait, no: $ \cot(-\pi/2)=\frac{\cos(-\pi/2)}{\sin(-\pi/2)}=\frac{0}{-1}=0 $? Wait, $ \sin(-\pi/2)= -1 $, $ \cos(-\pi/2)=0 $, so $ \cot(-\pi/2)=0 $. Wait, actually, at $ x = -\pi/2 $, $ \cot x = 0 $. Let's check the interval $ (-\pi, 0) $: take $ x = -\pi/4 $. $ \cot(-\pi/4)=\frac{\cos(-\pi/4)}{\sin(-\pi/4)}=\frac{\sqrt{2}/2}{-\sqrt{2}/2}=-1 $. So in $ (-\pi, 0) $, let's see the quadrant: $ -\pi < x < 0 $ is equivalent to $ \pi < x + 2\pi < 2\pi $ (fourth quadrant), where $ \cos x>0 $, $ \sin x<0 $, so $ \cot x=\frac{\cos x}{\sin x}<0 $. Wait, maybe better to look at the standard graph: $ y = \cot x $ has vertical asymptotes at $ x = 0, \pi, -\pi $, etc. In the interval $ (0, \pi) $, the graph goes from $ +\infty $ (near $ x=0^+ $) to $ -\infty $ (near $ x=\pi^- $), passing through $ (\pi/2, 0) $. In the interval $ (-\pi, 0) $, it goes from $ -\infty $ (near $ x=-\pi^+ $) to $ +\infty $ (near $ x=0^- $), passing through $ (-\pi/2, 0) $. Now let's analyze the graphs: - Top-left (blue): Asymptotes at $ -\pi, 0, \pi $. In $ (0, \pi) $, the graph is negative near $ 0 $ and positive near $ \pi $? Wait, no. Wait, the blue graph: between $ 0 $ and $ \pi $, the left part (near 0) is below the x-axis (negative), right part (near $ \pi $) is above? No, that doesn't match $ \cot x $. - Top-right (green): Asymptotes at $ -\pi, 0, \pi $. Passes through the origin? $ \cot 0 $ is undefined, so it can't pass through (0,0). So green is out. - Bottom-left (orange): Passes through the origin? $ \cot 0 $ is undefined, so orange is out (since it goes through (0,0)). - Bottom-right (red): Asymptotes at $ -\pi, 0, \pi $. Let's check the interval $ (0, \pi) $: near $ x=0^+ $, $ \cot x $ approaches $ +\infty $ (since $ \sin x $ is small positive, $ \cos x $ is near 1, so $ \cot x $ is large positive). Near $ x=\pi^- $, $ \sin x $ is small positive, $ \cos x $ is near -1, so $ \cot x $ is large negative. Wait, no: $ x $ near $ \pi^- $ (just less than $ \pi $), $ \sin x $ is positive (small), $ \cos x $ is negative (near -1), so $ \cot x = \cos x / \sin x $ is negative (large negative), so it goes from $ +\infty $ (near 0) to $ -\infty $ (near $ \pi $) in $ (0, \pi) $. In $ (-\pi, 0) $: near $ x=-\pi^+ $ (just more than $ -\pi $), $ \sin x $ is negative (small), $ \cos x $ is negative (near -1), so $ \cot x = \cos x / \sin x $ is positive (large positive). Near $ x=0^- $ (just less than 0), $ \sin x $ is negative (small), $ \cos x $ is positive (near 1), so $ \cot x = \cos x / \sin x $ is negative (large negative). Wait, maybe I got the sign wrong. Let's take $ x = \pi/4 $ (in $ (0, \pi/2) $): $ \cot(\pi/4)=1 $ (positive). $ x = 3\pi/4 $ (in $ (\pi/2, \pi) $): $ \cot(3\pi/4)= -1 $ (negative). So in $ (0, \pi) $, $ \cot x $ is positive in $ (0, \pi/2) $, negative in $ (\pi/2, \pi) $, passing through $ (\pi/2, 0) $. In $ (-\pi, 0) $, $ x = -\pi/4 $ (equivalent to $ 7\pi/4 $, fourth quadrant): $ \cot(-\pi/4)= -1 $ (negative). $ x = -3\pi/4 $ (equivalent to $ 5\pi/4 $, third quadrant): $ \cot(-3\pi/4)=1 $ (positive). So in $ (-\pi, 0) $, $ \cot x $ is negative in $ (-\pi/2, 0) $, positive in $ (-\pi, -\pi/2) $, passing through $ (-\pi/2, 0) $. Now looking at the bottom-right (red) graph: - Near $ x=0^+ $: the graph is going up (towards $ +\infty $)? Wait, no, the red graph on the right side of $ x=0 $ (between 0 and $ \pi $): as $ x $ approaches $ 0^+ $, the graph is near the top (positive large), and as $ x $ approaches $ \pi^- $, it's near the bottom (negative large), passing through $ (\pi/2, 0) $? Wait, no, the red graph in $ (0, \pi) $: let's see the y-axis: at $ x=0 $, it's an asymptote. The red graph on the right of 0 (between 0 and $ \pi $): the curve is on the right side of the asymptote at 0, going from top (near 0) down to bottom (near $ \pi $), passing through the x-axis? Wait, maybe I made a mistake. Wait, the correct graph of $ y = \cot x $ has vertical asymptotes at $ x = n\pi $, and in each period $ (n\pi, (n+1)\pi) $, it decreases from $ +\infty $ to $ -\infty $, passing through $ x = n\pi + \pi/2 $ (where it crosses the x-axis). So in $ (0, \pi) $, it goes from $ +\infty $ (x→0⁺) to $ -\infty $ (x→π⁻), crossing the x-axis at $ \pi/2 $. In $ (-\pi, 0) $, it goes from $ -\infty $ (x→-π⁺) to $ +\infty $ (x→0⁻), crossing the x-axis at $ -\pi/2 $. Now let's check the bottom-right (red) graph: - Asymptotes at $ -\pi, 0, \pi $. In $ (0, \pi) $: the graph is on the right side of the asymptote at 0 (x=0), going from top (positive) down to bottom (negative) as x approaches $ \pi $, which matches $ \cot x $ (since at $ x=\pi/2 $, it should cross the x-axis). Wait, does the red graph cross the x-axis? Let's see: the red graph in $ (0, \pi) $ seems to cross the x-axis (since there's a point where it's on the x-axis between 0 and $ \pi $). Similarly, in $ (-\pi, 0) $, it crosses the x-axis between $ -\pi $ and 0. Wait, maybe the bottom-right graph is the correct one? Wait, no, wait the top-left (blue) graph: let's check again. Wait, maybe I confused the direction. Let's recall: $ y = \tan x $ has asymptotes at $ \pi/2 + n\pi $, and $ y = \cot x $ at $ n\pi $. The graph of $ \cot x $ is similar to $ -\tan(x - \pi/2) $. Alternatively, let's look at the standard graph: $ y = \cot x $ has vertical asymptotes at 0, $ \pi $, $ -\pi $, etc. In the interval $ (0, \pi) $, the function is positive when x is in $ (0, \pi/2) $ and negative when x is in $ (\pi/2, \pi) $, with a zero at $ \pi/2 $. In $ (-\pi, 0) $, it's negative when x is in $ (-\pi/2, 0) $ and positive when x is in $ (-\pi, -\pi/2) $, with a zero at $ -\pi/2 $. Now let's analyze each graph: 1. Top-left (blue): Asymptotes at $ -\pi, 0, \pi $. In $ (0, \pi) $: the left part (near 0) is below the x-axis (negative), right part (near $ \pi $) is above (positive). That would mean it's increasing from negative to positive, which is the opposite of $ \cot x $ (which decreases from positive to negative in $ (0, \pi) $). So blue is out. 2. Top-right (green): Passes through the origin (0,0), but $ \cot 0 $ is undefined (asymptote at x=0), so green is out. 3. Bottom-left (orange): Passes through the origin (0,0), same issue as green: $ \cot 0 $ is undefined, so orange is out. 4. Bottom-right (red): Asymptotes at $ -\pi, 0, \pi $. In $ (0, \pi) $: the graph is on the right side of the asymptote at 0 (x=0), going from top (positive) down to bottom (negative) as x approaches $ \pi $, which matches $ \cot x $ (decreasing from $ +\infty $ to $ -\infty $ in $ (0, \pi) $, crossing the x-axis at $ \pi/2 $). In $ (-\pi, 0) $: it's on the left side of the asymptote at 0 (x=0), going from bottom (negative) up to top (positive) as x approaches 0 from the left, which matches $ \cot x $ (increasing from $ -\infty $ to $ +\infty $ in $ (-\pi, 0) $, crossing the x-axis at $ -\pi/2 $). Also, it doesn't pass through the origin, which is correct because $ \cot 0 $ is undefined. So the bottom-right (red) graph is the correct one. Wait, but let's confirm with another approach: $ \cot x = \frac{1}{\tan x} $, so where $ \tan x $ is positive, $ \cot x $ is positive, and where $ \tan x $ is negative, $ \cot x $ is negative. $ \tan x $ is positive in $ (0, \pi/2) $ and $ (\pi, 3\pi/2) $, negative in $ (\pi/2, \pi) $ and $ (3\pi/2, 2\pi) $ (or $ (-\pi/2, 0) $ and $ (-\pi, -\pi/2) $ for negative x). So $ \cot x $ is positive in $ (0, \pi/2) $ and $ (\pi, 3\pi/2) $, negative in $ (\pi/2, \pi) $ and $ (-\pi/2, 0) $. So in $ (0, \pi) $, $ \cot x $ is positive then negative, which matches the bottom-right graph (positive near 0, negative near $ \pi $). In $ (-\pi, 0) $, $ \cot x $ is negative near 0 (since $ (-\pi/2, 0) $ is negative) and positive near $ -\pi $ (since $ (-\pi, -\pi/2) $ is positive), which also matches the bottom-right graph (negative near 0, positive near $ -\pi $). So the correct graph is the bottom-right one (red). # Answer: The graph on the Bottom-Right (the red graph)

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Explanation:

Step1: Recall properties of \( y = \cot x \)

Step2: Analyze the asymptotes and sign

Answer: