Question

which function represents y = -tan x ?

Explanation:

Step1: Recall the graph of \( y = \tan x \)

The function \( y = \tan x \) has vertical asymptotes at \( x = \frac{\pi}{2} + k\pi \) (where \( k \) is an integer), passes through the origin \((0,0)\), and in the interval \((-\frac{\pi}{2}, \frac{\pi}{2})\), it is increasing from \(-\infty\) to \(+\infty\).

Step2: Analyze the transformation for \( y = -\tan x \)

The negative sign reflects the graph of \( y = \tan x \) over the \( x \)-axis. So, in the interval \((-\frac{\pi}{2}, \frac{\pi}{2})\), the function \( y = -\tan x \) should be decreasing from \(+\infty\) to \(-\infty\) (opposite of the increasing behavior of \( y = \tan x \)).

Step3: Compare with the given graphs

- The red graph (top - left) has the same increasing behavior as \( y=\tan x \) (since it's increasing near the origin), so it's \( y = \tan x \), not \( y=-\tan x \).
- The orange graph (top - right): Let's check the behavior near the origin. For \( y = -\tan x \), when \( x = 0 \), \( y = 0 \). But the orange graph seems to have a different sign behavior (maybe not the reflection). Wait, let's check the green graph (bottom - left). Wait, no, let's re - check. Wait, the green graph (bottom - left): Wait, no, the bottom - left green graph: Wait, actually, the bottom - left green graph? Wait, no, the top - right orange graph? Wait, no, let's look at the bottom - left graph (green). Wait, no, the correct one: The graph of \( y = -\tan x \) should be a reflection of \( y=\tan x \). The standard \( y = \tan x \) has a "positive" slope near the origin (increasing), so \( y=-\tan x \) should have a "negative" slope near the origin (decreasing). Looking at the four graphs:

The red graph (top - left) is increasing near the origin (like \( y = \tan x \)). The orange graph (top - right): Let's check the direction. The green graph (bottom - left): Wait, the bottom - left green graph: When we look at the region near \( x = 0 \), the green graph is decreasing (since as \( x \) increases from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \), \( y \) decreases from \( +\infty \) to \( -\infty \)), which is the behavior of \( y=-\tan x \). Wait, no, wait the bottom - left graph: Wait, the axes: The x - axis has marks at \( -\pi,0,\pi \). The vertical asymptotes of \( y = \tan x \) are at \( x=-\frac{\pi}{2},\frac{\pi}{2},\frac{3\pi}{2},-\frac{3\pi}{2} \), etc. So between \( -\pi \) and \( \pi \), the asymptotes are at \( -\frac{\pi}{2} \) and \( \frac{\pi}{2} \). So the graph of \( y = \tan x \) has two branches between \( -\pi \) and \( \pi \): one between \( -\pi \) and \( -\frac{\pi}{2} \), one between \( -\frac{\pi}{2} \) and \( \frac{\pi}{2} \), and one between \( \frac{\pi}{2} \) and \( \pi \). For \( y = \tan x \), in the interval \( (-\frac{\pi}{2},\frac{\pi}{2}) \), it goes from \( -\infty \) to \( +\infty \) (increasing). For \( y=-\tan x \), in the interval \( (-\frac{\pi}{2},\frac{\pi}{2}) \), it goes from \( +\infty \) to \( -\infty \) (decreasing).

Looking at the four graphs:

- Top - left (red): Increasing near \( x = 0 \) (like \( y=\tan x \)).
- Top - right (orange): Let's see the direction. It seems to be a hyperbola - like but not the reflection.
- Bottom - left (green): Decreasing near \( x = 0 \) (since as \( x \) increases from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \), \( y \) decreases from \( +\infty \) to \( -\infty \)), which is the behavior of \( y = -\tan x \).
- Bottom - right (blue): This looks like a different function (maybe a secant or cosecant - like, but no, the blue graph has more asymptotes, which is not the case for tangent which has asymptotes at \( x=\frac{\pi}{2}+k\pi \)).

Wait, actually, the correct graph is the bottom - left (green) graph? Wait, no, wait the original \( y = \tan x \) passes through \( (0,0) \), has vertical asymptotes at \( x=\pm\frac{\pi}{2},\pm\frac{3\pi}{2},\cdots \). The graph of \( y = -\tan x \) will also pass through \( (0,0) \), have the same vertical asymptotes, but be a reflection over the \( x \)-axis. So the red graph is \( y = \tan x \) (increasing near 0), the green graph (bottom - left) is \( y=-\tan x \) (decreasing near 0). Wait, but in the options, the bottom - left is the green graph. Wait, but the user's graph: Let me re - check the labels. The four graphs:

Top - left: red, top - right: orange, bottom - left: green, bottom - right: blue.

The key is: \( y = \tan x \) is increasing on \( (-\frac{\pi}{2},\frac{\pi}{2}) \), so \( y=-\tan x \) is decreasing on \( (-\frac{\pi}{2},\frac{\pi}{2}) \). So we need to find the graph that is decreasing near \( x = 0 \) (between the asymptotes \( x=-\frac{\pi}{2} \) and \( x=\frac{\pi}{2} \)).

Looking at the red graph (top - left): as \( x \) increases from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \), \( y \) increases from \( -\infty \) to \( +\infty \) (so \( y = \tan x \)).

The green graph (bottom - left): as \( x \) increases from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \), \( y \) decreases from \( +\infty \) to \( -\infty \) (so \( y=-\tan x \)).

Wait, but the bottom - left graph is the green one? Wait, no, in the user's image, the bottom - left graph (green) has the curve decreasing through the origin. So the correct graph is the bottom - left (green) graph. But wait, maybe I made a mistake. Wait, the top - right (orange) graph: Let's check the value at \( x=\frac{\pi}{4} \). For \( y = -\tan x \), at \( x=\frac{\pi}{4} \), \( y=- 1 \). For \( y = \tan x \), at \( x=\frac{\pi}{4} \), \( y = 1 \). So in the red graph (top - left), at \( x=\frac{\pi}{4} \), \( y \) is positive (so \( y=\tan x \)). In the green graph (bottom - left), at \( x=\frac{\pi}{4} \), \( y \) is negative (so \( y = -\tan x \)). So the bottom - left (green) graph is the graph of \( y=-\tan x \).

Answer:

The bottom - left (green) graph