Question

write the following tangent function.

Explanation:

Step1: Recall Tangent Function Form

The general form of a tangent function is \( y = A\tan(Bx - C) + D \). We need to determine \( A \), \( B \), \( C \), \( D \) from the graph.

Step2: Check Vertical Shift (\( D \))

The graph passes through the origin (or symmetric around \( y=0 \)), so \( D = 0 \).

Step3: Determine Period

The standard tangent function \( y = \tan(x) \) has a period of \( \pi \). Let's check the period of the given graph. The distance between two consecutive vertical asymptotes (or key points) should help. Wait, looking at the graph, let's see the x - axis labels. Wait, maybe the function is \( y=\tan(2x) \)? Wait, no, let's check the phase shift. Wait, the standard tangent function \( y = \tan(x) \) has vertical asymptotes at \( x=\frac{\pi}{2}+k\pi \). But in the graph, let's see the key points. Wait, maybe the function is \( y = \tan(2x) \)? Wait, no, let's re - examine. Wait, the general form: period of \( y = \tan(Bx) \) is \( \frac{\pi}{|B|} \). Let's check the graph. Wait, maybe the function is \( y=\tan(2x) \)? Wait, no, let's see the x - values. Wait, the graph seems to have a period of \( \frac{\pi}{2} \)? Wait, no, maybe I made a mistake. Wait, the standard tangent function \( y = \tan(x) \) has a period of \( \pi \), with vertical asymptotes at \( x=\frac{\pi}{2}+k\pi \) and passes through \( (0,0) \), \( (\frac{\pi}{4},1) \), \( (-\frac{\pi}{4}, - 1) \). Looking at the given graph, let's check the point \( (\frac{\pi}{4},1) \)? Wait, no, maybe the function is \( y = \tan(2x) \)? Wait, no, let's see the x - axis. Wait, the graph has a period of \( \frac{\pi}{2} \)? Wait, no, let's calculate \( B \). If the period is \( \frac{\pi}{2} \), then \( \frac{\pi}{|B|}=\frac{\pi}{2}\implies |B| = 2 \). But wait, maybe the function is \( y=\tan(x) \)? No, wait, the graph's key points: let's check the x - value at which \( y = 1 \). Wait, maybe the function is \( y=\tan(2x) \)? Wait, no, let's start over.

Wait, the standard tangent function is \( y = \tan(x) \), which has a period of \( \pi \), vertical asymptotes at \( x=\frac{\pi}{2}+k\pi \), and passes through \( (0,0) \), \( (\frac{\pi}{4},1) \), \( (-\frac{\pi}{4}, - 1) \). Looking at the given graph, let's check the x - axis labels. The x - axis has \( \pm\frac{\pi}{4},\pm\frac{\pi}{2},\pm\frac{3\pi}{4},\pm\pi \). Let's check the point \( (\frac{\pi}{4},1) \): if \( y = \tan(x) \), at \( x=\frac{\pi}{4} \), \( y = 1 \), which matches the standard tangent function. Wait, but the graph in the picture, does it match \( y=\tan(x) \)? Wait, maybe the graph is of \( y = \tan(x) \). Wait, but let's confirm. The standard tangent function \( y=\tan(x) \) has the following properties:
- Domain: \( \{x|x eq\frac{\pi}{2}+k\pi,k\in\mathbb{Z}\} \)
- Range: \( (-\infty,\infty) \)
- Passes through \( (0,0) \), \( (\frac{\pi}{4},1) \), \( (-\frac{\pi}{4}, - 1) \)
- Vertical asymptotes at \( x=\frac{\pi}{2}+k\pi,k\in\mathbb{Z} \)

Looking at the given graph, it seems to have the same shape as \( y = \tan(x) \). Wait, but maybe there is a vertical stretch or compression. Wait, if we check the point \( (\frac{\pi}{4},1) \), if the function is \( y = \tan(x) \), then at \( x = \frac{\pi}{4} \), \( y=1 \), which matches the standard tangent function. So maybe the function is \( y=\tan(x) \). Wait, but let's check the vertical asymptotes. The vertical asymptotes of \( y = \tan(x) \) are at \( x=\frac{\pi}{2}+k\pi \), which seems to match the graph's asymptotes (looking at the x - axis labels like \( \frac{\pi}{2},\frac{3\pi}{4},\pi,-\frac{\pi}{2},-\frac{3\pi}{4},-\pi \)). So the tangent function here is \( y = \tan(x) \). Wait, but maybe I made a mistake. Wait, another way: the general form \( y = A\tan(Bx - C)+D \). Since the graph is centered at \( y = 0 \) (no vertical shift, \( D = 0 \)), no phase shift (\( C = 0 \)), and if we assume \( A = 1 \), \( B = 1 \), then \( y=\tan(x) \).

Answer:

\( y = \tan(x) \)