Which graph best represents the function $y={\tan }^{-1}x$ (inverse tangent)?

• A. An increasing S-shaped curve passing through $(0,0)$ with horizontal asymptotes $y=\frac{\pi }{2}$ and $y=-\frac{\pi }{2}$.
• B. A U-shaped parabola opening upward with vertex at $(0,0)$ and symmetry about the $y$-axis.
• C. A decreasing S-shaped curve passing through $(0,0)$ with horizontal asymptotes $y=\frac{\pi }{2}$ and $y=-\frac{\pi }{2}$.
• D. A repeating curve with vertical asymptotes at $x=\frac{\pi }{2}+k\pi$ and zeros at $x=k\pi$.

Answer: (A)

Verified step by step guidance

1. Recall that the function given is the inverse tangent function, denoted as \(y = \tan^{-1} x\), also called arctangent. This function returns the angle whose tangent is \(x\).
2. Understand the general shape of \(y = \tan^{-1} x\): it is an increasing function that passes through the origin \((0,0)\) because \(\tan^{-1}(0) = 0\).
3. Identify the horizontal asymptotes of the inverse tangent function. As \(x \to +\infty\), \(y\) approaches \(\frac{\pi}{2}\), and as \(x \to -\infty\), \(y\) approaches \(-\frac{\pi}{2}\). These are the horizontal asymptotes.
4. Note that the graph is S-shaped (sigmoid-like), increasing smoothly from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\), crossing the origin, and never repeating or oscillating.
5. Compare these properties with the given options: the correct graph is the increasing S-shaped curve passing through \((0,0)\) with horizontal asymptotes at \(y = \frac{\pi}{2}\) and \(y = -\frac{\pi}{2}\).