Question

write the equation of the function graphed above.

Explanation:

Step1: Determine the general form of the inverse tangent function

The general form of the inverse tangent function is \(y = A\tan^{- 1}(Bx - C)+D\).

Step2: Find the vertical stretch factor \(A\)

Since the graph has no vertical stretch (the range of the basic \(y = \tan^{-1}(x)\) is \((-\frac{\pi}{2},\frac{\pi}{2})\) and the given graph seems to follow the normal range behavior for the key - point analysis), \(A = 1\).

Step3: Find the horizontal compression factor \(B\)

The period of \(y=\tan^{-1}(x)\) is not really in the traditional sense (as its domain is \((-\infty,\infty)\) and range is \((-\frac{\pi}{2},\frac{\pi}{2})\)), but for the transformation formula \(y = \tan^{-1}(Bx)\), when \(x = 0\), \(y = 0\). Let's use a key - point. For \(y=\tan^{-1}(x)\), when \(x = 1\), \(y=\frac{\pi}{4}\). In our graph, if we assume a point - by - point correspondence (using the fact that the shape is similar to \(y = \tan^{-1}(x)\) and no horizontal shift is observed from the origin crossing), \(B = 1\).

Step4: Check for horizontal and vertical shifts

Since the graph passes through the origin \((0,0)\), \(C = 0\) (no horizontal shift, because for \(y=\tan^{-1}(B(x - C))\), when \(x = C\), \(y = 0\) in the un - stretched case) and \(D = 0\) (no vertical shift, as \(y=\tan^{-1}(x)+D\) would shift the graph up or down. If \(D
eq0\), the graph would not pass through \((0,0)\)).

Answer:

\(y=\tan^{-1}(x)\)