How do you find the asymptotes of $y = {tan}^{- 1} (x - 1) + \frac{π}{2}$ $y = tan^-1(x-1) + pi/2$ ?

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How do you find the asymptotes of $y = {tan}^{- 1} (x - 1) + \frac{π}{2}$ $y = tan^-1(x-1) + pi/2$ ?

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Answer 1 · 2016-02-15T16:48:58

y = 0 and y = $π$ $pi$ .

Explanation:

Inverse tangent varies from - $\frac{π}{2}$ $pi/2$ to + $\frac{π}{2}$ $pi/2$ . Correspondingly, (x - 1) varies from - infinity to + infinity and . y varies from 0 to $π$ $pi$
.(1, $\frac{π}{2}$ $pi/2$ ) is the center of this curve that extends to infinity in both +x and -x directions, tending to touch the asymptotes, y =0 and y = $π$ $pi$ ,. at infinity. . ,

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How do you find the asymptotes of $y = {tan}^{- 1} (x - 1) + \frac{π}{2}$ $y = tan^-1(x-1) + pi/2$ ?

1 Answer

Explanation:

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How do you find the asymptotes of $y = {tan}^{- 1} (x - 1) + \frac{π}{2}$ $y = tan^-1(x-1) + pi/2$ ?