How do you find the amplitude and period of a function $y = - tan (x - \frac{π}{2})$ $y = -tan(x - pi/2)$ ?

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Answer 1 · 2016-01-10T09:15:15

An explanation is given below.

The $A tan (B (x - C)) + D$ $Atan(B(x-C))+D$ function has $no amplitude.$ $color(red)"no amplitude."$
The period is given by $\frac{π}{B}$ $pi/B$

$y = - tan (x - \frac{π}{2})$ $y=-tan(x-pi/2)$
On comparing with $A tan (B (x - C)) + D$ $Atan(B(x-C))+D$
We can see $B = 1$ $B=1$

The period is $\frac{π}{1} = π$ $pi/1 = pi$

The period of the given function is $π$ $pi$

How do you find the amplitude and period of a function y = -tan(x - pi/2)y=−tan(x−π2)y = -tan(x - pi/2)?