# How do you find the amplitude, period, and shift for y=1 +sec ( (x-pi)/3)?

Nov 20, 2017

Amplitude: none
Period: $6 \pi$
Phase: $- \pi$

#### Explanation:

Given a trigonometric function of the form

$y = A \sec \left(B x - C\right) + D$

Amplitude: Since the $\sec$ function does not have a maximum or minimum, there is not amplitude.

Period: $\frac{2 \pi}{\left\mid B \right\mid}$

Phase (i.e., horizontal) shift: $\frac{C}{B}$

Vertical shift: $\left\mid D \right\mid$

Given $y = 1 + \sec \left(\frac{x - \pi}{3}\right)$

$y = \sec \left(\frac{x - \pi}{3}\right) + 1$

$y = \sec \left(\frac{1}{3} x - \frac{\pi}{3}\right) + 1$

By the formula,

Period: $\frac{2 \pi}{1 / 3} = 6 \pi$

Phase shift: $\frac{- \pi / 3}{1 / 3} = - \pi$