How do you find the amplitude, period, and shift for $y = 1 + sec (\frac{x - π}{3})$ $y=1 +sec ( (x-pi)/3)$ ?

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Answer 1 · 2017-11-20T17:06:02

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

Given a trigonometric function of the form

$y = A sec (B x - C) + D$ $y=Asec(Bx - C) + D$

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

Period: $\frac{2 π}{| B |}$ $(2pi)/abs(B)$

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

Vertical shift: $| D |$ $abs(D)$

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

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

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

By the formula,

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

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

How do you find the amplitude, period, and shift for y=1+sec(x−π3)y=1 +sec ( (x-pi)/3)?