How do you find the amplitude, period and graph #y=1/3 sectheta#?

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Answer 1 · 2018-06-20T04:11:39

As below.

#y = A sec (Bx - C) + D# is the standard form

#"Given : " y = (1/3) sec theta#

#A = (1/30, B = 1#

#Amplitude = |A| = 1/3#

#"Period " = (2pi) / |B| = (2pi) / 1 = 2pi#

#"Phase Shift " = -C / B = 0#

#"Vertical Shift " = D = 0#

graph{(1/3) sec x [-10, 10, -5, 5]}