How do you find the amplitude, period, vertical and phase shift and graph #y=4+5sec[1/3(theta+(2pi)/3)]#?

1 Answer
Mar 19, 2018

As below.

Explanation:

Standard form of equation : #Asec(Bx - C) + D#

Given Equation : #y = 4 +5 sec [(1/3)(theta + ((2pi)/3)]#

Rewiting in the standard form,

#y = 5 sec [(theta / 3) + ((2pi)/9)] + 4#

#A = 5, B = 1/3, C = (2pi)/9, D = 4#

#Amplitude = A = "None"#, Function #sec( theta)# doesn't have an amplitude.

#"Period" = P = (2pi) / |B| = ((2pi) / (1/3)) = 6pi#

#"Phase Shift" = (-C) / B = ((-2pi)/9) / (1/3) = -(2pi)/3#

#"Vertical Shift" = D = 4#

graph{5 sec((x/3) + ((2pi)/9)) + 4 [-10, 10, -5, 5]}