How do you find the period, phase and vertical shift of #y=2sec(1/2(theta-90^circ))#?

Question

1478 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-04T09:40:28

As described below.

Standard form of equation is #y = A sec (Bx - C) + D#

#y = 2 sec (1/2(theta - 90^@))#

#y = 2 sec ((theta / 2) - 45^@)#

#y = 2 sec ((theta / 2 ) - (pi/4))#

#"Amplitude " = |A| = "None for secant function"#

#"Period " = pi / |B| = (2pi) / (1/2) = 4pi#

#"Phase Shift " = -C / B =( pi/4) / (1/2) = pi/2 " to the right"#

#"Vertical Shift " = D = 0#

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