How do you graph #y= 1 + 3 sec(x/2)#?

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Answer 1 · 2018-05-06T11:34:43

As below.

#y = 3sec (x/2) + 1#

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

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

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

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

#"Vertical Shift " = D = 1#

graph{3 sec (x/2) + 1 [-10, 10, -5, 5]}