How do you graph $y = sec(3x + pi/2)$ ?

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Answer 1 · 2018-03-30T08:54:28

As below.

$y = sec(3x + (pi/2))$

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

$Amplitude = A = color(red)(" None") " for secant"$

$"Period " = (2pi) / |B| = (2pi) / 3$

$"Phase Shift " = -C / B = -(pi/2) / 3 = -pi/6, " " pi/6 " to the left"$

$"Vertical Shift " = D = 0$

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

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