How do you graph $y=2sec(3(x-pi/6))+1$ ?

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Answer 1 · 2018-06-24T02:58:39

As below.

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

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

$A = 2, B = 3, C = -pi/2, D = 1$

$Amplitude = |A| = "NONE for secant function"$

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

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

$"Vertical Shift " = D = 1$

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

How do you graph y=2sec(3(x-pi/6))+1y=2sec(3(x-pi/6))+1?