How do you graph $y = sec (3 x + \frac{π}{2})$ $y = sec(3x + pi/2)$ ?

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How do you graph $y = sec (3 x + \frac{π}{2})$ $y = sec(3x + pi/2)$ ?

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Answer 1 · 2018-03-30T08:54:28

As below.

Explanation:

$y = sec (3 x + (\frac{π}{2}))$ $y = sec(3x + (pi/2))$

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

$A m p l i t u d e = A = None for secant$ $Amplitude = A = color(red)(" None") " for secant"$

$Period = \frac{2 π}{| B |} = \frac{2 π}{3}$ $"Period " = (2pi) / |B| = (2pi) / 3$

$Phase Shift = - \frac{C}{B} = - \frac{\frac{π}{2}}{3} = - \frac{π}{6}, \frac{π}{6} to the left$ $"Phase Shift " = -C / B = -(pi/2) / 3 = -pi/6, " " pi/6 " to the left"$

$Vertical Shift = D = 0$ $"Vertical Shift " = D = 0$

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

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How do you graph $y = sec (3 x + \frac{π}{2})$ $y = sec(3x + pi/2)$ ?

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How do you graph y=sec(3x+π2)y = sec(3x + pi/2)?

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How do you graph $y = sec (3 x + \frac{π}{2})$ $y = sec(3x + pi/2)$ ?