How do you graph $y = - 2 sec 2 (x + π) + 3$ $y=-2sec2(x+pi)+3$ ?

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Answer 1 · 2018-06-16T09:53:07

As below.

$y = - 2 sec (2 x + 2 π) + 3$ $y = - 2 sec (2x + 2pi) + 3$

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

$A = - 2, B = 2, C = - 2 π, D = 3$ $A = -2, B = 2, C = -2pi, D = 3$

$A m p l i t u d e = | A | = None for secant function$ $Amplitude = |A| = "None for secant function"$

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

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

$Vertical Shift = D = 3$ $"Vertical Shift " = D = 3$

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

How do you graph y=−2sec2(x+π)+3y=-2sec2(x+pi)+3?