How do you graph #y = 3 sec(2x)#? Trigonometry Graphing Trigonometric Functions Graphing Tangent, Cotangent, Secant, and Cosecant 1 Answer sankarankalyanam May 6, 2018 As below. Explanation: #y = A sec (Bx - C) + D " is the standard form of secant function"# #"Given : " y = 3 sec (2x)# #"Amplitude is NONE for a secant function"# #"Period + (2pi) / |B| = (2pi) / 2 = pi# #"Phase shift " = - C / B = 0# #Vertical shift " = D = 0# graph{3 sec (2x) [-10, 10, -5, 5]} Answer link Related questions How do you find the asymptotes for the cotangent function? How do you graph tangent and cotangent functions? How do you Sketch the graph of #y=-2+cot(1/3)x# over the interval #[0, 6pi]#? How do you graph #y=-3tan(x-(pi/4))# over the interval #[-pi, 2pi]#? How do you sketch a graph of #h(x)=5+frac{1}{2} \sec 4x# over the interval #[0,2pi]#? What is the amplitude, period and frequency for the function #y=-1+\frac{1}{3} \cot 2x#? How do you graph #y=tan(2x+pi/4)#? What is the domain of #y = tan(x) + 2#? How do you graph #csc(x-pi/2)#? How do you graph #y=sec(x+4)#? See all questions in Graphing Tangent, Cotangent, Secant, and Cosecant Impact of this question 6644 views around the world You can reuse this answer Creative Commons License