How do you sketch the graph of $y=2sec3x$ ?

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Answer 1 · 2018-03-07T07:17:25

As below

Standard form of equation is $y = a sec (bx - c) + d$

$y = 2 sec 3x$

$a = 2, b = 3, c = 0, d = 0$

Period $= (2pi) / |b| = (2pi) / 2 = pi$

Phase shift $= c / b = 0$

Phase shift 0 to the right

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

How do you sketch the graph of y=2sec3xy=2sec3x?