How do you graph $y=1+2secx$ ?

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Answer 1 · 2018-03-29T02:23:01

As below.

Standard form is $y = a sec (Bx -C) + D$

Given equation is $y = 2 sec (x) + 1$

$Amplitude = " None, for secant function"$

$"Period " = (2pi) / |B| = 2pi / 1 = 2pi$

$" Phase shift = (-C) / B = 0$

$"Vertical Shift " = D = 1$

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

How do you graph y=1+2secxy=1+2secx?