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

Question

3368 views around the world

You can reuse this answer
Creative Commons License

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]}