How do you graph $y=sec(x-pi/4)$ ?

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Answer 1 · 2018-03-12T01:39:59

As below.

$y = asec(bx-c)+d$ is the standard form.

$y = sec(x - pi/4)$

Amplitude $a =$ none.

Period $= (2pi)/|b| = (2pi) / 1 = 2pi$

Phase shift $= -c / b =( (pi)/4) / 1 = pi / 4$ $color(white)(aaa)(pi/4)$ to the right.

$Vertical shift$ = d = 0#

graph{sec(x - pi/4) [-10, 10, -5, 5]}

How do you graph y=sec(x-pi/4)y=sec(x-pi/4)?