How do you graph $y= 1 + 3 sec(x/2)$ ?

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Answer 1 · 2018-05-06T11:34:43

As below.

$y = 3sec (x/2) + 1$

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

$Amplitude = |A| = " None, for secant function"$

$"Period " = (2pi) / |B| = (2pi) / (1/2) = 4pi$

$"Phase Shift " = -C / B = 0$

$"Vertical Shift " = D = 1$

graph{3 sec (x/2) + 1 [-10, 10, -5, 5]}

How do you graph y= 1 + 3 sec(x/2)y= 1 + 3 sec(x/2)?