Question

How do you find the amplitude, period, vertical and phase shift and graph `y=4+5sec[1/3(theta+(2pi)/3)]`?

Answer 1

As below.

Explanation:

Standard form of equation : `Asec(Bx - C) + D`

Given Equation : `y = 4 +5 sec [(1/3)(theta + ((2pi)/3)]`

Rewiting in the standard form,

`y = 5 sec [(theta / 3) + ((2pi)/9)] + 4`

`A = 5, B = 1/3, C = (2pi)/9, D = 4`

`Amplitude = A = "None"`, Function `sec( theta)` doesn't have an amplitude.

`"Period" = P = (2pi) / |B| = ((2pi) / (1/3)) = 6pi`

`"Phase Shift" = (-C) / B = ((-2pi)/9) / (1/3) = -(2pi)/3`

`"Vertical Shift" = D = 4`

graph{5 sec((x/3) + ((2pi)/9)) + 4 [-10, 10, -5, 5]}