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

Mar 19, 2018

As below.

#### Explanation:

Standard form of equation : $A \sec \left(B x - C\right) + D$

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

Rewiting in the standard form,

$y = 5 \sec \left[\left(\frac{\theta}{3}\right) + \left(\frac{2 \pi}{9}\right)\right] + 4$

$A = 5 , B = \frac{1}{3} , C = \frac{2 \pi}{9} , D = 4$

$A m p l i t u \mathrm{de} = A = \text{None}$, Function $\sec \left(\theta\right)$ doesn't have an amplitude.

$\text{Period} = P = \frac{2 \pi}{|} B | = \left(\frac{2 \pi}{\frac{1}{3}}\right) = 6 \pi$

$\text{Phase Shift} = \frac{- C}{B} = \frac{\frac{- 2 \pi}{9}}{\frac{1}{3}} = - \frac{2 \pi}{3}$

$\text{Vertical Shift} = D = 4$

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