# How do you find the amplitude, period and graph y=1/3 sectheta?

Jun 20, 2018

As below.

#### Explanation:

$y = A \sec \left(B x - C\right) + D$ is the standard form

$\text{Given : } y = \left(\frac{1}{3}\right) \sec \theta$

A = (1/30, B = 1

$A m p l i t u \mathrm{de} = | A | = \frac{1}{3}$

$\text{Period } = \frac{2 \pi}{|} B | = \frac{2 \pi}{1} = 2 \pi$

$\text{Phase Shift } = - \frac{C}{B} = 0$

$\text{Vertical Shift } = D = 0$

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