# How do you find the period, phase and vertical shift of y=2sec(1/2(theta-90^circ))?

May 4, 2018

As described below.

#### Explanation:

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

$y = 2 \sec \left(\frac{1}{2} \left(\theta - {90}^{\circ}\right)\right)$

$y = 2 \sec \left(\left(\frac{\theta}{2}\right) - {45}^{\circ}\right)$

$y = 2 \sec \left(\left(\frac{\theta}{2}\right) - \left(\frac{\pi}{4}\right)\right)$

$\text{Amplitude " = |A| = "None for secant function}$

$\text{Period } = \frac{\pi}{|} B | = \frac{2 \pi}{\frac{1}{2}} = 4 \pi$

$\text{Phase Shift " = -C / B =( pi/4) / (1/2) = pi/2 " to the right}$

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

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