# How do you find the period, phase and vertical shift of y=1/2csc3(theta-45^circ)+1?

Nov 27, 2017

To find the period, look at k in the equation $y = a \csc k \left(\theta - d\right) + c$
The phase and vertical shift are also right in the equation as d and c.

#### Explanation:

Period:
$p e r i o d = \frac{2 \pi}{k}$

In the equation, look for k, which in this case, is 3.

So:
$p e r i o d = \frac{2 \pi}{3}$

In degrees, it would be
${180}^{o} / \left(\pi r a d\right) = r a d$

$= {180}^{o} / \left(\pi r a d\right) \cdot \frac{2 \pi}{3}$

$= \frac{{180}^{o} \cdot 3}{2}$

$= {270}^{o}$

Phase Shift:
The phase shift is also in the equation too, in which case you look for d. In this equation d is ${45}^{o}$

Since it is a negative ${45}^{o}$, it is then said that the graph moves ${45}^{o}$ to the left.

Vertical Shift:
The vertical shift is the last part of the equation: c, which is in this equation, 1 . If it is positive, it goes up. Negative shifts down. In this equation, the graph will shift up 1 unit.

Please correct me if I am wrong!