Question

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

Answer 1

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

Explanation:

Period:
`period=(2pi)/k`

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

So:
`period= (2pi)/3`

In degrees, it would be
`180^o/(pi rad)=rad`

`=180^o/(pi rad) * (2pi)/3`

`=(180^o*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!