# I need help solving these 2 questions. can you please explain to me step by step. I have a test coming up soon. y= 2sec(2x-pi/2)+1 y= 2cot(x+pi/6) I also need to graph these

*print*Print*list*Cite

### 1 Answer

(1) `y=2sec(2x-pi/2)+1 `

The general form is y=a sec[b(x-h)]+k

a performs a vertical dilation, b is a horizontal dilation (affects the period), h performs a horizontal translation, and k performs a vertical translation. These transformations are done on the base function y=sec x.

Rewrite as: `y=2 sec[2(x-pi/4)]+1 `

From the base function y= sec x:

There is a vertical stretch of factor 2.

There is a horizontal compression of factor 2. Since the period of the secant is 2pi, the period of this graph is pi.

There is a horizontal translation of ` pi/4 ` units to the right.

There is a vertical shift up 1 unit.

The graph :

(2) `y=2cot(x+pi/6) `

There is a vertical stretch of factor 2 and a horizontal translation (phase shift) of `pi/6 ` units to the left. There is no horizontal dilation (no change in period) and no vertical translation. The base function is y=cot x .

The graph:

**Sources:**