Thread: Graph y=4-3/4sec(3x-∏)

Having some trouble with the phase shift of this problem.
Graph y=4-3/4sec(3x-∏)
I graphed cos first but the p.s. keeps me up at night. Any walk throughs would be appreciated!

Thanks

Is it

I would go one step at a time. If y=f(x) then

f(3x) is f(x) compressed by a factor of 3 (in other words you'll complete 3 cycles in every 2pi radians)
f(x-pi) is f(x) shifted pi units to the right on the x axis
(3/4)f(x) is a vertical "stretch" of 3/4. For trig graphs this only really changes the extrema - in this case f(x) will vary between -3/4 and 3/4
-f(x) is f(x) inverted/flipped in the x axis
4-f(x) is f(x) shifted up the y axis by 4 units.

If you confirm the equation in question I will graph these steps for you using software

Sorry for Double post check down

Its the second equation.
The equation had no parenthesis around the -3/4 (amplitude)
The trouble I am having is dividing the P.S. into four equal parts.
P.S. -d/b= ∏/3 (Start) Period 2∏/b=2∏/3 (End)
What are the points in between?

Your phase shift is . Therefore if you draw the graph of (I don't know if this is correct since I still don't know if sec(x) is in the numerator or denominator but cos was mentioned in the OP)

From the graph of cos(3x) your phase shift will be units in the positive direction of the x-axis. Essentially it maps each value of to a value of (I introduced theta to make it easier to understand)

Spoiler:

edit: with regards to the image the phase difference should be pi/3

Sec(x) is in the numerator
4 is the demoninator

What are the x-axis values in between -2/pi to 2/pi and 2/pi to pi?