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 $\displaystyle y = 4 - \dfrac{3}{4 \sec(3x-\pi)} \text { or } y = 4-\dfrac{3\sec(3x-\pi)}{4}$
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
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 $\displaystyle \frac{\pi}{3}$. Therefore if you draw the graph of $\displaystyle \cos(3x)$ (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 $\displaystyle \frac{\pi}{3}$ units in the positive direction of the x-axis. Essentially it maps each value of $\displaystyle \theta$ to a value of $\displaystyle \left(\theta, \theta+\frac{\pi}{3}\right)$ (I introduced theta to make it easier to understand)
Spoiler:
edit: with regards to the image the phase difference should be pi/3