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

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- Dec 21st 2010, 07:33 AMlinipopentGraph 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 - Dec 21st 2010, 07:40 AMe^(i*pi)
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 - Dec 21st 2010, 11:18 AMlinipopent
Sorry for Double post check down

- Dec 21st 2010, 11:33 AMlinipopent
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? - Dec 22nd 2010, 08:26 AMe^(i*pi)
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 - Dec 22nd 2010, 10:02 AMlinipopent
Sec(x) is in the numerator

4 is the demoninator - Dec 24th 2010, 10:08 AMlinipopent
What are the x-axis values in between -2/pi to 2/pi and 2/pi to pi?