Sin, cos, sec, and csc graphs. HELP!!

[Youhaveemily]

So I'm new to this and I don't know if this is how you ask a question or not so I'm going to continue this anyways. I'm having trouble with graphing these graphs and I don't really understand what I'm doing. I have this equation: y= -2cos(2x+(pi/3))-1

now are these qualities about the graph correct?
Amplitude:2 (flips)
period: pi
vertical translation: 1 (down)
phase shift: pi/3 (left)

also, on the graph I don't understand what points come before pi on the graph. Like where would pi/3 be on the graph? Or 3pi/4, anything like that. What are the points?

 

[MarkFL]

You did an excellent job of posting! You clearly stated the problem (with proper bracketing characters too) and showed your work! We couldn't ask for anything better. :cool:

Everything looks good except for the phase shift. If we write the function as:

$\displaystyle y=-2\cos\left(2\left(x+\dfrac{\pi}{6} \right) \right)-1$

We see the phase shift is $\displaystyle \dfrac{\pi}{6}$ units to the left.

To graph this now, consider that the period is $\displaystyle \pi$ units. So, we could consider the period $\displaystyle \left[-\dfrac{\pi}{6},-\dfrac{\pi}{6}+\pi \right]=\left[-\dfrac{\pi}{6},\dfrac{5\pi}{6} \right]$.

Now, we know the curve varies sinusoidally between $\displaystyle y=-1\pm2$ or $\displaystyle -3\le y\le1$.

At the beginning of the period, we are at the minimum value, at 1/4 of the way through the period we are at equilibrium, halfway through we are at the maximum, three quarters of the way through we are back at equilibrium, and at the end, we have returned to the minimum. So, we simply begin at $\displaystyle x=-\dfrac{\pi}{6}$ and increment by one-quarter period, i.e., $\displaystyle \dfrac{\pi}{4}$ to get the following points on the curve:

$\displaystyle \left(-\dfrac{\pi}{6},-3 \right)$

$\displaystyle \left(-\dfrac{\pi}{6}+\dfrac{\pi}{4},-1 \right)=\left(\dfrac{\pi}{12},-1 \right)$

$\displaystyle \left(-\dfrac{\pi}{6}+\dfrac{2\pi}{4},1 \right)=\left(\dfrac{\pi}{3},1 \right)$

$\displaystyle \left(-\dfrac{\pi}{6}+\dfrac{3\pi}{4},-1 \right)=\left(\dfrac{7\pi}{12},-1 \right)$

$\displaystyle \left(-\dfrac{\pi}{6}+\dfrac{4\pi}{4},-3 \right)=\left(\dfrac{5\pi}{6},-3 \right)$

Plot these points, and connect them with a sinusoidal curve, and you have a reasonable sketch of the curve over 1 period.

 

[dsk2]

Hey Youhaveemily,

Type y= -2cos(2x+(pi/3))-1 in the google search bar. You will see the plot. Isnt it cool!

Now, change the amplitude, phase, translation amount and see what you get. Try to vary them gradually by both increasing and decreasing the values. Change the signs and so on. Delete part of the equation and see how it affects the plot. Play around.

Then you will understand how each constant affects the plot and what it means. Read about the formal definitions for a better understanding.

Cheers,
Sai.