## anonymous 3 years ago how do I graph $y=-3cos(3x)$? I know that 3 is the amplitude I know that the graph will be upside down. I know that the period is $\frac{2\pi}{3}$

1. stamp

Copy the image and begin by marking your amplitude, the highest and lowest point of the wave.|dw:1360290026921:dw|

2. anonymous

|dw:1360290074055:dw|

3. stamp

now label $(\frac{2\pi}{3},f(\frac{2\pi}{3}))$this point is the end of your period.

4. anonymous

|dw:1360290246963:dw|

5. stamp

$f(x)=-3cos(3x)$$f(\frac{3\pi}{2})=-3cos(3(\frac{3\pi}{2}))=-3cos(9\pi/2)$

6. anonymous

it's $\frac{2\pi}{3}$ for the period

7. stamp

@JenniferSmart1 My fault. You evaluated it correctly, I would have gotten 0 for f(3pi/2) while f(2pi/3) is -3. Ok, so now can you finish the graph from here? So far you have done an excellent job.

8. anonymous

|dw:1360290496676:dw|

9. anonymous

?

10. stamp

Could we label the highest point (x, f(x)) if we were asked?

11. anonymous

Let's see $( \frac{\pi}{3} , 3)$ because $\frac{2\pi}{3\cdot2}$

12. stamp

Ok! I think your graph seems pretty complete, well done. Here is the actual graph for reference http://www.wolframalpha.com/input/?i=-3cos%283x%29

13. anonymous

I'm guessing because its half way between 2pi/3

14. anonymous

Thanks!