is my graph of y = -2cos(2x+[pi/2])+1 correct?

[r75x]

is my graph of y = -2cos(2x+[pi/2])+1 correct?

amplitude = 2
period = (2pi)/(2) = pi
phase shift = -pi/2 * 1/2 = -pi/4
vertical shift = 1

I know that it's a bad sketch, but thanks for checking!

 

[tkhunny]

Looks like you got everything except the negative sign out in front.

 

[r75x]

oh, should my graph be flipped?

 

[tkhunny]

Right...

I like to figure out at least one point of reference. A nice min or max is often useful.

I know that cos(x) = -1 (a minimum) for x = pi. Then, 2x + pi/2 = pi ==> x = pi/4 ==> The graph should be at it's maximum at x = pi/4. That MAY BE what you have drawn. I really can't tell.

In any case, if your intent was to draw exactly one period of the function, you should end up where you started. \(\displaystyle f(-\pi/4)=f(3\pi/4)=-1\) You defintely don't have that.

 

[tkhunny]

You can also think on the Range a piece at a time, without worrying about period or phase shift.

f(x) = cos(x) [-1,1]
f(x) = 2*cos(x) [-2,2]
f(x) = -2*cos(x) [-2,2]
f(x) = -2*cos(x)+1 [-1,3]

Let me rephrase my original comments. Your words and equations appear quite reasonable and complete. The graph is not nearly so good as "bad sketch".