from the diagram, i can say that the

amplitude,ais 3 but the ans given is 1.5.

y intercept, b is 0 but the ans given is 1.5

and how to determine then?? the ans for n is 4.

i need the above ans before continue to b question..tq

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- Oct 26th 2009, 02:27 PMnikkamplitude for cos graph
from the diagram, i can say that the

amplitude,*a*is 3 but the ans given is 1.5.

y intercept, b is 0 but the ans given is 1.5

and how to determine the*n*?? the ans for n is 4.

i need the above ans before continue to b question..tq - Oct 26th 2009, 02:32 PMANDS!
I had a big thing typed out but lost it.

Long story short - their answers are correct. This graph has been flipped and shifted horizontally. Go back to what the graph of cosine looks like; if you still need help let me know. - Oct 26th 2009, 02:36 PMnikk
do u mean the attached grah already flipped and shifted..

i need your help on this

the for your quick reply - Oct 26th 2009, 02:43 PMANDS!
Well, yes the attached graph has already been flipped and shifted. . .but you already know that from the equation they give you (assuming they aren't being jerks):

-acos(nx)+b

If you work from scratch and start applying the transformations:

-cos(x): Graph of cosine is flipped (so 0,1 becomes 0,-1)

-acos(x): Graph of cosine is stretched/compressed by a factor of a.

-acos(nx): Graph of cosines period is adjusted from $\displaystyle 2\pi$ to $\displaystyle \frac{2\pi}{n}$

-acos(nx)+b: Graph of cosine is shifted vertically by b units.

You then end up with the graph you are looking at right now. It is up to you however to, using the values of points on this graph as well as an understanding of how to calculate amplitude, period and shifting, figure out what a, b, c and n are. - Oct 26th 2009, 02:58 PMnikk
tq for the above information and i will try to work out on it..

tq - Oct 26th 2009, 03:15 PMDebsta
How are you going with the problem? Thought I'd help.

The best way I think is to first draw in the equilibrium line (sometimes called mean value line) basically its the horizontal line through the middle of the graph. Thats at y=1.5. That's what b is.

a is the amplitude which is the max vertical distance between the equilibrium line and the curve, so a = 1.5. (The neg sign on a simply means the graph has been flipped)

The period of the function is = 2pi/n. On your graph the period is clearly 0.5pi. Therefore solve 2pi/n =0.5pi to give n=4.

Hope this helps.