1. ## determaining trig equations

when refering to a graph...how do you find the horizontial translation? i know you use sin or cosine but i dont understand the method very clearly. can someone clarify this?

2. Originally Posted by Happy1234
when refering to a graph...how do you find the horizontial translation? i know you use sin or cosine but i dont understand the method very clearly. can someone clarify this?
If you have $y = a \sin (bx + c) + d$ the horizontal shift is given by d. Similarly for cos.

3. thanks for responding..however that wasnt exactly my question. iunderstand that d is equal to the H.T. but if from interpreting a graph i have to find the equation. how do i find Horizotial translation....

4. Originally Posted by mr fantastic
If you have $y = a \sin (bx + c) + d$ the horizontal shift is given by d. Similarly for cos.
Originally Posted by Happy1234
thanks for responding..however that wasnt exactly my question. iunderstand that d is equal to the H.T.
actually, d gives the vertical shift

the horizontal shift is given by: $- \frac cb$ in this form...

i prefer the form: $y = a \sin k(x - b) + c$

but if from interpreting a graph i have to find the equation. how do i find Horizotial translation....
well, know where the graphs should begin and track where they shifted to. for instance, you know that sin(0) = 0. so if you have a function that you are interpreting to be a sine graph and you see that sin(0) = 1, say, you know the shift will be $- \frac {\pi}2$ since you would see that $\sin \left( - \frac {\pi}2 \right) = 0$ for that graph. things like that

5. Originally Posted by Happy1234
thanks for responding..however that wasnt exactly my question. iunderstand that d is equal to the H.T. but if from interpreting a graph i have to find the equation. how do i find Horizotial translation....
First off: Whoops ... my mistake. I gave the vertical shift ...... The horizontal shift is given by -c/b.

As far as looking at a graph is concerned, you can get the horizontal shift by looking at where a simple turning point should be without any shift (for y = sin x there's a max turning point at 90 degrees, for y = cos x there's a max turning point at x = 0) and then seeing how far it has been shifted ....

6. Originally Posted by mr fantastic
As far as looking at a graph is concerned, you can get the horizontal shift by looking at where a simple turning point should be without any shift (for y = sin x there's a max turning point at 90 degrees, for y = cos x there's a max turning point at x = 0) and then seeing how far it has been shifted ....
that is perhaps a better way to look for it

7. thanks so much!