Trig functino help - tiny question

[K.ourt]

kay, so I have this equation: y=2sin3(x+(pi/6))-1 that I have to take through a table of transformations.

I can get from y=sin(x) to y=2sin(x), but from there I cant get to y+2sin3(x).

It would be a horizontal stretch by a factor of (1/3), right? Meaning I would multiply the x values by (1/3).

So for (pi/2) x (1/3) would I get like 6pi? But that doesnt check..

 

[daon]

kay, so I have this equation: y=2sin3(x+(pi/6))-1 that I have to take through a table of transformations.

I can get from y=sin(x) to y=2sin(x), but from there I cant get to y+2sin3(x).

It would be a horizontal stretch by a factor of (1/3), right? Meaning I would multiply the x values by (1/3).

So for (pi/2) x (1/3) would I get like 6pi? But that doesnt check..

For cosine the period will change to: \(\displaystyle \frac{2 \pi}{3}\)

Period change is found by (Old period)/(what is multiplying the angle)

 

[soroban]

Hello, K.ourt!

Let's take baby-steps . . .

I have this equation: \(\displaystyle y\:=\:2\cdot\sin\left[3\left(x\,+\,\frac{\pi}{6}\right)\right]\,-\,1\)

that I have to take through a table of transformations.

We already know the graph of: \(\displaystyle \,y\:=\:\sin x\)

Now, we start "inside" the function.

The \(\displaystyle \,+\,\frac{\pi}{6}\) moves the graph \(\displaystyle \frac{\pi}{6}\) units to the left.

The \(\displaystyle \,3\) compresses the cycle by a factor of \(\displaystyle \frac{1}{3}\).

The \(\displaystyle \,2\) gives it an amplitude of 2.

And the \(\displaystyle \,-1\) lowers the graph one unit.