Graphing Trig Function: Amplitude 4, Period (2\pi/3), Range [-4,4]

[Dasysaaaa]

I need some help graphing this trig function.

f(x)=4sin(3x-2)

When Graphing, the points should be over 2 periods and 9 points.

I have:
-Amplitude: 4
-Period=(2\pi/3)
-Range=[-4,4]

I need help on:
-Graphing the points over 2 periods and 9 periods (parent function and f(x))
-Table of 9 points (f(x) and parent function)
-Increments (f(x) and parent function)

Thank You in advance if You can help me!(Blush)

 

[MarkFL]

I agree with what you've done so far regarding the amplitude, period and range. Now, if I were going to sketch a graph of $f(x)$, I would write it in the form:

\(\displaystyle f(x)=4\sin\left(3\left(x-\frac{2}{3}\right)\right)\)

In this form, we can see that $f$ is shifted \(\displaystyle \frac{2}{3}\) units to the right compared to $y=4\sin(3x)$. Now, if I was going to graph $f$ over just one period, I would choose the domain:

\(\displaystyle \left[\frac{2}{3},\frac{2}{3}(\pi+1)\right]\)

On this domain, the sinusoid described by $f$ will begine at 0, movie to to 4, then back down to 0, continue down to -4, then move back to to 0, completing 1 cycle.

I would divide this domain into 4 subdivisions of equal width corresponding to the extrema and equilibria (that is, the zero, maximum and minimum values for $f$). This gives us the $x$-values:

\(\displaystyle x=\frac{2}{3}+\frac{k}{4}\cdot\frac{2}{3}\pi=\frac{2}{3}+\frac{k}{6}\pi\) where $k\in\{0,1,2,3,4\}$

Putting all this together, this gives us the 5 points:

\(\displaystyle \left(\frac{2}{3},0\right),\,\left(\frac{2}{3}+\frac{1}{6}\pi,4\right),\,\left(\frac{2}{3}+\frac{1}{3}\pi,0\right),\,\left(\frac{2}{3}+\frac{1}{2}\pi,-4\right),\,\left(\frac{2}{3}+\frac{2}{3}\pi,0\right)\)

If you continue for another period, this would give you 4 more points for a total of nine...can you continue?