Pessimist Singularitarian

#2
December 18th, 2016,
04:11
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?