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

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The discussion focuses on graphing the trigonometric function f(x)=4sin(3x-2) with an amplitude of 4, a period of (2π/3), and a range of [-4,4]. Participants emphasize the importance of graphing over two periods and identifying nine key points, including the function's maximum, minimum, and zero crossings. A suggested approach involves rewriting the function to highlight its phase shift and determining the appropriate domain for one period. The conversation includes calculating specific x-values for the graph and encourages further exploration to complete the nine points needed. Overall, the thread provides guidance on effectively visualizing the function's behavior.
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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)
 
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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:

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

In this form, we can see that $f$ is shifted $$\frac{2}{3}$$ 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:

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

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:

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

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

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

If you continue for another period, this would give you 4 more points for a total of nine...can you continue?
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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