Chapter 4.4, Problem 96E

Solution

To identify: The A, B and C The graphs of the sine and cosine functions are waveforms like the figure below. By correctly labeling the coordinates of points A, B, and C, you will get the graph of the function given.

  $y=3\sin \left(2x-\pi \right)$ A is the first x-intercept on the right of the y-axis.

  

The point of A, B and C $\left(\frac{\pi }{2},0\right)$ , $\left(\frac{\pi }{4},2\right)$ , $\left(\frac{3\pi }{4},0\right)$ respectively

Given information

The given equation is $y=3\sin \left(2x-\pi \right)$ A is the first x-intercept on the right of the y-axis.

  

Explanation:

Consider the given information

The graph of sinusoidal function of the from

  $y=a\sin \left(b\left(x-h\right)\right)+kory=a\cos \left(b\left(x-h\right)\right)+k$ have the following characteristics

  $\begin{array}{c}amplitude=\left|a\right|\\ period=\frac{2\pi }{\left|b\right|}\end{array}$

Applying this concept to the given function $y=3\sin \left(2x-\pi \right)$

We have $a=3andb=2$

So

  $\begin{array}{c}amplitude=\left|3\right|\\ =3\end{array}$

  $\begin{array}{c}period=\frac{2\pi }{\left|2\right|}\\ =\frac{2\pi }{2}\\ =\pi \end{array}$

The amplitude and period of the sinusoidal function are $3and\pi$ respectively The sinusoidal function $\frac{\pi }{2}$ unit to the right

Based from the given we found out that point A is the x − intercept (y = 0) on the right of

the y − axis Since the sinusoid is shifted by $\frac{\pi }{2}$ units to the right then the coordinate of point A is given by $\left(\frac{\pi }{2},0\right)$

Hence point A lies at $\left(\frac{\pi }{2},0\right)$

Use the computed amplitude and period from step 1 to determine the coordinate of the point B.

Notice that point B lies at the maximum and at $\frac{1}{4}$ of a period hence the coordinate of point B is given by

  $\begin{array}{c}\left(x+\frac{p}{4},a\right)\\ \left(\frac{\pi }{2}+\frac{\pi }{4},3\right)\\ \left(\frac{3\pi }{4},3\right)\end{array}$

Hence point B lies at $\left(\frac{3\pi }{4},3\right)$

Notice that the cycle ends at point C which lies at the x − axis $\left(y=0\right)$ Using this information and the computed period from step 1 we have

  $\begin{array}{c}\left(x+p,0\right)\\ \left(\frac{\pi }{2}+\pi ,0\right)\\ \left(\frac{3\pi }{2},0\right)\end{array}$

Hence point C lies at $\left(\frac{3\pi }{2},0\right)$

The point of A, B and C $\left(\frac{\pi }{2},0\right)$ , $\left(\frac{\pi }{4},2\right)$ , $\left(\frac{3\pi }{4},0\right)$ respectively