Chapter 4.5, Problem 43E

To determine

To sketch : the graph of the function including two full periods

Explanation of Solution

Given information: $y=3\sin \left(x+\pi \right)$

Concept Involved:

Amplitude:

The constant factor $a$ in $y=a\sin x$ and $y=a\cos x$ as a scaling factor − a vertical stretch or vertical shrink of the basic curve. When $\left|a\right|>1,$ the basic curve is stretched, and when $0<\left|a\right|<1,$ the basic curve is shrunk. The result is that the graphs of $y=a\sin x$ and $y=a\cos x$ range between $–a$ and $a$ instead of between $-1$ and $1$ . The absolute value of $a$ is the amplitude of the function. The range of the function for $a>0$ is $-a\le y\le a$ .

The amplitude of $y=a\sin x$ and $y=a\cos x$ represents half the distance between the maximum and the minimum values of the function and is given by $Amplitude=\left|a\right|$ .

Period:

Let $b$ be a positive real number.

The period $y=a\sin bx$ and $y=a\cos bx$ is given by $Period=2\pi /b$

Note that when $0<b<1,$ the period of $y=a\sin bx$ is greater than $2\pi$ and represents a horizontal stretch of the basic curve. Similarly, when $b>1,$ the period of $y=a\sin bx$ is less than $2\pi$ and represents a horizontal shrink of the basic curve. These two statements are also true for $y=a\cos bx$ . When b is negative, rewrite the function using the identify $\sin \left(-x\right)=-\sin xor\cos \left(-x\right)=\cos x$ .

Horizontal Translation:

The constant c in the equation $y=a\sin \left(bx-c\right)$ and $y=a\cos \left(bx-c\right)$ results in horizontal translations (shifts) of the basics curves. For example, compare the graphs of $y=a\sin bx$ & $y=a\sin \left(bx-c\right)$ . The graph of $y=a\sin \left(bx-c\right)$ completes one cycle from $bx-c=0$ to $bx-c=2\pi$ . Solve for x to find that interval for one cycle is $\stackrel{Leftendpoint}{\overbrace{c/b}}\le x\le \stackrel{Rightendpoint}{\overbrace{c/b+\underset{Period}{\underbrace{2\pi /b}}}}$ . This implies that the period of $y=a\sin \left(bx-c\right)$ is $2\pi /b$ , and the graph of $y=a\sin bx$ is shifted by an amount $c/b$ . $PhaseShift=c/b$ .

Vertical Translation:

The constant $d$ in the equation $y=a\sin \left(bx-c\right)+d$ and $y=a\cos \left(bx-c\right)+d$ results in vertical translations of the basic curves. The shift is $d$ units up for $d>0$ and $d$ units down for $d<0.$ In other words, the graph oscillates about the horizontal line $y=d$ instead of about the x- axis.

Calculation:

Rewrite the given function $y=3\sin \left(x-\left(-\pi \right)\right)$ in the form $y=a\sin \left(bx-c\right)+d$

  $y=3\sin \left(1x-\left(-\pi \right)\right)+0$

Identify $a,b,c,\&d$ of our given function by comparing it with $y=a\sin \left(bx-c\right)+d$

  $a=3$ ; $b=1$ ; $c=-\pi$ ; $d=0$

Identify Amplitude, Period, Phase shift {Horizontal Shift}, Vertical shift& mid-line

Amplitude is $\left|a\right|$	Period is $2\pi /b$	Phase shift is $c/b$	Vertical Shift is $d$	Mid-line of function
$\left|3\right|=3$	$\frac{2\pi }{1}=2\pi$	$\begin{array}{l}\frac{c}{b}=\frac{-\pi }{1}\\ =-\pi \end{array}$	$d=0$	$y=0$

Graph:

  

Interpretation:

The graph of the function $y=3\sin \left(x+\pi \right)$ has one complete cycle in the interval $\left(-\pi ,\pi \right)$ .It has a midline $y=0$ , has amplitude as 3, maximum height of the function 3 andminimum height of the function is -3. The start point {Left end} of the graph would be $-\pi$ & the end point of one complete cycle is $\pi$ .