/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 104 Use a graphing utility to graph ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 104

Use a graphing utility to graph two periods of the function. $$y=3 \sin (2 x-\pi)+5$$

Short Answer

Expert verified
The function \(y=3 \sin (2 x-\pi)+5\) is graphed by understanding the amplitude, period, phase shift and vertical shift of the function and applying these modifications to the basic sinusoidal graph. The final graph depicts a sine function with amplitude of 3, a period of \(\pi\), a phase shift of \(\pi\), and an upward shift of 5 units.

Step by step solution

01

Identify & Understand the Function Components

The function given is \(y=3 \sin (2 x-\pi)+5\). Here, '3' is the amplitude, '2' is the frequency which determines the period of the function, \(-\pi\) is the phase shift and '5' is the vertical shift. The effect of these components is as follows: Amplitude ('3') will scale the function vertically. Frequency ('2') will adjust the period of the function. It is inversely proportional to the period or the distance of one full cycle of the wave. Phase shift (\(-\pi\)) will move the function horizontally on the x-axis. And finally, the vertical shift ('5') will move the function up or down along the y-axis.
02

Graphing the Basic Function

Before modifications, graph a basic sine function which has an amplitude of 1, no phase shift, no vertical shift and a period of \(2\pi\). It starts from the origin (0,0), peaks at \(\pi/2\), goes to zero at \(\pi\), reaches its minimum at \(3\pi/2\), and goes back to zero at \(2\pi\), completing one period. Now, realize that the modifications will alter this standard pattern of the sine wave.
03

Apply the Amplitude

This function's amplitude is 3. So, the maximum and minimum values of the function, which were 1 and -1, respectively, will now be 3 and -3. Change the range on your graph to reflect this.
04

Apply the Period

The period of the function can be calculated as \(2\pi / |2| = \pi\). Thus, each cycle of our function will now be completed over a range of \(\pi\) instead of \(2\pi\). Adjust your graph to depict this.
05

Apply the Phase Shift

The phase shift of the function is -蟺, which will move our function 蟺 units to the right. Incorporate this shift into your graph.
06

Apply the Vertical Shift

Lastly, add the vertical shift of 5 units upwards. All points from the previously drawn graph should be shifted up by 5 units.
07

Plot the Graph

Now plot the final graph with the graphing utility using the parameters and adjustments defined in the previous steps. Remember to plot two periods of the function.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Determine whether each statement makes sense or does not make sense, and explain your reasoning. Because \(y=\sin x\) has an inverse function if \(x\) is restricted to \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right],\) they should make restrictions easier to remember by also using \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) as the restriction for \(y=\cos x\).

Use a sketch to find the exact value of each expression. $$ \csc \left[\cos ^{-1}\left(-\frac{\sqrt{3}}{2}\right)\right] $$

Determine the domain and the range of each function. $$ f(x)=\cos ^{-1}(\cos x) $$

For years, mathematicians were challenged by the following problem: What is the area of a region under a curve between two values of \(x ?\) The problem was solved in the seventeenth century with the development of integral calculus. Using calculus, the area of the region under \(y=\frac{1}{x^{2}+1},\) above the \(x\) -axis, and between \(x=a\) and \(x=b\) is \(\tan ^{-1} b-\tan ^{-1} a\). Use this result, shown in the figure, to find the area of the region under \(y=\frac{1}{x^{2}+1}\) above the \(x\) -axis, and between the values of a and b given in Exercises \(97-98\). (GRAPH CANNOT COPY) \(a=0\) and \(b=2\)

Determine the domain and the range of each function. $$ f(x)=\sin ^{-1}(\cos x) $$
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Pure Maths

Read Explanation

Statistics

Read Explanation

Calculus

Read Explanation

Discrete Mathematics

Read Explanation

Probability and Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.