/*! 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 48 Sketch the graph of the function... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 48

Sketch the graph of the function. (Include two full periods.) $$y=-10 \cos \frac{\pi x}{6}$$

Short Answer

Expert verified
First, plot the amplitude of the function, which is determined by the absolute value of the coefficient of the cosine function (10 in this case). Then, calculate the period of the function, which is determined by the reciprocal of the absolute value of the coefficient of \(x\) (which equates to 12). Finally, sketch the graph of the function for two complete cycles, remembering that for a cosine function, the function starts from its highest point at \(x=0\), goes down to the lowest point and back up to the starting point at \(x=12\), and then repeats this cycle until \(x=24\).

Step by step solution

01

Plotting the Amplitude

Begin by drawing the amplitude of the function. This is determined by the coefficient of the cosine function, which in this case is \(-10\). This means that the function will peak at \(10\) and valley at \(-10\) as cosine function ranges from \(-1\) to \(1\). Draw horizontal lines at \(y=10\) and \(y=-10\) to represent this amplitude.
02

Calculating the Period and Determining the Interval

The period of a cosine function is usually \(2\pi\). However, in this case, due to the coefficient of \(x\) in the cosine function which is \(\frac{\pi}{6}\), the period will be \(\frac{2\pi}{\frac{\pi}{6}}=12\). Meaning, the function will complete a full cycle from peak to peak (or valley to valley) in an interval of \(12\) units on the x-axis. Also, two full periods should be plotted, this means the x values will range from 0 to \(2*12=24\). Divide this interval into equal parts for easy plotting. For example, \(0, 6, 12, 18, 24\) can serve as key points.
03

Plotting the Function

Normally, a cosine wave starts from its highest point, goes down to its lowest point and comes back up to its highest point in one period. In this case, the highest point is 10 (at \(x=0\) and \(x=24\)), midway it hits the lowest point, which is \(-10\) (at \(x=12\)). It also crosses the x-axis at \(x=6\) and \(x=18\). With these key points, you can sketch the graph of the function, making sure it covers two full periods from \(x=0\) to \(x=24\).
04

Repeat the Cycle

Repeat the pattern for the second cycle from \(x=12\) to \(x=24\). Also, mark the key points: peak at \(x=24\), crosses the x-axis at \(x=18\) and valley at \(x=12\)

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

The numbers of hours \(H\) of daylight in Denver, Colorado, on the 15 th of each month are: \(1(9.67), 2(10.72), \quad 3(11.92), \quad 4(13.25)\) \(5(14.37), \quad 6(14.97), \quad 7(14.72), \quad 8(13.77), \quad 9(12.48)\) \(10(11.18), \quad 11(10.00), \quad 12(9.38) . \quad\) The month is represented by \(t,\) with \(t=1\) corresponding to January. A model for the data is \(H(t)=12.13+2.77 \sin \left(\frac{\pi t}{6}-1.60\right)\) (a) Use a graphing utility to graph the data points and the model in the same viewing window. (b) What is the period of the model? Is it what you expected? Explain. (c) What is the amplitude of the model? What does it represent in the context of the problem? Explain.

Write the function in terms of the sine function by using the identity $$A \cos \omega t+B \sin \omega t=\sqrt{A^{2}+B^{2}} \sin \left(\omega t+\arctan \frac{A}{B}\right).$$ Use a graphing utility to graph both forms of the function. What does the graph imply? $$f(t)=3 \cos 2 t+3 \sin 2 t$$

Graph the functions \(f\) and \(g .\) Use the graphs to make a conjecture about the relationship between the functions. $$f(x)=\sin x+\cos \left(x+\frac{\pi}{2}\right), \quad g(x)=0$$

Use a graphing utility to graph the function. Use the graph to determine the behavior of the function as \(x \rightarrow c\) (a) \(x \rightarrow\left(\frac{\pi}{2}\right)^{+}\) (b) \(x \rightarrow\left(\frac{\pi}{2}\right)^{-}\) (c) \(x \rightarrow\left(-\frac{\pi}{2}\right)^{+}\) (d) \(x \rightarrow\left(-\frac{\pi}{2}\right)^{-}\) $$f(x)=\sec x$$

Use a graphing utility to graph the function. Use the graph to determine the behavior of the function as \(x \rightarrow c\) (a) \(x \rightarrow\left(\frac{\pi}{2}\right)^{+}\) (b) \(x \rightarrow\left(\frac{\pi}{2}\right)^{-}\) (c) \(x \rightarrow\left(-\frac{\pi}{2}\right)^{+}\) (d) \(x \rightarrow\left(-\frac{\pi}{2}\right)^{-}\) $$f(x)=\tan x$$
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Geometry

Read Explanation

Statistics

Read Explanation

Calculus

Read Explanation

Pure Maths

Read Explanation

Applied Mathematics

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.