/*! 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 91 Begin by graphing the absolute v... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 91

Begin by graphing the absolute value function, \(f(x)=|x| .\) Then use transformations of this graph to graph the given function. $$h(x)=2|x+4|$$

Short Answer

Expert verified
The graph of \(h(x)=2|x+4|\) is like the graph of the absolute value function \(f(x)=|x|\), but stretched vertically due to the factor of 2 and shifted four units to left due to the \(+4\) inside the absolute value function, placing the vertex at the point (-4,0).

Step by step solution

01

Graph the Absolute Value Function

Start by drawing the graph of \(f(x)=|x|\). This function is shaped like a 'V' with the point of the 'V' located at the point (0,0). This point is called the vertex. The line goes upwards from this point in both positive and negative direction of x-axis forming a 'V' shape.
02

Apply the Scaling Factor

The next transformation to apply to the graph of \(f(x)=|x|\) is the scaling factor of 2 in the function \(h(x)=2|x+4|\). This will cause a vertical stretch of the graph of \(f(x)=|x|\). The 'V' becomes narrower or steeper.
03

Apply the Horizontal Shift

The final transformation is the horizontal shift caused by \(+4\) inside the absolute value function in \(h(x)=2|x+4|\). This shifts the vertex of the 'V' four units to the left (due to positive sign inside the absolute value function, which always shifts opposite to the traditional direction). The vertex will now be at the point (-4,0).

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 function $$f(x)=-0.00002 x^{3}+0.008 x^{2}-0.3 x+6.95$$ models the number of annual physician visits, \(f(x),\) by a person of age \(x .\) Graph the function in a [0,100,5] by [0,40,2] viewing rectangle. What does the shape of the graph indicate about the relationship between one's age and the number of annual physician visits? Use the \([\mathrm{TABLE}]\) or minimum function capability to find the coordinates of the minimum point on the graph of the function. What does this mean?

Complete the square and write the equation in standard form. Then give the center and radius of each circle and graph the equation. $$x^{2}+y^{2}+6 x+2 y+6=0$$

Determine whether each statement makes sense or does not make sense, and explain your reasoning. To avoid sign errors when finding \(h\) and \(k,\) I place parentheses around the numbers that follow the subtraction signs in a circle's equation.

Will help you prepare for the material covered in the next section. Write an algebraic expression for the fare increase if a 200 dollars plane ticket is increased to \(x\) dollars.

Complete the square and write the equation in standard form. Then give the center and radius of each circle and graph the equation. $$x^{2}+y^{2}+x+y-\frac{1}{2}=0$$
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Decision Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Calculus

Read Explanation

Geometry

Read Explanation

Mechanics Maths

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.