/*! 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 20 Use transformations of graphs to... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Use transformations of graphs to sketch the graphs of \(y_{1}, y_{2},\) and \(y_{3}\) by hand. Check by graphing in an appropriate viewing window of your calculator. $$y_{1}=3-|x|, \quad y_{2}=3-|3 x|, \quad y_{3}=3-\left|\frac{1}{3} x\right|$$

Short Answer

Expert verified
Vertical shift for \(y_1\), horizontal compression for \(y_2\), and horizontal stretch for \(y_3\).

Step by step solution

01

Understand the Base Function

The base function here is the absolute value function, which is given by \( y = |x| \). It forms a 'V' shape with its vertex at the origin (0, 0). The graph is symmetric with respect to the y-axis.
02

Graph the Original Function

With the base function \( y = |x| \), we start by plotting the basic V-shaped graph. It has a vertex at the origin (0,0) and opens upwards.
03

Apply Vertical Transformation to Graph \( y_1 = 3 - |x| \)

The expression \( y_1 = 3 - |x| \) represents a vertical shift of the base graph upwards by 3 units. Plot the graph with the vertex now at (0,3) instead of (0,0).
04

Apply Horizontal Scaling to Graph \( y_2 = 3 - |3x| \)

In \( y_2 = 3 - |3x| \), the \( |3x| \) term suggests a horizontal compression by a factor of 3. This means the graph of \( y = |x| \) will be squeezed towards the y-axis. The vertex remains at (0,3) since this transformation doesn't affect the y-intercept.
05

Apply Horizontal Stretching to Graph \( y_3 = 3 - \left|\frac{1}{3}x\right| \)

For \( y_3 = 3 - \left|\frac{1}{3}x\right| \), the \( \left|\frac{1}{3}x\right| \) represents a horizontal stretch by a factor of 3 of the base graph. The graph becomes wider. Again, the vertex stays at (0,3), maintaining the vertical shift.
06

Verify with Graphing Calculator

Using a graphing calculator, enter the equations \( y_1 = 3 - |x| \), \( y_2 = 3 - |3x| \), and \( y_3 = 3 - \left|\frac{1}{3}x\right| \). Check in an appropriate viewing window to ensure the graphs correspond to the transformations described.

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 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Absolute Value Function
The absolute value function is a fundamental tool in mathematics that is represented by the equation \( y = |x| \). This function generates a distinctive 'V' shaped graph with its vertex at the origin (0,0). The key feature of this graph is its symmetry about the y-axis, meaning it reflects evenly on both sides. The graph increases linearly as it moves away from the vertex in both directions. It is crucial to understand this base function as it serves as the starting point for more complex graph transformations.
  • The vertex is at (0,0).
  • The graph opens upwards.
  • The slope is 1 in both directions.
This understanding forms a foundation for learning vertical shifts and horizontal scaling of functions.
Vertical Shift
A vertical shift involves moving the graph of a function up or down without altering its shape. In the context of the absolute value function, consider the graph \( y_1 = 3 - |x| \). This equation shows a vertical shift because the entire graph moves upwards by 3 units. This results in the vertex moving from (0,0) to (0,3).
A vertical shift is straightforward:
  • Adding a constant to the function (e.g., \(+3 \)) moves the graph up.
  • Subtracting a constant moves the graph down.
The shape of the graph itself doesn't change; it simply relocates higher or lower on the y-axis. Understand this as a simple yet powerful transformation that can redefine the position of any function’s graph.
Horizontal Scaling
Horizontal scaling changes the width of the graph by stretching or compressing it along the x-axis. In the function \( y_2 = 3 - |3x| \), we see a horizontal compression. This compresses the graph towards the y-axis due to the factor of 3 multiplied inside the absolute value, meaning that what would normally happen at \( x = 3 \) now occurs at \( x = 1 \). This squeezes the graph into a more narrow shape.
Contrast this with the function \( y_3 = 3 - \left|\frac{1}{3}x\right| \), which represents a horizontal stretch. Here, the graph extends horizontally due to the factor of \( \frac{1}{3} \) inside the absolute value, so effects that occur at \( x = 1 \) in the original graph now occur at \( x = 3 \).
Key things to remember about horizontal scaling:
  • Multiplying the input by a factor greater than 1 compresses the graph.
  • Multiplying by a fraction stretches the graph.
Both transformations maintain the y-intercept which is affected by vertical shifts, such as moving to the point (0,3).
Graphing Calculator
Using a graphing calculator can be an effective way to visualize graph transformations like vertical shifts and horizontal scaling. With these tools, you can observe how equations like \( y_1 = 3 - |x| \), \( y_2 = 3 - |3x| \), and \( y_3 = 3 - \left|\frac{1}{3}x\right| \) transform from their base forms.
Here's how you can utilize a graphing calculator:
  • Enter each function into the calculator to see how the graph is affected by the transformations. Watch how the graph moves or stretches.
  • Adjust the viewing window to better observe these changes. Ensure the domain and range are set wide enough to capture all transformations.
Seeing these transformations on a calculator empowers learners to grasp how different algebraic manipulations affect graphical representation. It helps solidify their understanding by connecting algebraic theory to visual graphs.

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

Solve each equation or inequality graphically. $$|2 x+7|=6 x-1$$

Solve each problem. Systolic blood pressure is the maximum pressure produced by each heartbeat. Both low blood pressure and high blood pressure are cause for medical concern. Therefore, health care professionals are interested in a patient's "pressure difference from normal," or \(P_{d}\). If 120 is considered a normal systolic pressure, \(P_{d}=|P-120|,\) where \(P\) is the patient's recorded systolic pressure. For example, a patient with a systolic pressure \(P\) of 113 would have a pressure difference from normal of \(P_{d}=|P-120|=|113-120|=|-7|=7\) (a) Calculate the \(P_{d}\) value for a woman whose actual systolic pressure is 116 and whose normal value should be 125 (b) If a patient's \(P_{d}\) value is 17 and the normal pressure for his sex and age should be \(120,\) what are the two possible values for his systolic blood pressure?

Determine the difference quotient \(\frac{f(x+h)-f(x)}{h}\) (where \(h \neq 0\) ) for each function \(f\). Simplify completely. $$f(x)=x^{2}+2 x$$

For each pair of functions, (a) find \((f+g)(x),(f-g)(x),\) and \((f g)(x)\) (b) give the domains of the functions in part (a); (c) find \(\frac{f}{8}\) and give its domain; (d) find \(f \circ g\) and give its domain; and (e) find \(g \circ f\) and give its domain. Do not use a calculator. $$f(x)=\sqrt{x^{2}+3}, g(x)=x+1$$

When a thermal inversion layer is over a city, pollutants cannot rise vertically, but are trapped below the layer and must disperse horizontally. Assume that a factory smokestack begins emitting a pollutant at 8 A.M. and that the pollutant disperses horizontally over a circular area. Let \(t\) represent the time in hours since the factory began emitting pollutants \((t=0\) represents 8 A.M.) and assume that the radius of the circle of pollution is $$r(t)=2 t \quad \text{miles}.$$ The area of a circle of radius \(r\) is represented by $$\mathscr{A}(r)=\pi r^{2}$$ (a) Find \((\mathscr{A} \circ r)(t)\) (b) Interpret \((\mathscr{A} \circ r)(t)\) (c) What is the area of the circular region covered by the layer at noon? (d) Support your result graphically.

See all solutions

Recommended explanations on Math Textbooks

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.