/*! 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 3 Sketch the graph of $$ |x-3|... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 3

Sketch the graph of $$ |x-3|+1 $$

Short Answer

Expert verified
The graph is a V-shape with vertex at \((3,1)\), translated 3 units right and 1 unit up from the origin.

Step by step solution

01

Identify the basic function

The basic function to consider in this case is the absolute value function, which is written as \(|x|\). The given function is \(|x-3|+1\), which is a transformation of \(|x|\).
02

Identify the transformations

Notice the transformations applied to the basic absolute value function. The term \(x-3\) indicates a horizontal shift, moving the graph 3 units to the right. The \(+1\) outside the absolute value function indicates a vertical shift, moving the graph 1 unit up.
03

Plot the vertex

Since the graph is moved 3 units to the right and 1 unit up, the new vertex of the function will be at the point \((3,1)\).
04

Sketch the V-shape

The absolute value function has a distinctive V-shape. Starting from the vertex \((3, 1)\), draw two lines: one line going up and to the left with a slope of -1, and another line going up and to the right with a slope of 1.
05

Label the graph

Label the vertex \((3, 1)\) and the axes on the graph. Ensure that the shape clearly shows the V-formation stemming from the vertex.

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!

Key Concepts

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

Transformations of Functions
Transformations of functions are adjustments made to the basic graphs. These changes help create new graphs from basic ones. The key transformations include:
  • Horizontal Shifts
  • Vertical Shifts
  • Reflections
  • Stretching and Shrinking
In our example, the function \(|x-3| + 1\) is a transformed version of the basic absolute value function \(|x|\). By understanding how each transformation works, we can easily sketch new functions.
Graphing Absolute Value Functions
An absolute value function generally looks like \(|x|\) and forms a V-shape. This graph is symmetrical around the y-axis and has its vertex (lowest point) at \(0, 0\). The general form of an absolute value function is \(|x-h| + k\), where specific transformations shape the graph.
  • Vertex: The lowest or highest point of the V-shape, determined by the transformations.
  • Slopes: Two lines going outward from the vertex, both having slopes of 1 or -1 depending on direction.
By recognizing these aspects, graphing absolute value functions becomes simpler.
Horizontal and Vertical Shifts
Shifting is moving the graph without changing its shape. It can be horizontal (left or right) or vertical (up or down).
Horizontal Shift:
If the function is \|x-h|\, the graph of \|x|\ shifts \(h\) units to the right if \(h > 0\), and \(h\) units to the left if \(h < 0\). For example, \|x-3|\ shifts the graph 3 units to the right.

Vertical Shift:
If the function is \|x| + k\, the graph of \|x|\ shifts \(k\) units up if \(k > 0\), and \(k\) units down if \(k < 0\). For instance, the \(+1\) in \|x-3| + 1\ shifts the graph 1 unit up.
In our exercise, the graph \|x-3| + 1\ is transformed from \|x|\ by shifting it 3 units to the right and 1 unit up.

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

Sketch the following functions on the same axes: a \(f(r)=\pi r^{2}\) and \(g(r)=r^{2}\) with domain \(0 \leq r \leq 10\) b \(f(v)=\frac{v^{2}}{20}\) and \(g(v)=v^{2}\) with domain \(0 \leq v \leq 10\) c \(f(T)=\frac{10}{T}\) and \(g(T)=\frac{1}{T}\) with domain \(0

if [controlengineering] The steady-state error, \(e_{\mathrm{ss}}\), of a system is given by $$ e_{5 \mathrm{~s}}=\lim _{s \rightarrow 0} \frac{M}{1+G(s)} $$ where \(M\) is a constant. Given that $$ G(s)=\left(\frac{k_{1}+k_{2}}{s}\right) \times\left(\frac{1}{1+s \tau}\right) $$ where \(k_{1}, k_{2}\) and \(\tau\) are non-zero constants, evaluate \(e_{\mathrm{ss}}\). Employ algebra to find the limits of questions \(\mathbf{8}\) and \(\mathbf{9}\).

[ mechanics] The displacement, \(s(t)\), of a particle is given by $$ s(t)=t^{3}-t^{2}+5 $$

Let \(f(x)=x-1\) and \(g(x)=\sqrt{x}(x \geq 0)\) be functions. Express the following in terms of \(f\) and \(g\) : \(\mathbf{i} \sqrt{x}-1 \quad\) ii \(\sqrt{x-1} \quad\) iii \(\sqrt{x-1}+7\) iv \(\sqrt{x^{2}-2 x+1}\)

?[control engineering] A transfer function, \(G(s)\), of a system is given by $$ G(s)=\frac{10}{(s-2)} \times \frac{s}{\left(s^{2}+2 s-5\right)} $$ Simplify \(\frac{G(s)}{1+G(s) H(s)}\) where $$ H(s)=s+3 $$
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Probability and Statistics

Read Explanation

Logic and Functions

Read Explanation

Pure Maths

Read Explanation

Mechanics Maths

Read Explanation

Calculus

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.