assignment 1 number order and absolute value

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Course: 6th grade   >   Unit 5

• Absolute value examples

Intro to absolute value

• Meaning of absolute value
• Finding absolute values
• Absolute value review

Let's practice!

• Your answer should be
• an integer, like 6 ‍  
• a simplified proper fraction, like 3 / 5 ‍  
• a simplified improper fraction, like 7 / 4 ‍  
• a mixed number, like 1   3 / 4 ‍  
• an exact decimal, like 0.75 ‍  
• a multiple of pi, like 12   pi ‍   or 2 / 3   pi ‍  

The absolute value symbol

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Understand ordering and absolute value of rational numbers.

More specific topics in understand ordering and absolute value of rational numbers..

• Interpret statements of inequality as statements about the relative position of two numbers on a number line diagram.
• Understand the absolute value of a rational number as its distance from 0 on the number line; interpret absolute value as magnitude for a positive or negative quantity in a real-world situation.
• Distinguish comparisons of absolute value from statements about order.

Popular Tutorials in Understand ordering and absolute value of rational numbers.

What's an Inequality?

Inequalities come up all the time when you're working algebra problems. In this tutorial you'll learn what an inequality is, and you'll see all the common inequality symbols that you're likely to see :)

What Does Absolute Value Mean?

In this tutorial you'll see how you can think of absolute value in a very intuitive way. Let us know if you have any questions about it!

How Do You Compare Two Decimals Using a Number Line?

If you have two decimals and you want to know which is larger, you can use a number line to help you compare! Follow along with this tutorial to see how to use a number line to compare decimals.

What Does Absolute Value Mean In the Real World?

You may know how to calculate the absolute value of a number, but what are you really finding? This tutorial uses a real world example to help you gain a better understanding of absolute value.

How Do You Put Integers in Order Using a Number Line?

Putting numbers in order can help you better understand how the numbers are related. This tutorial shows you how to put positive and negative temperatures in order using a number line!

How Do You Compare Integers Using a Number Line?

Trying to figure out if a negative number is larger than another can be a little tricky. To make things easier, you could use a number line! This tutorial shows you how to use a number line to compare two negative numbers and determine which is larger.

What is the Number Line?

A number line is a way we can visually represent numbers. This tutorial gives you a great introduction to the number line and shows you how to graph numbers on the number line in order to compare them. Check it out!

How Do You Put Fractions and Decimals in Order?

Ordering numbers from least to greatest? Are the numbers in different forms? To make comparing easier, convert all the numbers to decimals. Then, plot those decimals on a number line and compare them! This tutorial shows you how!

Related Topics

Other topics in apply and extend previous understandings of numbers to the system of rational numbers. :.

• Understand that positive and negative numbers are used together to describe quantities having opposite directions or values (e.g., temperature above/below zero, elevation above/below sea level, credits/debits, positive/negative electric charge); use positive and negative numbers to represent quantities in real-world contexts, explaining the meaning of 0 in each situation.
• Understand a rational number as a point on the number line. Extend number line diagrams and coordinate axes familiar from previous grades to represent points on the line and in the plane with negative number coordinates.
• Solve real-world and mathematical problems by graphing points in all four quadrants of the coordinate plane. Include use of coordinates and absolute value to find distances between points with the same first coordinate or the same second coordinate.
• Terms of Use

3.6 Absolute Value Functions

Learning objectives.

In this section, you will:

• Graph an absolute value function.
• Solve an absolute value equation.

Until the 1920s, the so-called spiral nebulae were believed to be clouds of dust and gas in our own galaxy, some tens of thousands of light years away. Then, astronomer Edwin Hubble proved that these objects are galaxies in their own right, at distances of millions of light years. Today, astronomers can detect galaxies that are billions of light years away. Distances in the universe can be measured in all directions. As such, it is useful to consider distance as an absolute value function. In this section, we will continue our investigation of absolute value functions .

Understanding Absolute Value

Recall that in its basic form f ( x ) = | x | , f ( x ) = | x | , the absolute value function is one of our toolkit functions. The absolute value function is commonly thought of as providing the distance the number is from zero on a number line. Algebraically, for whatever the input value is, the output is the value without regard to sign. Knowing this, we can use absolute value functions to solve some kinds of real-world problems.

Absolute Value Function

The absolute value function can be defined as a piecewise function

Using Absolute Value to Determine Resistance

Electrical parts, such as resistors and capacitors, come with specified values of their operating parameters: resistance, capacitance, etc. However, due to imprecision in manufacturing, the actual values of these parameters vary somewhat from piece to piece, even when they are supposed to be the same. The best that manufacturers can do is to try to guarantee that the variations will stay within a specified range, often ±1%, ± 5%, ±1%, ± 5%, or ± 10% . ± 10% .

Suppose we have a resistor rated at 680 ohms, ± 5 % . ± 5 % . Use the absolute value function to express the range of possible values of the actual resistance.

We can find that 5% of 680 ohms is 34 ohms. The absolute value of the difference between the actual and nominal resistance should not exceed the stated variability, so, with the resistance R R in ohms,

Students who score within 20 points of 80 will pass a test. Write this as a distance from 80 using absolute value notation.

Graphing an Absolute Value Function

The most significant feature of the absolute value graph is the corner point at which the graph changes direction. This point is shown at the origin in Figure 2 .

Figure 3 shows the graph of y = 2 | x – 3 | + 4. y = 2 | x – 3 | + 4. The graph of y = | x | y = | x | has been shifted right 3 units, vertically stretched by a factor of 2, and shifted up 4 units. This means that the corner point is located at ( 3 , 4 ) ( 3 , 4 ) for this transformed function.

Writing an Equation for an Absolute Value Function Given a Graph

Write an equation for the function graphed in Figure 4 .

The basic absolute value function changes direction at the origin, so this graph has been shifted to the right 3 units and down 2 units from the basic toolkit function. See Figure 5 .

We also notice that the graph appears vertically stretched, because the width of the final graph on a horizontal line is not equal to 2 times the vertical distance from the corner to this line, as it would be for an unstretched absolute value function. Instead, the width is equal to 1 times the vertical distance as shown in Figure 6 .

From this information we can write the equation

Note that these equations are algebraically equivalent—the stretch for an absolute value function can be written interchangeably as a vertical or horizontal stretch or compression. Note also that if the vertical stretch factor is negative, there is also a reflection about the x-axis.

If we couldn’t observe the stretch of the function from the graphs, could we algebraically determine it?

Yes. If we are unable to determine the stretch based on the width of the graph, we can solve for the stretch factor by putting in a known pair of values for x x and f ( x ) . f ( x ) .

Now substituting in the point (1, 2)

Write the equation for the absolute value function that is horizontally shifted left 2 units, is vertically reflected, and vertically shifted up 3 units.

Do the graphs of absolute value functions always intersect the vertical axis? The horizontal axis?

Yes, they always intersect the vertical axis. The graph of an absolute value function will intersect the vertical axis when the input is zero.

No, they do not always intersect the horizontal axis. The graph may or may not intersect the horizontal axis, depending on how the graph has been shifted and reflected. It is possible for the absolute value function to intersect the horizontal axis at zero, one, or two points (see Figure 7 ).

Solving an Absolute Value Equation

In Other Type of Equations , we touched on the concepts of absolute value equations. Now that we understand a little more about their graphs, we can take another look at these types of equations. Now that we can graph an absolute value function, we will learn how to solve an absolute value equation. To solve an equation such as 8 = | 2 x − 6 | , 8 = | 2 x − 6 | , we notice that the absolute value will be equal to 8 if the quantity inside the absolute value is 8 or -8. This leads to two different equations we can solve independently.

Knowing how to solve problems involving absolute value functions is useful. For example, we may need to identify numbers or points on a line that are at a specified distance from a given reference point.

An absolute value equation is an equation in which the unknown variable appears in absolute value bars. For example,

Solutions to Absolute Value Equations

For real numbers A A and B B , an equation of the form | A | = B , | A | = B , with B ≥ 0 , B ≥ 0 , will have solutions when A = B A = B or A = − B . A = − B . If B < 0 , B < 0 , the equation | A | = B | A | = B has no solution.

Given the formula for an absolute value function, find the horizontal intercepts of its graph .

• Isolate the absolute value term.
• Use | A | = B | A | = B to write A = B A = B or −A = B , −A = B , assuming B > 0. B > 0.
• Solve for x . x .

Finding the Zeros of an Absolute Value Function

For the function f ( x ) = | 4 x + 1 | − 7 , f ( x ) = | 4 x + 1 | − 7 , find the values of x x such that f ( x ) = 0. f ( x ) = 0.

The function outputs 0 when x = 3 2 x = 3 2 or x = − 2. x = − 2. See Figure 8 .

For the function f ( x ) = | 2 x − 1 | − 3 , f ( x ) = | 2 x − 1 | − 3 , find the values of x x such that f ( x ) = 0. f ( x ) = 0.

Should we always expect two answers when solving | A | = B ? | A | = B ?

No. We may find one, two, or even no answers. For example, there is no solution to 2 + | 3 x − 5 | = 1. 2 + | 3 x − 5 | = 1.

Access these online resources for additional instruction and practice with absolute value.

• Graphing Absolute Value Functions
• Graphing Absolute Value Functions 2

3.6 Section Exercises

How do you solve an absolute value equation?

How can you tell whether an absolute value function has two x -intercepts without graphing the function?

When solving an absolute value function, the isolated absolute value term is equal to a negative number. What does that tell you about the graph of the absolute value function?

How can you use the graph of an absolute value function to determine the x -values for which the function values are negative?

Describe all numbers x x that are at a distance of 4 from the number 8. Express this set of numbers using absolute value notation.

Describe all numbers x x that are at a distance of 1 2 1 2 from the number −4. Express this set of numbers using absolute value notation.

Describe the situation in which the distance that point x x is from 10 is at least 15 units. Express this set of numbers using absolute value notation.

Find all function values f ( x ) f ( x ) such that the distance from f ( x ) f ( x ) to the value 8 is less than 0.03 units. Express this set of numbers using absolute value notation.

For the following exercises, find the x - and y -intercepts of the graphs of each function.

f ( x ) = 4 | x − 3 | + 4 f ( x ) = 4 | x − 3 | + 4

f ( x ) = − 3 | x − 2 | − 1 f ( x ) = − 3 | x − 2 | − 1

f ( x ) = − 2 | x + 1 | + 6 f ( x ) = − 2 | x + 1 | + 6

f ( x ) = − 5 | x + 2 | + 15 f ( x ) = − 5 | x + 2 | + 15

f ( x ) = 2 | x − 1 | − 6 f ( x ) = 2 | x − 1 | − 6

f ( x ) = | − 2 x + 1 | − 13 f ( x ) = | − 2 x + 1 | − 13

f ( x ) = − | x − 9 | + 16 f ( x ) = − | x − 9 | + 16

For the following exercises, graph the absolute value function. Plot at least five points by hand for each graph.

y = | x − 1 | y = | x − 1 |

y = | x + 1 | y = | x + 1 |

y = | x | + 1 y = | x | + 1

For the following exercises, graph the given functions by hand.

y = | x | − 2 y = | x | − 2

y = − | x | y = − | x |

y = − | x | − 2 y = − | x | − 2

y = − | x − 3 | − 2 y = − | x − 3 | − 2

f ( x ) = − | x − 1 | − 2 f ( x ) = − | x − 1 | − 2

f ( x ) = − | x + 3 | + 4 f ( x ) = − | x + 3 | + 4

f ( x ) = 2 | x + 3 | + 1 f ( x ) = 2 | x + 3 | + 1

f ( x ) = 3 | x − 2 | + 3 f ( x ) = 3 | x − 2 | + 3

f ( x ) = | 2 x − 4 | − 3 f ( x ) = | 2 x − 4 | − 3

f ( x ) = | 3 x + 9 | + 2 f ( x ) = | 3 x + 9 | + 2

f ( x ) = − | x − 1 | − 3 f ( x ) = − | x − 1 | − 3

f ( x ) = − | x + 4 | − 3 f ( x ) = − | x + 4 | − 3

f ( x ) = 1 2 | x + 4 | − 3 f ( x ) = 1 2 | x + 4 | − 3

Use a graphing utility to graph f ( x ) = 10 | x − 2 | f ( x ) = 10 | x − 2 | on the viewing window [ 0 , 4 ] . [ 0 , 4 ] . Identify the corresponding range. Show the graph.

Use a graphing utility to graph f ( x ) = − 100 | x | + 100 f ( x ) = − 100 | x | + 100 on the viewing window [ − 5 , 5 ] . [ − 5 , 5 ] . Identify the corresponding range. Show the graph.

For the following exercises, graph each function using a graphing utility. Specify the viewing window.

f ( x ) = − 0.1 | 0.1 ( 0.2 − x ) | + 0.3 f ( x ) = − 0.1 | 0.1 ( 0.2 − x ) | + 0.3

f ( x ) = 4 × 10 9 | x − ( 5 × 10 9 ) | + 2 × 10 9 f ( x ) = 4 × 10 9 | x − ( 5 × 10 9 ) | + 2 × 10 9

For the following exercises, solve the inequality.

If possible, find all values of a a such that there are no x - x - intercepts for f ( x ) = 2 | x + 1 | + a . f ( x ) = 2 | x + 1 | + a .

If possible, find all values of a a such that there are no y y -intercepts for f ( x ) = 2 | x + 1 | + a . f ( x ) = 2 | x + 1 | + a .

Real-World Applications

Cities A and B are on the same east-west line. Assume that city A is located at the origin. If the distance from city A to city B is at least 100 miles and x x represents the distance from city B to city A, express this using absolute value notation.

The true proportion p p of people who give a favorable rating to Congress is 8% with a margin of error of 1.5%. Describe this statement using an absolute value equation.

Students who score within 18 points of the number 82 will pass a particular test. Write this statement using absolute value notation and use the variable x x for the score.

A machinist must produce a bearing that is within 0.01 inches of the correct diameter of 5.0 inches. Using x x as the diameter of the bearing, write this statement using absolute value notation.

The tolerance for a ball bearing is 0.01. If the true diameter of the bearing is to be 2.0 inches and the measured value of the diameter is x x inches, express the tolerance using absolute value notation.

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• Publication date: Dec 21, 2021
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• Section URL: https://openstax.org/books/college-algebra-2e/pages/3-6-absolute-value-functions

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4.2: Absolute Value

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• David Arnold
• College of the Redwoods

Now that we have the fundamentals of piecewise-defined functions in place, we are ready to introduce the absolute value function. First, let’s state a trivial reminder of what it means to take the absolute value of a real number.

In a sense, the absolute value of a number is a measure of its magnitude, sans (without) its sign. Thus,

\[|7|=7 \qquad \text { and } \qquad |-7| =7 \nonumber \]

Here is the formal definition of the absolute value of a real number.

Definition: Absolute Value

To find the absolute value of any real number, first locate the number on the real line.

The absolute value of the number is defined as its distance from the origin.

For example, to find the absolute value of 7, locate 7 on the real line and then find its distance from the origin.

To find the absolute value of −7, locate −7 on the real line and then find its distance from the origin.

Some like to say that taking the absolute value “produces a number that is always positive.” However, this ignores an important exception, that is,

\[|0|=0 \nonumber \]

Thus, the correct statement is “the absolute value of any real number is either positive or it is zero,” i.e., the absolute value of a real number is “not negative.”2 Instead of using the phrase “not negative,” mathematicians prefer the word “nonnegative.” When we take the absolute value of a number, the result is always nonnegative; that is, the result is either positive or zero. In symbols,

\[|x| \geq 0 \text { for all real numbers } x \nonumber \]

This makes perfect sense in light of Definition 2. Distance is always nonnegative.

However, the discussion above is not of sufficient depth to handle more sophisticated problems involving absolute value.

A Piecewise Definition of Absolute Value

Because absolute value is intimately connected with distance, mathematicians and scientists find it an invaluable tool for measurement and error analysis. However, we will need a formulaic definition of the absolute value if we want to use this tool in a meaningful way. We need to develop a piecewise definition of the absolute value function, one that will define the absolute value for any arbitrary real number x.

We begin with a few observations. Remember, the absolute value of a number is always nonnegative (positive or zero).

• If a number is negative, negating that number will make it positive. | − 5| = −(−5) = 5, and similarly, | − 12| = −(−12) = 12. Thus, if x < 0 (if x is negative), then |x| = −x.
• If x = 0, then |x| = 0.
• If a number is positive, taking the absolute value of that number will not change a thing.

\[|5|=5, \text { and similarly, }|12|=12 \nonumber \]

Thus, if \(x>0\) (if \(x\) is positive), then \(|x|=x\).

We can summarize these three cases with a piecewise definition.

\[|x|=\left\{\begin{array}{ll}{-x,} & {\text { if } x<0} \\ {0,} & {\text { if } x=0} \\ {x,} & {\text { if } x>0}\end{array}\right. \nonumber \]

It is the first line in our piecewise definition (4) that usually leaves students scratching their heads. They might say “I thought absolute value makes a number positive (or zero), yet you have \(|x| = −x\); that is, you have the absolute value of x equal to a negative x.” Try as they might, this seems contradictory. Does it seem so to you?

However, there is no contradiction. If x < 0, that is, if x is a negative number, then −x is a positive number, and our intuitive notion of absolute value is not dissimilar to that of our piecewise definition (4). For example, if x = −8, then −x = 8, and even though we say “negative x,” in this case −x is a positive number.

If this still has you running confused, consider the simple fact that x and −x must have “opposite signs.” If one is positive, the other is negative, and vice versa. Consequently,

• if x is positive, then −x is negative, but
• if x is negative, then −x is positive.

Let’s summarize what we’ve learned thus far.

Summarizing the Definition on a Number Line

We like to use a number line to help summarize the definition of the absolute value of x.

Some remarks are in order for this summary on the number line.

• We first draw the real line then mark the “critical value” for the expression inside the absolute value bars on the number line. The number zero is a critical value for the expression x, because x changes sign as you move from one side of zero to the other.
• To the left of zero, x is a negative number. We indicate this with the minus sign below the number line. To the right of zero, x is a positive number, indicated with a plus sign below the number line.
• Above the number line, we simplify the expression |x|. To the left of zero, x is a negative number (look below the line), so |x| = −x. Note how the result −x is placed above the line to the left of zero. Similarly, to the right of zero, x is a positive number (look below the line), so |x| = x. Note how the result x is placed above the line to the right of zero.

In the piecewise definition of |x| in (4), note that we have three distinct pieces, one for each case discussed above. However, because |0| = 0, we can include this case with the piece |x| = x, if we adjust the condition to include zero.

\[|x|=\left\{\begin{array}{ll}{-x,} & {\text { if } x<0} \\ {x,} & {\text { if } x \geq 0}\end{array}\right. \nonumber \]

Note that this piecewise definition agrees with our discussion to date.

• In the first line of equation (6), if x is a negative number (i.e., if \(x < 0\)), then the absolute value must change x to a positive number by negating. That is, |x| = −x.
• In the second line of equation (6), if x is positive or zero (i.e., if \(x \geq 0\)), then there’s nothing to do except remove the absolute value bars. That is, |x| = x.

Because |0| = −0, we could just as well include the case for zero on the left, defining the absolute value with

\[|x|=\left\{\begin{array}{ll}{-x,} & {\text { if } x \leq 0} \\ {x,} & {\text { if } x>0}\end{array}\right. \nonumber \]

However, in this text we will always include the critical value on the right, as shown in Definition 5.

Constructing Piecewise Definitions

Let’s see if we can determine piecewise definitions for other expressions involving absolute value.

Example \(\PageIndex{1}\)

Determine a piecewise definition for |x − 2|.

First, set the expression inside the absolute value bars equal to zero and solve for x.

\[\begin{aligned} x-2 &=0 \\ x &=2 \end{aligned} \nonumber \]

Note that x − 2 = 0 at x = 2. This is the “critical value” for this expression. Draw a real line and mark this critical value of x on the line. Place the expression x − 2 below the line at its left end.

Next, determine the sign of x − 2 for values of x on each side of 2. This is easily done by “testing” a point on each side of 2 in the expression x − 2.

• Take x = 1, which lies to the left of the critical value 2 on our number line. Substitute this value of x in the expression x − 2, obtaining

\[x-2=1-2=-1 \nonumber \]

which is negative. Indeed, regardless of which x-value you pick to the left of 2, when inserted into the expression x − 2, you will get a negative result (you should check this for other values of x to the left of 2). We indicate that the expression x − 2 is negative for values of x to the left of 2 by placing a minus (−) sign below the number line to the left of 2.

• Next, pick x = 3, which lies to the right of the critical value 2 on the number line. Substitute this value of x into the expression x − 2, obtaining

\[x-2=3-2=1 \nonumber \]

which is positive. Indeed, regardless of which x-value you pick to the right of 2, when inserted into the expression x − 2, you will get a positive result (you should check this for other values of x to the right of 2). We indicate that the expression x − 2 is positive for values of x to the right of 2 by placing a plus (+) sign below the number line to the right of 2 (see the number line above).

The next step is to remove the absolute value bars from the expression |x−2|, depending on the sign of x − 2.

• To the left of 2, the expression x − 2 is negative (note the minus sign (−) below the number line), so |x − 2| = −(x − 2). That is, we have to negate x − 2 to make it positive. This is indicated by placing −(x − 2) above the line to the left of 2.

• To the right of 2, the expression x − 2 is positive (note the plus sign (+) below the line), so |x − 2| = x − 2. That is, we simply remove the absolute value bars because the quantity inside is already positive. This is indicated by placing x − 2 above the line to the right of 2 (see the number line above).

We can use this last number line summary to construct a piecewise definition of the expression |x − 2|.

\[|x-2|=\left\{\begin{array}{ll}{-(x-2),} & {\text { if } x<2,} \\ {x-2,} & {\text { if } x \geq 2}\end{array}=\left\{\begin{array}{ll}{-x+2,} & {\text { if } x<2} \\ {x-2,} & {\text { if } x \geq 2}\end{array}\right.\right. \nonumber \]

Our number line and piecewise definition agree: |x − 2| = −(x − 2) to the left of 2 and |x − 2| = x − 2 to the right of 2. Further, note how we’ve included the critical value of 2 “on the right” in our piecewise definition.

Let’s summarize the method we followed to construct the piecewise function above.

Constructing a Piecewise Definition for Absolute Value

When presented with the absolute value of an algebraic expression, perform the following steps to remove the absolute value bars and construct an equivalent piecewise definition.

• Take the expression that is inside the absolute value bars, and set that expression equal to zero. Then solve for x. This value of x is called a “critical value.” (Note: The expression inside the absolute value bars could have more than one critical value. We will not encounter such problems in this text.)
• Place your critical value on a number line.
• Place the expression inside the absolute value bars below the number line at the left end.
• Test the sign of the expression inside the absolute value bars by inserting a value of x from each side of the critical value and marking the result with a plus (+) or minus (−) sign below the number line.
• Place the original expression, the one including the absolute value bars, above the number line at the left end.
• Use the sign of the expression inside the absolute value bars (indicated by the plus and minus signs below the number line) to remove the absolute value bars, placing the results above the number line on each side of the critical value.
• Construct a piecewise definition that mimics the results on the number line.

Let’s apply this technique to another example.

Example \(\PageIndex{2}\)

Determine a piecewise definition for |3 − 2x|.

Step 1 : First set the expression inside the absolute value bars equal to zero and solve for x.

\[\begin{aligned} 3-2 x &=0 \\ x &=3 / 2 \end{aligned} \nonumber \]

Note that 3 − 2x = 0 at x = 3/2. This is the “critical value” for this expression.

Steps 2 and 3 : Draw a number line and mark this critical value on the line. The next step requires that we place the expression inside the absolute value bars, namely 3 − 2x, underneath the line at its left end.

Step 4 : Next, determine the sign of 3 − 2x for values of x on each side of 3/2. This is easily done by “testing” a point on each side of 3/2 in the expression 3 − 2x.

• Take x = 1, which lies to the left of 3/2. Substitute this value of x into the expression 3 − 2x, obtaining \[3-2 x=3-2(1)=1 \nonumber \] which is positive. Indicate this result by placing a plus sign (+) below the number line to the left of 3/2.

• Next, pick x = 2, which lies to the right of 3/2. Substitute this value of x into the expression 3 − 2x, obtaining \[3-2 x=3-2(2)=-1 \nonumber \] which is negative. Indicate this result by placing a negative sign (−) below the line to the right of 3/2 (see the number line above).

Steps 5 and 6 : Place the original expression, namely |3 − 2x|, above the number line at the left end. The next step is to remove the absolute value bars from the expression |3 − 2x|.

• To the left of 3/2, the expression 3 − 2x is positive (note the plus sign (+) below the number line), so |3−2x| = 3−2x. Indicate this result by placing the expression 3 − 2x above the number line to the left of 3/2.

• To the right of 3/2, the expression 3−2x is negative (note the minus sign (−) below the numberline), so |3−2x| = −(3−2x). That is, we have to negate 3−2x to make it positive. This is indicated by placing the expression −(3 − 2x) above the line to the right of 3/2 (see the number line above).

Step 7 : We can use this last number line summary to write a piecewise definition for the expression |3 − 2x|.

\[|3-2 x|=\left\{\begin{array}{ll}{3-2 x,} & {\text { if } x<3 / 2 .} \\ {-(3-2 x),} & {\text { if } x \geq 3 / 2}\end{array}=\left\{\begin{array}{ll}{3-2 x,} & {\text { if } x<3 / 2} \\ {-3+2 x,} & {\text { if } x \geq 3 / 2}\end{array}\right.\right. \nonumber \]

Again, note how we’ve included the critical value of 3/2 “on the right.”

Drawing the Graph of an Absolute Value Function

Now that we know how to construct a piecewise definition for an expression containing absolute value bars, we can use what we learned in the previous section to draw the graph.

Example \(\PageIndex{3}\)

Sketch the graph of the function f(x) = |3 − 2x|.

In Example \(\PageIndex{2}\), we constructed the following piecewise definition.

\[f(x)=|3-2 x|=\left\{\begin{array}{ll}{3-2 x,} & {\text { if } x<3 / 2} \\ {-3+2 x,} & {\text { if } x \geq 3 / 2}\end{array}\right. \nonumber \]

We now sketch each piece of this function.

• If x < 3/2, then f(x) = 3 − 2x (see equation (10)). This is a ray, starting at x = 3/2 and extending to the left. At x = 3/2,

\[f(3 / 2)=3-2(3 / 2)=3-3=0 \nonumber \]

Thus, the endpoint of the ray is located at (3/2, 0).

Next, pick a value of x that lies to the left of 3/2. At x = 0,

\[f(0)=3-2(0)=3-0=3 \nonumber \]

Thus, a second point on the ray is (0, 3).

A table containing the two evaluated points and a sketch of the accompanying ray are shown in Figure \(\PageIndex{1}\). Because f(x) = 3 − 2x only if x is strictly less than 3/2, the point at (3/2, 0) is unfilled.

• If x ≥ 3/2, then f(x) = −3 + 2x (see equation (10)). This is a ray, starting at x = 3/2 and extending to the right. At x = 3/2, \[f(3 / 2)=-3+2(3 / 2)=-3+3=0 \nonumber \]

Next, pick a value of x that lies to the right of 3/2. At x = 3, \[f(3)=-3+2(3)=-3+6=3 \nonumber \]

Thus, a second point on the ray is (3, 3). A table containing the two evaluated points and a sketch of the accompanying ray are shown in Figure \(\PageIndex{2}\). Because f(x) = −3 + 2x for all values of x that are greater than or equal to 3/2, the point at (3/2, 0) is filled in this plot.

• To sketch the graph of f(x) = |3 − 2x|, we need only combine the two pieces from Figures \(\PageIndex{1}\) and \(\PageIndex{2}\). The result is shown in Figure \(\PageIndex{3}\).

Note the “V-shape” of the graph. We will refer to the point at the tip of the “V” as the vertex of the absolute value function.

In Figure \(\PageIndex{3}\), the equation of the left-hand branch of the “V” is y = 3 − 2x. An alternate approach to drawing this branch is to note that its graph is contained in the graph of the full line y = 3 − 2x, which has slope −2 and y-intercept at (0, 3). Thus, one could draw the full line using the slope and y-intercept, then erase that part of the line that lies to the right of x = 3/2. A similar strategy would work for the right-hand branch of y = |3 − 2x|.

Using Transformations

Consider again the basic definition of the absolute value of x.

\[f(x)=|x|=\left\{\begin{array}{ll}{-x,} & {\text { if } x<0} \\ {x,} & {\text { if } x \geq 0}\end{array}\right. \nonumber \]

Some basic observations are:

• If x < 0, then f(x) = −x. This ray starts at the origin and extends to the left with slope −1. Its graph is pictured in Figure \(\PageIndex{4}\)(a).
• If \(x \geq 0\), then f(x) = x. This ray starts at the origin and extends to the right with slope 1. Its graph is pictured in Figure \(\PageIndex{4}\)(b).
• We combine the graphs in Figures \(\PageIndex{4}\)(a) and \(\PageIndex{4}\)(b) to produce the graph of f(x) = |x| in Figure \(\PageIndex{4}\)(c).

You should commit the graph of f(x) = |x| to memory. Things to note:

• The graph of f(x) = |x| is “V-shaped.”
• The vertex of the graph is at the point (0, 0).
• The left-hand branch has equation y = −x and slope −1.
• The right-hand branch has equation y = x and slope 1.
• Each branch of the graph of f(x) = |x| forms a perfect 45◦ angle with the x-axis.

Now that we know how to draw the graph of f(x) = |x|, we can use the transformations we learned in Chapter 2 (sections 5 and 6) to sketch a number of simple graphs involving absolute value.

Example \(\PageIndex{4}\)

Sketch the graph of f(x) = |x − 3|.

First, sketch the graph of y = f(x) = |x|, as shown in Figure \(\PageIndex{5}\)(a). Note that if f(x) = |x|, then

\[y=f(x-3)=|x-3| \nonumber \]

To sketch the graph of y = f(x − 3) = |x − 3|, shift the graph of y = f(x) = |x| three units to the right, producing the result shown in Figure \(\PageIndex{5}\)(b).

We can check this result using the graphing calculator. Load the function f(x) = |x − 3| into Y1 in the Y= menu on your graphing calculator as shown in Figure \(\PageIndex{6}\)(a). Push the MATH button, right-arrow to the NUM menu, then select 1:abs( (see Figure \(\PageIndex{6}\)(b)) to enter the absolute value in Y1. Push the ZOOM button, then select 6:ZStandard to produce the image shown in Figure \(\PageIndex{6}\)(c).

Let’s look at another simple example.

Example \(\PageIndex{5}\)

Sketch the graph of f(x) = |x| − 4.

First, sketch the graph of y = f(x) = |x|, as shown in Figure \(\PageIndex{7}\)(a). Note that if f(x) = |x|, then \[y=f(x)-4=|x|-4 \nonumber \]

To sketch the graph of y = f(x) − 4 = |x| − 4, shift the graph of y = f(x) = |x| downward 4 units, producing the result shown in Figure \(\PageIndex{5}\)(b).

Let’s look at one final example.

Example \(\PageIndex{6}\)

Sketch the graph of f(x) = −|x| + 5. State the domain and range of this function.

• First, sketch the graph of y = f(x) = |x|, as shown in Figure \(\PageIndex{8}\)(a).
• Next, sketch the graph of y = −f(x) = −|x|, which is a reflection of the graph of y = f(x) = |x| across the x-axis and is pictured in Figure \(\PageIndex{8}\)(b).
• Finally, we will want to sketch the graph of y = −f(x) + 5 = −|x| + 5. To do this, we shift the graph of y = −f(x) = −|x| in Figure \(\PageIndex{8}\)(b) upward 5 units to produce the result in Figure \(\PageIndex{8}\)(c).

To find the domain of f(x) = −|x| + 5, project all points on the graph onto the x-axis, as shown in Figure \(\PageIndex{9}\)(a). Thus, the domain of f is \((-\infty, \infty)\). To find the range, project all points on the graph onto the y-axis, as shown in Figure \(\PageIndex{9}\)(b). Thus, the range is \((-\infty, 5]\).

For each of the functions in Exercises 1 - 8 , as in Examples 7 and 8 in the narrative, mark the “critical value” on a number line, then mark the sign of the expression inside of the absolute value bars below the number line. Above the number line, remove the absolute value bars according to the sign of the expression you marked below the number line. Once your number line summary is finished, create a piecewise definition for the given absolute value function.

Exercise \(\PageIndex{1}\)

f(x) = |x+1|

\[f(x)=\left\{\begin{array}{ll}{-x-1,} & {\text { if } x<-1} \\ {x+1,} & {\text { if } x \ge -1} \nonumber \end{array}\right. \nonumber \]

Exercise \(\PageIndex{2}\)

f(x) = |x−4|

Exercise \(\PageIndex{3}\)

g(x) = |4−5x|

\[g(x)=\left\{\begin{array}{ll}{4-5x,} & {\text { if } x<\frac{4}{5}} \\ {-4+5x,} & {\text { if } x \ge \frac{4}{5}} \nonumber \end{array}\right. \nonumber \]

Exercise \(\PageIndex{4}\)

g(x) = |3−2x|

Exercise \(\PageIndex{5}\)

h(x) = |−x−5|

\[h(x)=\left\{\begin{array}{ll}{-x-5,} & {\text { if } x<-5} \\ {x+5,} & {\text { if } x \ge -5} \nonumber \end{array}\right. \nonumber \]

Exercise \(\PageIndex{6}\)

h(x) = |−x−3|

Exercise \(\PageIndex{7}\)

f(x) = x+|x|

\[f(x)=\left\{\begin{array}{ll}{0,} & {\text { if } x<0} \\ {2x,} & {\text { if } x \ge 0} \nonumber \end{array}\right. \nonumber \]

Exercise \(\PageIndex{8}\)

\(f(x) = \frac{|x|}{x}\)

For each of the functions in Exercises 9 - 16 , perform each of the following tasks.

• Create a piecewise definition for the given function, using the technique in Exercises 1 - 8 and Examples 7 and 8 in the narrative.
• Following the lead in Example 9 in the narrative, use your piecewise definition to sketch the graph of the given function on a sheet of graph paper. Please place each exercise on its own coordinate system.

Exercise \(\PageIndex{9}\)

f(x) = |x−1|

\[f(x)=\left\{\begin{array}{ll}{-x+1,} & {\text { if } x<1} \\ {x-1,} & {\text { if } x \ge 1} \nonumber \end{array}\right. \nonumber \]

Exercise \(\PageIndex{10}\)

f(x) = |x+2|

Exercise \(\PageIndex{11}\)

g(x) = |2x−1|

\[g(x)=\left\{\begin{array}{ll}{-2x+1,} & {\text { if } x<\frac{1}{2}} \\ {2x-1,} & {\text { if } x \ge \frac{1}{2}} \nonumber \end{array}\right. \nonumber \]

Exercise \(\PageIndex{12}\)

g(x) = |5−2x|

Exercise \(\PageIndex{13}\)

h(x) = |1−3x|

\[h(x)=\left\{\begin{array}{ll}{1-3x,} & {\text { if } x<\frac{1}{3}} \\ {-1+3x,} & {\text { if } x \ge \frac{1}{3}} \nonumber \end{array}\right. \nonumber \]

Exercise \(\PageIndex{14}\)

h(x) = |2x+1|

Exercise \(\PageIndex{15}\)

f(x) = x−|x|

\[f(x)=\left\{\begin{array}{ll}{2x,} & {\text { if } x<0} \\ {0,} & {\text { if } x \ge 0} \nonumber \end{array}\right. \nonumber \]

Exercise \(\PageIndex{16}\)

f(x) = x+|x−1|

Exercise \(\PageIndex{17}\)

Use a graphing calculator to draw the graphs of y = |x|, y = 2|x|, y = 3|x|, and y = 4|x| on the same viewing window. In your own words, explain what you learned in this exercise.

Multiplying by a factor of a > 1, as in y = a|x|, stretches the graph of y = |x| vertically by a factor of a. The higher the value of a, the more it stretches vertically.

Exercise \(\PageIndex{18}\)

Use a graphing calculator to draw the graphs of y = |x|, y = (1/2)|x|, y = (1/3)|x|, and y = (1/4)|x| on the same viewing window. In your own words, explain what you learned in this exercise.

Exercise \(\PageIndex{19}\)

Use a graphing calculator to draw the graphs of y = |x|, y = |x−2|, y =|x−4|, and y = |x−6| on the same viewing window. In your own words, explain what you learned in this exercise.

Subtracting a positive value of a, as in y = |x−a|, shifts the graph a units to the right.

Exercise \(\PageIndex{20}\)

Use a graphing calculator to draw the graphs of y = |x|, y = |x+2|, y = |x+4|, and y = |x+6| on the same view- ing window. In your own words, explain what you learned in this exercise.

In Exercises 21 - 36 , perform each of the following tasks. Feel free to check your work with your graphing calculator, but you should be able to do all of the work by hand.

• Set up a coordinate system on a sheet of graph paper. Label and scale each axis. Create an accurate plot of the function y = |x| on your coordinate system and label this graph with its equation.
• Use the technique of Examples 12,13, and 14 in the narrative to help select the appropriate geometric transformations to transform the equation y = |x| into the form of the function given in the exercise. On the same coordinate system, use a different colored pencil or pen to draw the graph of the function resulting from your applied transformation. Label the resulting graph with its equation.
• Use interval notation to describe the domain and range of the given function.

Exercise \(\PageIndex{21}\)

f(x) = |−x|

The graphs of y = |x| and y = |−x| coincide. The domain is \((−\infty, \infty)\) and the range is \([0, \infty)\).

Exercise \(\PageIndex{22}\)

f(x) = −|x|

Exercise \(\PageIndex{23}\)

\(f(x) = \frac{1}{2}|x|\)

The domain is \((−\infty, \infty)\) and the range is \([0, \infty)\).

Exercise \(\PageIndex{24}\)

f(x) = −2|x|

Exercise \(\PageIndex{25}\)

f(x) = |x+4|

Exercise \(\PageIndex{26}\)

f(x) = |x−2|

Exercise \(\PageIndex{27}\)

f(x) = |x|+2

The domain is \((−\infty, \infty)\) and the range is \([2, \infty)\).

Exercise \(\PageIndex{28}\)

f(x) = |x|−3

Exercise \(\PageIndex{29}\)

f(x) = |x+3|+2

Exercise \(\PageIndex{30}\)

f(x) = |x−3|−4

Exercise \(\PageIndex{31}\)

f(x) = −|x−2|

The domain is \((−\infty, \infty)\) and the range is \((−\infty, 0]\).

Exercise \(\PageIndex{32}\)

f(x) = −|x|−2

Exercise \(\PageIndex{33}\)

f(x) = −|x|+4

The domain is \((−\infty, \infty)\) and the range is \((−\infty, 4]\).

Exercise \(\PageIndex{34}\)

f(x) = −|x+4| ​​​​​​

Exercise \(\PageIndex{35}\)

f(x) = −|x−1|+5

The domain is \((−\infty, \infty)\) and the range is \((−\infty, 5]\).

Exercise \(\PageIndex{36}\)

f(x) = −|x+5|+2

Absolute Value

Our Absolute Value lesson plan teaches students about absolute value, including what it is and what the symbols used to designate absolute zero are. Students also learn how the absolute value symbols are interpreted in equations and that absolute values are always positive because they measure distance.

Included with this lesson are some adjustments or additions that you can make if you’d like, found in the “Options for Lesson” section of the Classroom Procedure page. One of the optional additions to this lesson is to hand out index cards with problems and then have students place themselves on a number line for their answer or order themselves according to their answer from greatest to least.

Description

Additional information, what our absolute value lesson plan includes.

Lesson Objectives and Overview: Absolute Value defines absolute value, teaches the symbols used to designate absolute zero and how they are interpreted in equations, and explains that absolute values are always positive because they measure distance. At the end of the lesson, students will be able to understand absolute value and how to solve expressions using absolute value bars as grouping symbols. This lesson is for students in 5th grade and 6th grade.

Classroom Procedure

Every lesson plan provides you with a classroom procedure page that outlines a step-by-step guide to follow. You do not have to follow the guide exactly. The guide helps you organize the lesson and details when to hand out worksheets. It also lists information in the blue box that you might find useful. You will find the lesson objectives, state standards, and number of class sessions the lesson should take to complete in this area. In addition, it describes the supplies you will need as well as what and how you need to prepare beforehand. The only supplies you will need are decks of cards.

Options for Lesson

Included with this lesson is an “Options for Lesson” section that lists a number of suggestions for activities to add to the lesson or substitutions for the ones already in the lesson. To add to the lesson activity, you could h ave students increase the number of rows for the game or add in your own deck of cards with larger numbers. For an additional activity, you could  hand out index cards with problems and then have students place themselves on a number line for their answer or order themselves according to their answer from greatest to least. Finally, you can provide challenge questions using fractions or decimals.

Teacher Notes

The teacher notes page includes lines that you can use to add your own notes as you’re preparing for this lesson.

ABSOLUTE VALUE LESSON PLAN CONTENT PAGES

The Absolute Value lesson plan includes two pages of content. All numbers except zero have an absolute value. The symbol for absolute value is two straight lines around the number or expression. For example, |9| = 9 means the absolute value of 9 is 9, and |-8| = 8 means the absolute value of -8 is 8.

It’s important to pay attention to whether the negative is placed inside or outside of the absolute value bars. When it’s inside, the absolute value is positive. When it’s outside, the absolute value is negative.

Absolute value is the distance from zero on a number line. Absolute value is a measure of distance and is therefore always positive. The lesson includes two number lines to illustrate this. The absolute value of 0 is zero, because zero is not positive or negative.

We can also think of the absolute value bars as a grouping symbol, like parenthesis. When we use the order of operations to solve an expression or equation, we solve anything inside absolute value bars before doing any other operation. The lesson includes several examples that illustrate this. The first is: |-6 + 2| = |-4| = 4 .

If there is more than one set of absolute value bars, you complete the operations inside the sets of bars before completing any other operations. In the order of operations, the first set is grouping symbols. The lesson includes more examples to illustrate this. In one example, we’re solving the following problem using these steps: |-3 + 1| + |5 + 2| = |-4| + |7| = 4 + 7 = 11 .

The lesson closes with a few more examples to help students practice using these concepts.

ABSOLUTE VALUE LESSON PLAN WORKSHEETS

The Absolute Value lesson plan includes three worksheets: an activity worksheet, a practice worksheet, and a homework assignment. You can refer to the guide on the classroom procedure page to determine when to hand out each worksheet.

PLAYING ZERO ACTIVITY WORKSHEET

Students will complete the lesson activity in groups. To play the game, each player gets two cards, one face up and one down. All the students at the table can see the face up cards. Black cards are positive while red cards are negative. Each player can get additional cards by asking them to “hit me”. They can have up to four cards each and must use all of them. For each round, the person with the total closest to zero wins that round.

At the end of the game, students will take the absolute value of their “total” column for each hand and will add them together. The person closest to zero in the absolute value column wins the game.

ABSOLUTE VALUE PRACTICE WORKSHEET

The practice worksheet asks students to complete two short exercises. For the first, they will find the absolute value of different numbers. For the second, they will simplify expressions, showing their work.

MATCHING HOMEWORK ASSIGNMENT

For the homework assignment, students will match nine problems to their correct answers. They will then either find the absolute value or simplify each equation.

This lesson also includes a quiz that you can use to test students’ understanding of the lesson material. For the quiz, students will find the absolute value of six numbers and will then simplify six expressions, showing their work.

Worksheet Answer Keys

This lesson plan includes answer keys for the practice worksheet, the homework assignment, and the quiz.  If you choose to administer the lesson pages to your students via PDF, you will need to save a new file that omits these pages. Otherwise, you can simply print out the applicable pages and keep these as reference for yourself when grading assignments.

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Lesson 1: The Absolute Value

Table of Contents

Hi everyone! Read through the material below, watch the videos, and follow up with your instructor if you have questions.

Topic . This lesson covers Session 1: The Absolute Value.

Learning Outcomes.

• Solve Absolute Value equations
• Understand and use interval and inequality notation
• Solve Absolute Value Inequalities

WeBWorK . There are two WeBWorK assignments on today’s material:

• Interval Notation

Absolute Value Inequalities

Additional Video Resources.

• Videos on Absolute Value
• Videos on Interval Notation

Question of the Day: If $|x|=5$ then what does $x$ equal?

Background Regarding Numbers

VIDEO: Types of Numbers

Video by Irania Vazquez .

Absolute Value

Definition . The absolute value of a real number $c$, denoted by $|c|$, is the non-negative number which is equal in magnitude (or size) to $c$, i.e., is the number resulting from disregarding the sign:

$|c|=\left\{ \begin{array}{cl} c, & \text { if } c \text { is positive or zero } \\ -c, & \text { if } c \text { is negative } \end{array}$

• For which real numbers $x$ do you have $|x|=3$?
• Solve for $x:|3 x-4|=5$
• Solve for $x:\quad -2 \cdot|12+3 x|=-18$

VIDEO: Example 1 – Absolute Value Equations

Inequalities and Interval Notation

In this section we need to discuss two ideas:

First, the notion of ordering or inequalit y. For example:

$4<9$ reads as 4 is less than 9 $-3 \leq 2$ reads as -3 is less than or equal to 2 $\frac{7}{6}>1$ reads as $\frac{7}{6}$ is greater than 1 $2 \geq-3$ reads as 2 is greater than or equal to -3

The second is the called intervals – an interval is a set of numbers on the number line lying (for example) between two endpoints. We can give an interval in three ways: using inequalities, using interval notation, or by drawing a number line. It’s important to be able to describe the same interval in different ways. The following table shows how these three methods are connected:

To solve an inequality containing an absolute value, we use the following steps:

Step 1: Solve the corresponding equality. The solution of the equality divides the real number line into several subintervals.

Step 2: Using step 1, check the inequality for a number in each of the subintervals. This check determines the intervals of the solution set. Step 3: Check the endpoints of the intervals.

Example 2. Solve for $x$ :

• $|3 x-5| \geq 11$
• $|12-5 x| \leq 1$
• $|x+7|\leq -3$, $|x+7|\geq -3$

VIDEO: Example 2 – Absolute Value Inequalities and Interval Notation

Exit Question

Solve for $x: \quad|-3-2x| \geq 7$

In interval notation, the solution set is $S=(-\infty,-5]\cup [2,\infty)$

Good job! You are now ready to practice on your own – give the WeBWorK assignment a try. If you get stuck, try using the “Ask for Help” button to ask a question on the Q&A site.

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