Lesson Explainer: Absolute Value Functions | Nagwa Lesson Explainer: Absolute Value Functions | Nagwa

Lesson Explainer: Absolute Value Functions Mathematics

In this explainer, we will learn how to evaluate and graph absolute value functions and identify their domain and range.

Recall that the absolute value of a real number is its distance from 0 on the number line. For example, in the expression |βˆ’5| (which can be read as the absolute value of βˆ’5), the number βˆ’5 is shown within absolute value symbols. Since βˆ’5 is located 5 units from 0 on the number line, the value of the expression is 5. The value of the expression |5| (which can be read as the absolute value of 5) is 5 as well, because 5 is also located 5 units from 0 on the number line.

A distance is never negative, so the absolute value of a number will always be positive or 0. In addition to numbers, we can put algebraic expressions within absolute value symbols to define functions. These types of functions are called absolute value functions.

Definition: Absolute Value Function

An absolute value function is a function with a definition that contains an algebraic expression within absolute value symbols. An example of an absolute value function is 𝑔(π‘₯)=|π‘₯|, where π‘₯βˆˆβ„.

One way to graph an absolute value function is to begin by inputting values into the function and then recording the resulting outputs in a table of values. For the function 𝑔(π‘₯)=|π‘₯|, such a table is shown:

π‘₯βˆ’3βˆ’2βˆ’10123
𝑔(π‘₯)3210123

By plotting the ordered pairs in the table on the coordinate plane and drawing lines through them, we can create the v-shaped graph of 𝑦=𝑔(π‘₯) below:

We can see that the graph lies only on or above the π‘₯-axis. The arrows tell us that it extends infinitely upward and to both the left and right. This means that while the function can have any input value, its output values are only the nonnegative numbers. We can also see that the graph decreases to the left of the 𝑦-axis before touching the origin, and it then increases to the right of the 𝑦-axis. The portions to the left and to the right of the 𝑦-axis are mirror images of each other. Using our observations, we can describe various properties of the graph and of the function that it represents.

Properties: The Absolute Value Function 𝑔(π‘₯) = |π‘₯| and Its Graph

  • Vertex: (0,0)
  • Line of symmetry: π‘₯=0
  • Domain: ℝ, also written as (βˆ’βˆž,∞)
  • Range: 𝑔(π‘₯)β‰₯0, also written as [0,∞[
  • π‘₯-intercept: 0
  • 𝑦-intercept: 0

All absolute value functions that have a linear expression within the absolute value symbols have graphs that are v-shaped. However, the functions’ other properties may differ. Transforming the graph of the function 𝑔(π‘₯)=|π‘₯| by translating it, stretching or compressing it, or reflecting it across an axis can allow us to more easily sketch the graph of an absolute value function in a different form so that we can identify the domain and range of the function.

Definition: Transformations of the Graph of the Absolute Value Function 𝑔(π‘₯) = |π‘₯|

  • The graph of 𝑓(π‘₯)=𝑔(π‘₯βˆ’β„Ž) is a horizontal translation of the graph of 𝑔(π‘₯)=|π‘₯|. If β„Ž is positive, the translation is β„Ž units right. If β„Ž is negative, the translation is |β„Ž| units left. The domain and range of 𝑓(π‘₯) are the same as those of 𝑔(π‘₯).
  • The graph of 𝑓(π‘₯)=𝑔(π‘₯)+π‘˜ is a vertical translation of the graph of 𝑔(π‘₯)=|π‘₯|. If π‘˜ is positive, the translation is π‘˜ units up. If π‘˜ is negative, the translation is |π‘˜| units down. The domain of 𝑓(π‘₯) is the same as that of 𝑔(π‘₯), but the range is 𝑓(π‘₯)β‰₯π‘˜.
  • The graph of 𝑓(π‘₯)=π‘Žβ‹…π‘”(π‘₯) is a vertical stretch of the graph of 𝑔(π‘₯)=|π‘₯| by a scale factor of |π‘Ž| when |π‘Ž|>1 and a vertical compression by a scale factor of |π‘Ž| when 0<|π‘Ž|<1. If π‘Ž is negative, the graph of 𝑔(π‘₯) is also reflected over the π‘₯-axis, which means that the graph of 𝑓(π‘₯) will open downward instead of upward.
  • The graph of 𝑓(π‘₯)=𝑔(𝑏⋅π‘₯) is a horizontal compression of the graph of 𝑔(π‘₯)=|π‘₯| by a scale factor of 1|𝑏| when |𝑏|>1 and a horizontal stretch by a scale factor of 1|𝑏| when 0<|𝑏|<1. If 𝑏 is negative, the graph of 𝑔(π‘₯) is also reflected over the 𝑦-axis.

Some examples of absolute value functions in various forms and their graphs are in the problems that follow.

Example 1: Finding the Range of an Absolute Value Function Using a Graph

Find the range of the function 𝑓(π‘₯)=|βˆ’2π‘₯βˆ’2|.

Answer

We can see that the graph has a vertex at (βˆ’1,0), a line of symmetry of π‘₯=βˆ’1, an π‘₯-intercept of βˆ’1, and a 𝑦-intercept of 2. By inputting values of π‘₯ into the function 𝑓(π‘₯)=|βˆ’2π‘₯βˆ’2| and observing the resulting outputs, we can confirm the graph’s validity.

Let’s use the graph to help determine the domain and range of 𝑓(π‘₯). Recall that the domain of a function is the set of all possible input values, while the range is the set of all possible output values. Another way to say this is that the domain is all possible values of the independent variable, while the range is all possible values of the dependent variable.

The arrows tell us that the graph extends infinitely to both the left and the right, so we know that the domain must be the set of all real numbers, or ℝ. The arrows also tell us that the graph extends infinitely upward, but we can see that it lies only on or above the π‘₯-axis. In other words, the smallest value of 𝑦 is 0, but there is no largest value. Thus, we know that the range is 𝑓(π‘₯)β‰₯0, or [0,∞[. Any absolute value function in the form 𝑓(π‘₯)=|π‘šπ‘₯+𝑏| will have this range.

Now let’s look at the graph of another absolute value function and find the function’s domain and range. This time, we will consider the function’s graph to be a transformation of the graph of 𝑔(π‘₯)=|π‘₯|.

Example 2: Finding the Domain and Range of an Absolute Value Function Using a Graph

Determine the domain and the range of the function 𝑓(π‘₯)=|π‘₯+1|+1.

Answer

We know that the domain and range of 𝑔(π‘₯)=|π‘₯| remain the same after a horizontal translation to the left or right, so the domain of 𝑓(π‘₯)=𝑔(π‘₯+β„Ž) is the set of all real numbers, or ℝ, and the range is 𝑓(π‘₯)β‰₯0, or [0,∞[. However, although the domain of 𝑔(π‘₯)=|π‘₯| remains the same after a vertical translation up or down, its range does not. The range of 𝑓(π‘₯)=𝑔(π‘₯)+π‘˜ is 𝑓(π‘₯)β‰₯π‘˜, or [π‘˜,∞[.

Here, we can see that the graph is a horizontal translation 1 unit to the left and a vertical translation 1 unit up of the graph of 𝑔(π‘₯)=|π‘₯|. That is, 𝑓(π‘₯)=𝑔(π‘₯+1)+1. Thus, the domain of the function 𝑓(π‘₯)=|π‘₯+1|+1 is ℝ, and the range is 𝑓(π‘₯)β‰₯1, or [1,∞[.

Note

We will arrive at the same answers for the function’s domain and range by using the fact that the domain is the set of all possible input values, while the range is the set of all possible output values. The arrows tell us that the graph extends infinitely to both the left and the right, so, again, we know that the domain of the function is the set of all real numbers, or ℝ. The arrows also tell us that the graph extends infinitely upward, but we can see that the graph lies only at or above a value of 1 for 𝑦. That is, the smallest value of 𝑦 is 1, but there is no largest value. Thus, we can again see that the range of the function is 𝑓(π‘₯)β‰₯1, or [1,∞[.

Next, let’s look at how to find the domain of an absolute value function without being given a graph. We will construct the function’s graph by considering its relationship to that of 𝑔(π‘₯)=|π‘₯|.

Example 3: Finding the Domain of an Absolute Value Function

Given that π‘Ž is a constant, what is the domain of the function 𝑓(π‘₯)=|π‘₯+π‘Ž|?

Answer

As we can see in the figures below, the graph of 𝑓(π‘₯)=|π‘₯+π‘Ž| is a horizontal translation of the graph of 𝑔(π‘₯)=|π‘₯|. If π‘Ž is positive, the translation is π‘Ž units to the left, and if π‘Ž is negative, the translation is π‘Ž units to the right.

Recall that for both 𝑔(π‘₯)=|π‘₯| and 𝑓(π‘₯)=𝑔(π‘₯+β„Ž), the domain is ℝ, and the range is 𝑓(π‘₯)β‰₯0, or [0,∞[. In other words, the domain and range of 𝑓(π‘₯) are the same as those of 𝑔(π‘₯), regardless of whether the horizontal translation is left or right, and regardless of the number of units. This means the domain of the function 𝑓(π‘₯)=|π‘₯+π‘Ž| must be the set of all real numbers, or ℝ.

Note

We know that the domain of a function is the set of all possible input values, or all possible values of the independent variable, π‘₯. To determine the domain of the function 𝑓(π‘₯)=|π‘₯+π‘Ž|, we could ask ourselves, β€œWhat values of π‘₯ can we input into the function?” To help answer this question, let’s consider what the function would be for some different values of the constant π‘Ž.

First we will consider a negative value for π‘Ž. Let’s assume that π‘Ž=βˆ’1. When this is the case, the function becomes 𝑓(π‘₯)=|π‘₯βˆ’1|. Now let’s input some values into the function and observe whether or not an output value is produced: 𝑓(βˆ’1)=|βˆ’1βˆ’1|=|βˆ’2|=2,𝑓(0)=|0βˆ’1|=|βˆ’1|=1,𝑓12=|||12βˆ’1|||=|||βˆ’12|||=12,𝑓(1)=|1βˆ’1|=|0|=0.

We can see that all of the input values give a corresponding output value. Now let’s look at what happens when π‘Ž=0. In this case, the function becomes 𝑓(π‘₯)=|π‘₯|: 𝑓(βˆ’1)=|βˆ’1|=1,𝑓(0)=|0|=0,𝑓12=|||12|||=12,𝑓(1)=|1|=1.

Again, we can see that all of the input values give a corresponding output value. Finally, let’s consider a positive value of a by looking at the function when π‘Ž=1. Now, the function becomes 𝑓(π‘₯)=|π‘₯+1|: 𝑓(βˆ’1)=|βˆ’1+1|=|0|=0,𝑓(0)=|0+1|=|1|=1,𝑓12=|||12+1|||=|||32|||=32,𝑓(1)=|1+1|=|2|=2.

Just as before, there is an output value for each of the input values. In fact, because the absolute value function 𝑔(π‘₯)=|π‘₯| is defined for all real numbers π‘₯, no matter what constant π‘Ž represents, we will always be able to input any value into the function 𝑓(π‘₯)=|π‘₯+π‘Ž| and get an output value. That is, the output value will never be undefined. Therefore, we again see that the domain of the function 𝑓(π‘₯)=|π‘₯+π‘Ž| is the set of all real numbers, or ℝ.

In the example that follows, we will construct the graph of another absolute value function to help find its domain and range. This time, we will use not only translations of the graph of 𝑔(π‘₯)=|π‘₯| when graphing our function, but a stretch and a reflection of 𝑔(π‘₯)=|π‘₯| as well.

Example 4: Finding the Domain and Range of an Absolute Value Function

Find the domain and range of the function 𝑓(π‘₯)=βˆ’4|π‘₯βˆ’5|βˆ’1.

Answer

To help determine the domain and range of the function, let’s begin by constructing its graph. Given 𝑔(π‘₯)=|π‘₯|, we know that the function’s definition is in the form 𝑓(π‘₯)=π‘Žβ‹…π‘”(π‘₯+β„Ž)+π‘˜, where π‘Ž=βˆ’4, β„Ž=βˆ’5, and π‘˜=βˆ’1. Thus, we can obtain the graph of 𝑦=𝑓(π‘₯) by vertically stretching the graph of 𝑦=𝑔(π‘₯) by a scale factor of 4, reflecting it over the π‘₯-axis, translating it 5 units to the right, and then translating it 1 unit down as shown:

We can verify that our graph is correct by inputting three values of π‘₯ into the function and observing the resulting output values. This is not necessary as long as we are careful with our transformations, but it is useful for more complicated functions such as this one. The values we choose for π‘₯ will make the value of the expression inside the absolute value symbols (π‘₯βˆ’5), negative, 0, and positive respectively: 𝑓(4)=βˆ’4|4βˆ’5|βˆ’1=βˆ’4|βˆ’1|βˆ’1=βˆ’4(1)βˆ’1=βˆ’4βˆ’1=βˆ’5(π‘₯βˆ’5βˆ’1),𝑓(5)=βˆ’4|5βˆ’5|βˆ’1=βˆ’4|0|βˆ’1=βˆ’4(0)βˆ’1=0βˆ’1=βˆ’1(π‘₯βˆ’50),𝑓(6)=βˆ’4|6βˆ’5|βˆ’1=βˆ’4|1|βˆ’1=βˆ’4(1)βˆ’1=βˆ’4βˆ’1=βˆ’5(π‘₯βˆ’51).thevalueofisthevalueofisthevalueofis

We get the ordered pairs (4,βˆ’5), (5,βˆ’1), and (6,βˆ’5). By plotting these points on the coordinate plane, and using our understanding of absolute value functions, we can again create the graph of 𝑦=𝑓(π‘₯) for the function 𝑓(π‘₯)=βˆ’4|π‘₯βˆ’5|βˆ’1 below:

The arrows on our graph tell us that the graph extends infinitely to both the left and to the right, so we know that the domain of 𝑓(π‘₯)=βˆ’4|π‘₯βˆ’5|βˆ’1 must be the set of all real numbers, or ℝ. The arrows also tell us that the graph extends infinitely downward, but we can see that it lies only at or below a value of βˆ’1 for 𝑦. In other words, the largest value of 𝑦 is βˆ’1, but there is no smallest value. This means that the range of the function is 𝑓(π‘₯)β‰€βˆ’1, or ]βˆ’βˆž,βˆ’1].

Now let’s look at a final example involving a real-world problem.

Example 5: Solving a Word Problem Involving an Absolute Value Function

A body was moving with a uniform velocity of magnitude 5 cm/s from the point 𝐴 to the point 𝐢 passing through the point 𝐡 without stopping. The distance between the body and the point 𝐡 is given by 𝑑(𝑑)=5|8βˆ’π‘‘|, where 𝑑 is the time in seconds, and 𝑑 is the distance in cm. Determine the distance between the body and the point 𝐡 after 5 seconds and after 11 seconds.

Answer

In order to determine the distance between the body and the point 𝐡 after 5 seconds and after 11 seconds, we need to evaluate the function 𝑑(𝑑)=5|8βˆ’π‘‘| at 𝑑=5 and at 𝑑=11. In other words, we must substitute both 5 and 11 into the function for 𝑑 and calculate the resulting value of 𝑑(𝑑).

Recall that the absolute value of a number is always positive, so regardless of whether the value of the expression inside the absolute value symbols (8βˆ’π‘‘) is positive or negative, its absolute value will be positive. This ensures that the distance given by the function will be positive, because the product of 5 and another positive number will always be positive. First, let’s find 𝑑(5): 𝑑(5)=5|8βˆ’5|=5|3|=5(3)=15(8βˆ’π‘‘3).thevalueofis

Based on our calculations, we know that, after 5 seconds, the distance between the body and point 𝐡 was 15 centimetres.

Now let’s find 𝑑(11): 𝑑(11)=5|8βˆ’11|=5|βˆ’3|=5(3)=15(8βˆ’π‘‘βˆ’3).thevalueofis

Our calculations show us that the distance between the body and point 𝐡 was also 15 centimetres after 11 seconds. Notice that our two distances are the same. This tells us that after 5 seconds, the body must not have yet reached point 𝐡 and that after 11 seconds, it must have already passed point 𝐡.

Now let’s finish by recapping some key points.

Key Points

  • An absolute value function is a function with a definition that contains an algebraic expression within absolute value symbols.
  • The domain of all absolute value functions that are in the form 𝑓(π‘₯)=|π‘šπ‘₯+𝑏| is the set of all real numbers, or ℝ, while the range is 𝑓(π‘₯)β‰₯0, or [0,∞[.
  • The graph of 𝑓(π‘₯)=𝑔(π‘₯βˆ’β„Ž) is a horizontal translation of the graph of 𝑔(π‘₯)=|π‘₯|. If β„Ž is positive, the translation is β„Ž units right. If β„Ž is negative, the translation is |β„Ž| units left. The domain and range of 𝑓(π‘₯) are the same as those of 𝑔(π‘₯).
  • The graph of 𝑓(π‘₯)=𝑔(π‘₯)+π‘˜ is a vertical translation of the graph of 𝑔(π‘₯)=|π‘₯|. If π‘˜ is positive, the translation is π‘˜ units up. If π‘˜ is negative, the translation is |π‘˜| units down. The domain of 𝑓(π‘₯) is the same as that of 𝑔(π‘₯), but the range is 𝑓(π‘₯)β‰₯π‘˜.
  • The graph of 𝑓(π‘₯)=π‘Žβ‹…π‘”(π‘₯) is a vertical stretch of the graph of 𝑔(π‘₯)=|π‘₯| by a scale factor of |π‘Ž| when |π‘Ž|>1 and a vertical compression by a scale factor of |π‘Ž| when 0<|π‘Ž|<1. If π‘Ž is negative, the graph of 𝑔(π‘₯) is also reflected over the π‘₯-axis, which means that the graph of 𝑓(π‘₯) will open downward instead of upward.
  • The graph of 𝑓(π‘₯)=𝑔(𝑏⋅π‘₯) is a horizontal compression of the graph of 𝑔(π‘₯)=|π‘₯| by a scale factor of 1|𝑏| when |𝑏|>1 and a horizontal stretch by a scale factor of 1|𝑏| when 0<|𝑏|<1. If 𝑏 is negative, the graph of 𝑔(π‘₯) is also reflected over the 𝑦-axis.

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