Which Absolute Value Function Defines This Graph

Which Absolute Value Function Defines This Graph.

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

Think that the
accented value
of a existent number is its distance from 0 on the number line. For example, in the expression
| βˆ’ 5 |
(which can be read equally the absolute value of
βˆ’ 5 ), the number
βˆ’ five
is shown within absolute value symbols. Since
βˆ’ v
is located five units from 0 on the number line, the value of the expression is v. The value of the expression
| 5 |
(which can be read every bit the absolute value of five) is 5 equally well, because 5 is also located 5 units from 0 on the number line.

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

One manner 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’ iii βˆ’ two βˆ’ 1 0 ane 2 3
𝑔 ( π‘₯ ) 3 2 1 0 one 2 3

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

We tin can see that the graph lies only on or to a higher place the
π‘₯ -axis. The arrows tell usa that it extends infinitely upward and to both the left and correct. This means that while the function tin can take any input value, its output values are only the nonnegative numbers. Nosotros can likewise see that the graph decreases to the left of the
𝑦 -centrality before touching the origin, and it and so increases to the correct of the
𝑦 -axis. The portions to the left and to the correct of the
𝑦 -centrality are mirror images of each other. Using our observations, we can depict various backdrop of the graph and of the function that information technology 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’ ∞ , ∞ )
  • 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. Yet, the functions’ other properties may differ. Transforming the graph of the office
𝑔 ( π‘₯ ) = | π‘₯ |
by translating it, stretching or compressing it, or reflecting it across an axis tin can allow united states to more easily sketch the graph of an accented value function in a different form so that we tin can place the domain and range of the function.

Definition: Transformations of the Graph of the Absolute Value Part 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 downwards. The domain of
    𝑓 ( π‘₯ )
    is the same as that of

    𝑔 ( π‘₯ ) ,
    merely the range is

    𝑓 ( π‘₯ ) β‰₯ π‘˜ .
  • The graph of
    𝑓 ( π‘₯ ) = π‘Ž β‹… 𝑔 ( π‘₯ )
    is a vertical stretch of the graph of
    𝑔 ( π‘₯ ) = | π‘₯ |
    by a scale gene of
    | π‘Ž |
    when
    | π‘Ž | > 1
    and a vertical compression past a scale factor of
    | π‘Ž |
    when

    0 < | π‘Ž | < i .
    If
    π‘Ž
    is negative, the graph of
    𝑔 ( π‘₯ )
    is also reflected over the
    π‘₯ -axis, which ways that the graph of
    𝑓 ( π‘₯ )
    will open downward instead of upward.
  • The graph of
    𝑓 ( π‘₯ ) = 𝑔 ( 𝑏 β‹… π‘₯ )
    is a horizontal compression of the graph of
    𝑔 ( π‘₯ ) = | π‘₯ |
    by a scale factor of
    ane | 𝑏 |
    when
    | 𝑏 | > i
    and a horizontal stretch by a scale cistron of
    one | 𝑏 |
    when

    0 < | 𝑏 | < 1 .
    If
    𝑏
    is negative, the graph of
    𝑔 ( π‘₯ )
    is also reflected over the
    𝑦 -axis.

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

Case 1: Finding the Range of an Absolute Value Role Using a Graph

Observe the range of the function

𝑓 ( π‘₯ ) = | βˆ’ 2 π‘₯ βˆ’ ii | .

Answer

Nosotros tin can meet that the graph has a vertex at

( βˆ’ 1 , 0 ) ,
a line of symmetry of

π‘₯ = βˆ’ 1 ,
an
π‘₯ -intercept of

βˆ’ one ,
and a
𝑦 -intercept of two. By inputting values of
π‘₯
into the function
𝑓 ( π‘₯ ) = | βˆ’ ii π‘₯ βˆ’ 2 |
and observing the resulting outputs, we can confirm the graph’s validity.

Allow’s use the graph to aid make up one’s mind the domain and range of

𝑓 ( π‘₯ ) .
Recall that the domain of a function is the set up 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 contained 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 correct, and so nosotros know that the domain must exist the set of all existent numbers, or

ℝ .
The arrows also tell u.s. that the graph extends infinitely upwardly, but we can run into that it lies simply on or above the
π‘₯ -axis. In other words, the smallest value of
𝑦
is 0, only there is no largest value. Thus, we know that the range is

𝑓 ( π‘₯ ) β‰₯ 0 ,
or

[ 0 , ∞ [ .
Any accented value function in the course
𝑓 ( π‘₯ ) = | π‘š π‘₯ + 𝑏 |
will have this range.

Now let’s look at the graph of some other absolute value role and find the function’due south 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 Office Using a Graph

Determine the domain and the range of the role

𝑓 ( π‘₯ ) = | π‘₯ + 1 | + i .

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 encounter that the graph is a horizontal translation i unit of measurement to the left and a vertical translation one unit up of the graph of

𝑔 ( π‘₯ ) = | π‘₯ | .
That is,

𝑓 ( π‘₯ ) = 𝑔 ( π‘₯ + 1 ) + i .
Thus, the domain of the role
𝑓 ( π‘₯ ) = | π‘₯ + 1 | + 1
is

ℝ ,
and the range is

𝑓 ( π‘₯ ) β‰₯ 1 ,
or

[ 1 , ∞ [ .

Note

We will arrive at the aforementioned answers for the office’southward domain and range by using the fact that the domain is the set of all possible input values, while the range is the prepare of all possible output values. The arrows tell us that the graph extends infinitely to both the left and the correct, so, again, we know that the domain of the part is the set of all existent numbers, or

ℝ .
The arrows also tell united states that the graph extends infinitely up, merely we can see that the graph lies merely at or higher up a value of one for

𝑦 .
That is, the smallest value of
𝑦
is 1, but there is no largest value. Thus, we can again run into that the range of the part is

𝑓 ( π‘₯ ) β‰₯ 1 ,
or

[ i , ∞ [ .

Side by side, let’due south expect at how to find the domain of an absolute value part without being given a graph. We volition construct the function’s graph past considering its relationship to that of

𝑔 ( π‘₯ ) = | π‘₯ | .

Case 3: Finding the Domain of an Absolute Value Part

Given that
π‘Ž
is a constant, what is the domain of the role

𝑓 ( π‘₯ ) = | π‘₯ + π‘Ž | ?

Answer

Every bit 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 correct.

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 every bit 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 fix of all real numbers, or

ℝ .

Annotation

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 make up one’s mind the domain of the role

𝑓 ( π‘₯ ) = | π‘₯ + π‘Ž | ,
nosotros could ask ourselves, β€œWhat values of
π‘₯
can nosotros input into the office?” To help answer this question, allow’s consider what the function would be for some unlike values of the constant

π‘Ž .

Kickoff we will consider a negative value for

π‘Ž .
Let’s assume that

π‘Ž = βˆ’ 1 .
When this is the instance, the function becomes

𝑓 ( π‘₯ ) = | π‘₯ βˆ’ 1 | .
Now permit’s input some values into the function and detect whether or not an output value is produced:
𝑓 ( βˆ’ ane ) = | βˆ’ 1 βˆ’ 1 | = | βˆ’ two | = two , 𝑓 ( 0 ) = | 0 βˆ’ 1 | = | βˆ’ 1 | = 1 , 𝑓 ο€Ό 1 2  = | | | i two βˆ’ 1 | | | = | | | βˆ’ ane 2 | | | = 1 ii , 𝑓 ( 1 ) = | 1 βˆ’ 1 | = | 0 | = 0 .

We tin can run across that all of the input values give a respective output value. Now let’south look at what happens when

π‘Ž = 0 .
In this instance, the function becomes

𝑓 ( π‘₯ ) = | π‘₯ | :

𝑓 ( βˆ’ i ) = | βˆ’ ane | = 1 , 𝑓 ( 0 ) = | 0 | = 0 , 𝑓 ο€Ό 1 2  = | | | one 2 | | | = 1 ii , 𝑓 ( 1 ) = | i | = one .

Over again, nosotros can run across that all of the input values give a corresponding output value. Finally, let’s consider a positive value of a past looking at the function when

π‘Ž = 1 .
Now, the office becomes

𝑓 ( π‘₯ ) = | π‘₯ + ane | :

𝑓 ( βˆ’ 1 ) = | βˆ’ 1 + one | = | 0 | = 0 , 𝑓 ( 0 ) = | 0 + 1 | = | 1 | = i , 𝑓 ο€Ό one 2  = | | | i 2 + 1 | | | = | | | 3 ii | | | = 3 2 , 𝑓 ( i ) = | 1 + i | = | ii | = 2 .

Just as before, there is an output value for each of the input values. In fact, because the accented value part
𝑔 ( π‘₯ ) = | π‘₯ |
is defined for all existent numbers

π‘₯ ,
no matter what constant
π‘Ž
represents, we volition always exist able to input whatsoever value into the role
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 accented value role to help find its domain and range. This time, nosotros will employ not only translations of the graph of
𝑔 ( π‘₯ ) = | π‘₯ |
when graphing our function, but a stretch and a reflection of
𝑔 ( π‘₯ ) = | π‘₯ |
besides.

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’ five | βˆ’ one .

Reply

To help make up one’s mind the domain and range of the office, let’s begin by amalgam its graph. Given

𝑔 ( π‘₯ ) = | π‘₯ | ,
we know that the function’s definition is in the grade

𝑓 ( π‘₯ ) = π‘Ž β‹… 𝑔 ( π‘₯ + β„Ž ) + π‘˜ ,
where

π‘Ž = βˆ’ four ,


β„Ž = βˆ’ v ,
and

π‘˜ = βˆ’ 1 .
Thus, we tin obtain the graph of
𝑦 = 𝑓 ( π‘₯ )
past vertically stretching the graph of
𝑦 = 𝑔 ( π‘₯ )
by a calibration factor of four, reflecting it over the
π‘₯ -centrality, translating it 5 units to the right, and then translating it i unit down as shown:

We tin verify that our graph is correct by inputting iii values of
π‘₯
into the function and observing the resulting output values. This is not necessary equally long as we are careful with our transformations, but it is useful for more complicated functions such as this 1. The values we choose for
π‘₯
will brand the value of the expression within the absolute value symbols

( π‘₯ βˆ’ v ) ,
negative, 0, and positive respectively:
𝑓 ( four ) = βˆ’ 4 | iv βˆ’ 5 | βˆ’ i = βˆ’ 4 | βˆ’ 1 | βˆ’ 1 = βˆ’ 4 ( 1 ) βˆ’ 1 = βˆ’ 4 βˆ’ 1 = βˆ’ 5 ( π‘₯ βˆ’ 5 βˆ’ 1 ) , 𝑓 ( v ) = βˆ’ 4 | 5 βˆ’ five | βˆ’ one = βˆ’ 4 | 0 | βˆ’ 1 = βˆ’ 4 ( 0 ) βˆ’ 1 = 0 βˆ’ 1 = βˆ’ 1 ( π‘₯ βˆ’ five 0 ) , 𝑓 ( 6 ) = βˆ’ four | half dozen βˆ’ 5 | βˆ’ ane = βˆ’ 4 | 1 | βˆ’ 1 = βˆ’ 4 ( one ) βˆ’ 1 = βˆ’ 4 βˆ’ i = βˆ’ 5 ( π‘₯ βˆ’ 5 1 ) . t h due east v a fifty u due east o f i s t h e v a 50 u due east o f i s t h due east five a l u e o f i s

We go the ordered pairs

( 4 , βˆ’ five ) ,


( 5 , βˆ’ ane ) ,
and

( 6 , βˆ’ 5 ) .
By plotting these points on the coordinate plane, and using our understanding of absolute value functions, nosotros tin once more create the graph of
𝑦 = 𝑓 ( π‘₯ )
for the function
𝑓 ( π‘₯ ) = βˆ’ 4 | π‘₯ βˆ’ 5 | βˆ’ 1
beneath:

The arrows on our graph tell us that the graph extends infinitely to both the left and to the correct, then we know that the domain of
𝑓 ( π‘₯ ) = βˆ’ 4 | π‘₯ βˆ’ v | βˆ’ 1
must exist the fix of all real numbers, or

ℝ .
The arrows too tell us that the graph extends infinitely down, but we can meet that information technology lies merely at or below a value of
βˆ’ ane
for

𝑦 .
In other words, the largest value of
𝑦
is

βˆ’ ane ,
simply there is no smallest value. This means that the range of the function is

𝑓 ( π‘₯ ) ≀ βˆ’ 1 ,
or

] βˆ’ ∞ , βˆ’ i ] .

At present allow’s look at a last example involving a real-earth problem.

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

A torso 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 | 8 βˆ’ 𝑑 | ,
where
𝑑
is the time in

seconds,
and
𝑑
is the distance in

cm.
Determine the distance between the body and the point
𝐡
later
5 seconds
and later on

11 seconds.

Reply

In order to decide the distance between the body and the point
𝐡
after
v seconds
and after

eleven seconds,
we demand to evaluate the function
𝑑 ( 𝑑 ) = 5 | 8 βˆ’ 𝑑 |
at
𝑑 = 5
and at

𝑑 = 1 1 .
In other words, we must substitute both five and 11 into the office for
𝑑
and calculate the resulting value of

𝑑 ( 𝑑 ) .

Call up that the accented value of a number is e’er positive, and then regardless of whether the value of the expression within the absolute value symbols
( 8 βˆ’ 𝑑 )
is positive or negative, its accented value will be positive. This ensures that the altitude given by the function will be positive, considering the product of v and another positive number will always be positive. First, permit’s observe

𝑑 ( v ) :

𝑑 ( 5 ) = 5 | 8 βˆ’ 5 | = 5 | iii | = 5 ( three ) = 1 5 ( 8 βˆ’ 𝑑 3 ) . t h east v a l u eastward o f i s

Based on our calculations, we know that, later

5 seconds,
the distance between the body and betoken
𝐡
was

15 centimetres.

Now let’due south notice

𝑑 ( 1 i ) :

𝑑 ( 1 ane ) = v | 8 βˆ’ 1 1 | = 5 | βˆ’ iii | = 5 ( 3 ) = 1 5 ( 8 βˆ’ 𝑑 βˆ’ iii ) . t h eastward v a l u e o f i due south

Our calculations show the states that the distance betwixt the body and point
𝐡
was too
15 centimetres
after

eleven seconds.
Notice that our two distances are the same. This tells us that after

five seconds,
the torso must not have yet reached point
𝐡
and that after

11 seconds,
it must accept already passed bespeak

𝐡 .

Now let’s end by recapping some primal points.

Key Points

  • An accented value function is a part with a definition that contains an algebraic expression inside absolute value symbols.
  • The domain of all absolute value functions that are in the class
    𝑓 ( π‘₯ ) = | π‘š π‘₯ + 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 downwardly. The domain of
    𝑓 ( π‘₯ )
    is the aforementioned equally that of

    𝑔 ( π‘₯ ) ,
    but the range is

    𝑓 ( π‘₯ ) β‰₯ π‘˜ .
  • The graph of
    𝑓 ( π‘₯ ) = π‘Ž β‹… 𝑔 ( π‘₯ )
    is a vertical stretch of the graph of
    𝑔 ( π‘₯ ) = | π‘₯ |
    by a scale factor of
    | π‘Ž |
    when
    | π‘Ž | > ane
    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 too reflected over the
    π‘₯ -axis, which means that the graph of
    𝑓 ( π‘₯ )
    volition open up downwardly instead of upward.
  • The graph of
    𝑓 ( π‘₯ ) = 𝑔 ( 𝑏 β‹… π‘₯ )
    is a horizontal compression of the graph of
    𝑔 ( π‘₯ ) = | π‘₯ |
    by a calibration factor of
    i | 𝑏 |
    when
    | 𝑏 | > 1
    and a horizontal stretch by a scale cistron of
    one | 𝑏 |
    when

    0 < | 𝑏 | < 1 .
    If
    𝑏
    is negative, the graph of
    𝑔 ( π‘₯ )
    is also reflected over the
    𝑦 -axis.

Which Absolute Value Function Defines This Graph

Source: https://www.nagwa.com/en/explainers/856195840620/