### Video Transcript

In this video, we will learn how to
evaluate and graph absolute value functions and identify their domain and range. We will begin by recalling what we
mean by an absolute value function. An absolute value function contains
an algebraic function within absolute value symbols. The absolute value of any number is
its distance from zero on the number line.

Letβs consider the function π of
π₯ which is equal to the absolute value, sometimes known as the modulus, of π₯. To graph this function, we will
choose some integer values of π₯ and find some ordered pairs. The absolute value of negative two
is equal to two as it is two away from zero on the number line. The absolute value of negative one
is equal to one as this is one away from zero on the number line. This leads us to the fact that the
absolute value of any number cannot be negative. We are not interested in the sign
and just the distance of the number from zero. The absolute value of zero, one,
and two are zero, one, and two, respectively.

We can then graph this function on
the coordinate plane where π¦ is equal to π of π₯. Our first coordinate or ordered
pair is negative two, two. We then have negative one, one. Our other three points are zero,
zero; one, one; and two, two. Connecting these points, we create
a V-shaped graph. This will be true for the absolute
value of any linear function of the form ππ₯ plus π. Letβs now consider some key points
or information from this graph. The vertex of the graph has
coordinates zero, zero. This is the minimum point of the
function the absolute value of π₯. The π¦-axis is a line of
symmetry. This means that the equation of the
line of symmetry is π₯ equals zero.

We know that the domain of any
function is those values that can be inputted into the function. As we can substitute any value of
π₯ into our function, the domain is the set of all real numbers. This can also be written as the
open interval negative β to the open interval β. The range is the outputs of the
function. It is all the values of π¦ or π of
π₯. These will all lie above or on the
π₯-axis. Therefore, π of π₯ is greater than
or equal to zero. Once again, this can be written as
an interval, from the closed interval zero to the open internal β. Finally, we see that the
π¦-intercept and π₯-intercept are both zero.

We will now look at a couple of
questions where we need to find the domain and range of an absolute value function
from its graph.

Find the range of the function π
of π₯ is equal to the absolute value of negative two π₯ minus two.

The range of a function π¦ equals
π of π₯ is the set of values π¦ takes for all values of π₯ within the domain of
π. For any graph drawn on the
coordinate plane, the range or output is all the set of values that π¦ takes. For the function π of π₯ which is
equal to the absolute value of negative two π₯ minus two, all our values of π¦ are
above or on the π₯-axis. We can therefore say that π¦ or π
of π₯ must be greater than or equal to zero. This can be written using interval
notation from the left-closed, right-open interval zero to β. We have a closed interval at zero
as the function can equal zero. This is denoted by the closed
circle or dot at negative one, zero. As π of π₯ can never actually
reach β, we use an open interval for the upper limit.

Determine the domain and range of
the function π of π₯ is equal to the absolute value of negative π₯ minus one plus
one.

The domain of any function is the
set of all possible input values. When the graph is drawn on the
coordinate plane, this is denoted by all the π₯-values that can be substituted into
the function. As we can substitute any value of
π₯ into our function, the domain is the set of all real values. This can also be written as the set
of values between the open interval negative β and the open interval β. The range of a function is the set
of all possible output values. On the coordinate plane, it is all
the values of π¦ or π of π₯. We can see on our graph that the
vertex or minimum point is at negative one, one. The range of the function π is
therefore from the left-closed, right-open interval one to β.

The solid circle or dot at negative
one, one means that we will use the closed interval as the value of one is included
in our range. The function π of π₯ which is the
absolute value of negative π₯ minus one plus one has a domain of all real values and
a range from one to β.

In our next two questions, we will
calculate the domain and range without being given a graph.

Given that π is a constant, what
is the domain of the function π of π₯ is equal to the absolute value of π₯ plus
π?

Letβs begin by recalling what the
graph of the function the absolute value of π₯ looks like. This is a V-shaped graph with
minimum points or vertex at zero, zero. We recall that the domain is the
set of input or π₯-values. This means that the domain of π of
π₯, the absolute value of π₯, is all real values. Whilst it is not mentioned in this
question, the range is the set of π¦ or output values. As the π¦-values of π of π₯ are
all greater than or equal to zero, the range of π of π₯ is from the left-closed,
right-open interval from zero to β.

Letβs now consider our function π
of π₯, which is equal to the absolute value of π₯ plus π. The function π of π₯ is a
horizontal translation of π of π₯ π units to the left. If π is a positive number, the
graph will shift or translate π units to the left. If π on the other hand was a
negative number, the graph would shift or translate to the right. As the graph has just shifted
horizontally, the range and domain have not altered. The domain of the function π of π₯
which is equal to the absolute value of π₯ plus π is the set of all real
values. This could also be written as the
set of values from negative β to β.

Find the domain and range of the
function π of π₯ is equal to negative four multiplied by the absolute value of π₯
minus five minus one.

Letβs begin by looking at what the
general function the absolute value of π₯ looks like. This is a V-shaped graph with
minimum point or vertex at zero, zero. We will now consider the
transformations that are made to obtain the function π of π₯. Letβs begin by considering the
absolute value of π₯ minus five. This is a translation five units to
the right. This function would therefore have
a minimum point or vertex at five, zero. Multiplying the absolute value of
π₯ minus five by negative four results in a stretch of scale factor negative
four. This means that the graph will be
four times steeper and also reflected in the π₯-axis.

Finally, we need to subtract one
from this function. This will result in a translation
one unit down in the π¦-direction. The function π of π₯ which is
equal to negative four multiplied by the absolute value of π₯ minus five minus one
is shown in green. We will now remove some of the
other graphs from the coordinate plane. The vertex or maximum point of π
of π₯ is at the coordinates five, negative one. We know that the domain of any
function is the set of π₯- or input values. As we can input any value into our
function π of π₯, the domain is the set of all real values from negative β to
β.

The range is the set of all output
or π¦-values. From the graph, we can see that
these are all values less than or equal to negative one. The range of π of π₯ is therefore
equal to the set of values on the left-open, right-closed interval from negative β
to negative one.

In the final question in this
video, we will evaluate an absolute value function by direct substitution.

A body was moving with a uniform
velocity of magnitude five centimeters per second from the point π΄ to the point πΆ
passing through the point π΅ without stopping. The distance between the body and
point π΅ is given by π of π‘ is equal to five multiplied by the absolute value of
eight minus π‘, where π‘ is the time in seconds and π is the distance in
centimeters. Determine the distance between the
body and the point π΅ after five seconds and after 11 seconds.

We are given a diagram which shows
the body that is about to move from point π΄ to point πΆ via point π΅ with a
velocity of five centimeters per second. Whilst there is a lot of
information in this question, the key point is that the function π of π‘ is equal
to five multiplied by eight minus π‘. π of π‘ is the distance of the
body from point π΅ after a given time. We need to calculate this distance
after five seconds and also after 11 seconds. After five seconds, π‘ is equal to
five. Therefore, we need to calculate π
of five.

This is equal to five multiplied by
the absolute value of eight minus five. Eight minus five is equal to three,
so we need to multiply five by the absolute value of three. As the absolute value of a number
is its distance from zero, the absolute value of three is three. As five multiplied by three is
equal to 15, the distance between the body and the point π΅ after five seconds is 15
centimeters.

We need to repeat this process when
π‘ equals 11. This means we need to calculate the
value of π of 11. This is equal to five multiplied by
the absolute value of eight minus 11. Eight minus 11 is equal to negative
three. As the absolute value of a number
is its distance from zero, the absolute value of negative three is also three. In fact, the absolute value of any
number will always be positive. Multiplying five by three once
again, we see that the distance between the body and point π΅ after 11 seconds is
also 15 centimeters.

In terms of our diagram, we can see
that after five seconds and 11 seconds, the body is the same distance away from
point π΅. After five seconds, it is still
approaching point π΅ from point π΄. And after 11 seconds, it is past
point π΅ and itβs heading towards Point πΆ.

We will now summarize the key
points from this video. In this video, we saw that the
absolute value of a number is its distance from zero. This means that the absolute value
of a number can never be negative. The domain of any function is the
set of π₯- or input values. We saw that when dealing with the
absolute value of linear functions of the form ππ₯ plus π, the domain was equal to
all real values from the open interval negative β to the open interval β. The range of a function is the set
of π¦- or output values. If π of π₯ is equal to the
absolute value of ππ₯ plus π, then the range will be from the closed interval zero
to the open interval β. The range will contain all values
of π¦ greater than or equal to zero.

We also saw in this video that the
transformations of the absolute value of linear functions sometimes alter the range
but never alter the domain. We saw that the function π of π₯
which is equal to π of π₯ minus β translates the function π of π₯ equals the
absolute value of π₯ horizontally. It shifts the graph β units to the
right. π of π₯ plus β would move the
graph β units to the left. Likewise, the function π of π₯
which is equal to π of π₯ plus π translates the function π of π₯ equals the
absolute value of π₯ vertically. This translation shifts the graph
π units up. As with the previous example, π of
π₯ minus π this time would shift the graph π units down.

Finally, we saw that the function
π of π₯ which is equal to π multiplied by π of π₯ stretches the function π of π₯
equals the absolute value of π₯. The value of π is the scale
factor. And if a is π negative number, the
graph is reflected in the π₯-axis. This means that it opens downwards
instead of upwards. The V-shaped graph is now upside
down. The last two transformations will
have an impact on the range of the absolute value function.