Lesson Video: Absolute Value Functions Mathematics

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.

Let’s consider the function 𝑓 of 𝑥 which is equal to the absolute value, sometimes known as the modulus, of 𝑥. To graph this function, we will choose some integer values of 𝑥 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 𝑦 is equal to 𝑓 of 𝑥. 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 𝑚𝑥 plus 𝑏. Let’s 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 𝑥. The 𝑦-axis is a line of symmetry. This means that the equation of the line of symmetry is 𝑥 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 𝑥 into our function, the domain is the set of all real numbers. This can also be written as the open interval negative ∞ to the open interval ∞. The range is the outputs of the function. It is all the values of 𝑦 or 𝑓 of 𝑥. These will all lie above or on the 𝑥-axis. Therefore, 𝑓 of 𝑥 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 ∞. Finally, we see that the 𝑦-intercept and 𝑥-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 𝑓 of 𝑥 is equal to the absolute value of negative two 𝑥 minus two.

The range of a function 𝑦 equals 𝑓 of 𝑥 is the set of values 𝑦 takes for all values of 𝑥 within the domain of 𝑓. For any graph drawn on the coordinate plane, the range or output is all the set of values that 𝑦 takes. For the function 𝑓 of 𝑥 which is equal to the absolute value of negative two 𝑥 minus two, all our values of 𝑦 are above or on the 𝑥-axis. We can therefore say that 𝑦 or 𝑓 of 𝑥 must be greater than or equal to zero. This can be written using interval notation from the left-closed, right-open interval zero to ∞. 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 𝑓 of 𝑥 can never actually reach ∞, we use an open interval for the upper limit.

Determine the domain and range of the function 𝑓 of 𝑥 is equal to the absolute value of negative 𝑥 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 𝑥-values that can be substituted into the function. As we can substitute any value of 𝑥 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 ∞ and the open interval ∞. The range of a function is the set of all possible output values. On the coordinate plane, it is all the values of 𝑦 or 𝑓 of 𝑥. We can see on our graph that the vertex or minimum point is at negative one, one. The range of the function 𝑓 is therefore from the left-closed, right-open interval one to ∞.

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 𝑓 of 𝑥 which is the absolute value of negative 𝑥 minus one plus one has a domain of all real values and a range from one to ∞.

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

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

Let’s begin by recalling what the graph of the function the absolute value of 𝑥 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 𝑥-values. This means that the domain of 𝑔 of 𝑥, the absolute value of 𝑥, is all real values. Whilst it is not mentioned in this question, the range is the set of 𝑦 or output values. As the 𝑦-values of 𝑔 of 𝑥 are all greater than or equal to zero, the range of 𝑔 of 𝑥 is from the left-closed, right-open interval from zero to ∞.

Let’s now consider our function 𝑓 of 𝑥, which is equal to the absolute value of 𝑥 plus 𝑎. The function 𝑓 of 𝑥 is a horizontal translation of 𝑔 of 𝑥 𝑎 units to the left. If 𝑎 is a positive number, the graph will shift or translate 𝑎 units to the left. If 𝑎 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 𝑓 of 𝑥 which is equal to the absolute value of 𝑥 plus 𝑎 is the set of all real values. This could also be written as the set of values from negative ∞ to ∞.

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

Let’s begin by looking at what the general function the absolute value of 𝑥 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 𝑓 of 𝑥. Let’s begin by considering the absolute value of 𝑥 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 𝑥 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 𝑥-axis.

Finally, we need to subtract one from this function. This will result in a translation one unit down in the 𝑦-direction. The function 𝑓 of 𝑥 which is equal to negative four multiplied by the absolute value of 𝑥 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 𝑓 of 𝑥 is at the coordinates five, negative one. We know that the domain of any function is the set of 𝑥- or input values. As we can input any value into our function 𝑓 of 𝑥, the domain is the set of all real values from negative ∞ to ∞.

The range is the set of all output or 𝑦-values. From the graph, we can see that these are all values less than or equal to negative one. The range of 𝑓 of 𝑥 is therefore equal to the set of values on the left-open, right-closed interval from negative ∞ 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 𝐴 to the point 𝐶 passing through the point 𝐵 without stopping. The distance between the body and point 𝐵 is given by 𝑑 of 𝑡 is equal to five multiplied by the absolute value of eight minus 𝑡, where 𝑡 is the time in seconds and 𝑑 is the distance in centimeters. Determine the distance between the body and the point 𝐵 after five seconds and after 11 seconds.

We are given a diagram which shows the body that is about to move from point 𝐴 to point 𝐶 via point 𝐵 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 𝑑 of 𝑡 is equal to five multiplied by eight minus 𝑡. 𝑑 of 𝑡 is the distance of the body from point 𝐵 after a given time. We need to calculate this distance after five seconds and also after 11 seconds. After five seconds, 𝑡 is equal to five. Therefore, we need to calculate 𝑑 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 𝐵 after five seconds is 15 centimeters.

We need to repeat this process when 𝑡 equals 11. This means we need to calculate the value of 𝑑 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 𝐵 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 𝐵. After five seconds, it is still approaching point 𝐵 from point 𝐴. And after 11 seconds, it is past point 𝐵 and it’s heading towards Point 𝐶.

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 𝑥- or input values. We saw that when dealing with the absolute value of linear functions of the form 𝑚𝑥 plus 𝑏, the domain was equal to all real values from the open interval negative ∞ to the open interval ∞. The range of a function is the set of 𝑦- or output values. If 𝑓 of 𝑥 is equal to the absolute value of 𝑚𝑥 plus 𝑏, then the range will be from the closed interval zero to the open interval ∞. The range will contain all values of 𝑦 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 𝑔 of 𝑥 which is equal to 𝑓 of 𝑥 minus ℎ translates the function 𝑓 of 𝑥 equals the absolute value of 𝑥 horizontally. It shifts the graph ℎ units to the right. 𝑓 of 𝑥 plus ℎ would move the graph ℎ units to the left. Likewise, the function 𝑔 of 𝑥 which is equal to 𝑓 of 𝑥 plus 𝑘 translates the function 𝑓 of 𝑥 equals the absolute value of 𝑥 vertically. This translation shifts the graph 𝑘 units up. As with the previous example, 𝑓 of 𝑥 minus 𝑘 this time would shift the graph 𝑘 units down.

Finally, we saw that the function 𝑔 of 𝑥 which is equal to 𝑎 multiplied by 𝑓 of 𝑥 stretches the function 𝑓 of 𝑥 equals the absolute value of 𝑥. The value of 𝑎 is the scale factor. And if a is 𝑎 negative number, the graph is reflected in the 𝑥-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.