Question Video: Identifying the Graph of an Absolute Value Function Mathematics • Third Year of Secondary School

Video Transcript

Which of the following is the graph of 𝑓 of 𝑥 is equal to 𝑥 multiplied by the absolute value of 𝑥 plus one? Options (A), (B), (C), (D), and (E).

In this question, we’re given a function 𝑓 of 𝑥, which involves the absolute value operation. We need to determine which of five given graphs is the graph of 𝑓 of 𝑥. And there’s many different ways of doing this. The easiest way is to eliminate options. And one way of doing this is to find the coordinates of some points which lie on the graph of 𝑦 is equal to 𝑓 of 𝑥. Let’s determine what the output of 𝑓 is when 𝑥 is equal to negative two.

We substitute 𝑥 is equal to negative two into the function 𝑓 of 𝑥 to get negative two multiplied by the absolute value of negative two plus one. And we can then simplify this. Negative two plus one is equal to negative one. So, we get negative two multiplied by the absolute value of negative one. And now we can recall taking the absolute value of a number removes its sign. So, the absolute value of negative one will be equal to one. So, we get that 𝑓 evaluated at negative two is equal to negative two. So, when 𝑥 is equal to negative two, the output of our function should be negative. It should be below the 𝑥-axis when 𝑥 is negative two.

However, we can see in options (A), (B), and (D) when 𝑥 is equal to negative two, the curve lies above the 𝑥-axis. So, it’s saying the outputs of these functions are positive. This doesn’t agree with our evaluation of 𝑓 at negative two. So, these cannot be the correct options. Let’s now clear some space and look at the other two options more closely.

Now that we have more space to look at these two options, we can notice that 𝑓 evaluated at negative two in both of these two curves is different. In option (C), if we look at the point on the curve with 𝑥-coordinate negative two, we can see that the output of the function, the 𝑦-coordinate of the point on this curve, is also negative two. However, this is not true in option (E). If we look at the point on the curve with 𝑥-coordinate negative two, we can see that the output value of this function, the 𝑦-coordinate of this point, is negative five. Therefore, the output of the function graphed in option (E) is negative five when 𝑥 is negative two. This cannot be the graph of 𝑓 of 𝑥 is equal to 𝑥 times the absolute value of 𝑥 plus one. And this is enough to conclude that the answer must be option (C), since all of the other graphs do not pass through the point with coordinates negative two, negative two.

And while this is a perfectly valid method of answering this question, there are a few problems with this method, the main one being we need to be given all five of the given options to use this method to answer the question. And it’s a much more useful skill to be able to sketch the graphs of functions. So, let’s also answer this question by sketching the graph of 𝑦 is equal to 𝑓 of 𝑥. There’s many different ways of doing this. However, let’s start by clearing some space.

We can then recall we can represent the absolute value function by using piecewise notation. The absolute value of 𝑥 is equal to 𝑥 whenever 𝑥 is greater than or equal to zero. And the absolute value of 𝑥 is equal to negative 𝑥 whenever our input values of 𝑥 are less than zero. This just tells us we output 𝑥 whenever 𝑥 is nonnegative. And if our input value of 𝑥 is negative, then we switch the sign of 𝑥. And we can use this to find a piecewise definition for our function 𝑓 of 𝑥. Let’s start by finding a piecewise definition of the absolute value of 𝑥 plus one.

To do this, let’s consider our piecewise definition of the absolute value of 𝑥. We want to take the absolute value of 𝑥 plus one. And there’s two ways of doing this. We can do this by using the definition of the absolute value function. And we would do this by noting when we input a nonnegative value, we want to output the same value. However, when we input a negative value, we want to switch the sign. However, this is not the only way. We could also just substitute 𝑥 plus one into this definition. This would give us that the absolute value of 𝑥 plus one is equal to 𝑥 plus one whenever 𝑥 plus one is greater than or equal to zero. And the absolute value of 𝑥 plus one is equal to negative one times 𝑥 plus one whenever 𝑥 plus one is less than zero.

We can simplify this definition slightly by rearranging our subdomains of the subfunctions. We can just subtract one from both sides of both inequalities. Doing this gives us a first subdomain of 𝑥 being greater than or equal to negative one and a second subdomain being 𝑥 is less than negative one. We can now use this to find a piecewise definition of 𝑓 of 𝑥, since 𝑓 of 𝑥 is just equal to 𝑥 multiplied by this piecewise function. And to multiply piecewise function by 𝑥, we just multiply all of its subfunctions by 𝑥. And doing this then gives us that 𝑓 of 𝑥 is equal to 𝑥 times 𝑥 plus one when 𝑥 is greater than or equal to negative one and 𝑓 of 𝑥 is equal to negative 𝑥 multiplied by 𝑥 plus one when 𝑥 is less than negative one.

And this is now a much easier form to sketch, since both of our subfunctions are factored quadratics. Now, all that’s left to do is sketch each of the subfunctions separately over their respective subdomains. And let’s start with the first subfunction 𝑥 multiplied by 𝑥 plus one for values of 𝑥 greater than or equal to negative one. We know that the graph of 𝑦 is equal to 𝑥 times 𝑥 plus one has two 𝑥-intercepts: one when 𝑥 is zero and one when 𝑥 is negative one. And since both of these values of 𝑥 are included in the subdomain of this subfunction, we can add both of these 𝑥-intercepts onto our graph.

Our graph will have 𝑥-intercepts at negative one and zero. In fact, we can now sketch this part of the graph, since our graph is a positive leading coefficient quadratic which has 𝑥-intercepts at negative one and zero. Remember though, we don’t sketch any 𝑥-coordinates below negative one since these are the values in our first subdomain. This gives us a sketch which looks like the following.

We can now do exactly the same to sketch the second subfunction. This time, we have a negative leading coefficient quadratic with 𝑥-intercepts at zero and negative one. This time, neither zero nor negative one are in the subdomain, since these values are not less than negative one. However, it is worth noting since negative one is a root, we know that this second subfunction will pass through the point with coordinates negative one, zero. So, we need to sketch a parabola with negative leading coefficient. So, we know it will open downwards, starting at the point with coordinates negative one, zero, which does not pass through the 𝑥-axis. We get a sketch which looks somewhat like the following. And now, we can see this exactly matches option (C).

Therefore, we were able to show of the five given options, option (C) represents the graph of 𝑓 of 𝑥 is equal to 𝑥 multiplied by the absolute value of 𝑥 plus one.