### Video Transcript

Which of the following is the range of π of π₯ is equal to negative three minus the absolute value of two minus π₯? Option (A) the left-closed, right-open interval from negative three to β. Option (B) the left-open, right-closed interval from negative β to zero. Option (C) the left-open, right-closed interval from negative β to negative three. Option (D) the set of real numbers minus the set including two. Or option (E) the set of real numbers.

In this question, weβre given a function π of π₯, and weβre asked to determine which of five given options is the range of this function. So letβs start by recalling what we mean by the range of a function. Itβs the set of all output values of the function given its domain. And thereβs several different ways we can determine the range of a function. One way of doing this is from a graph of the function. And to determine the range of π of π₯ in this way, weβre going to need to sketch the graph π¦ is equal to π of π₯. To do this, we can note that the equation of π of π₯ is similar to the equation of π of π₯ is the absolute value of π₯. In fact, we can perform a series of transformations of π of π₯ to get the graph of π¦ is equal to π of π₯.

To determine the transformations needed to transform π of π₯ into π of π₯, letβs start by rewriting π of π₯ in terms of π of π₯. First, weβll reorder the two terms of π of π₯. π of π₯ is equal to negative one times the absolute value of two minus π₯ minus three. We could now write this in terms of π of π₯. However, we can simplify this slightly by taking a factor of negative one outside the expression inside the absolute value symbol.

We notice the absolute value of negative one times π₯ minus two will be equal to the absolute value of π₯ minus two, because remember the absolute value of an expression only cares about its size, not its sign. Therefore, π of π₯ is equal to negative one times the absolute value of π₯ minus two minus three. And we can write this as negative one times π of π₯ minus two minus three.

And now that weβve written π of π₯ in this form, we can determine the series of transformations that will transform π¦ is equal to π of π₯ into π¦ is equal to π of π₯. We start with the coefficient of π. This is negative one, which means we need to reflect in the horizontal axis. Next, thereβs no coefficient of π₯, so we move on to subtracting two from the inputs. This means we translate the graph two units right. Finally, weβre subtracting three from the output values. This means we need to translate the graph three units down.

We can now sketch π¦ is equal to π of π₯ by applying these transformations to the graph of π¦ is equal to the absolute value of π₯. First, we need to reflect this in the horizontal axis. Doing this, we get the following. This is the graph π¦ is equal to negative the absolute value of π₯. After this, we need to translate the graph two units to the right. Translating the graph two units to the right gives us the following. Itβs the sketch of π¦ is equal to negative the absolute value of π₯ minus two. And whenever weβre sketching the graph of a function, it can be useful to include useful points on the graph.

For example, in π¦ is equal to negative the absolute value of π₯, the corner of the graph is at the origin. When we translate this two units right, it will be at π₯ is equal to two. This is the π₯-intercept of this graph. Finally, we need to translate this graph three units down. Translating the graph three units down gives us the graph of π¦ is equal to negative the absolute value of π₯ minus two minus three. And we can note that the points at the corner of our curve translate down three units to the coordinates two, negative three.

Weβre now ready to determine the range of our function from its graph. However, letβs start by clearing some space so we can work with a bigger diagram. To determine the range of π¦ is equal to π of π₯ from its graph, we start by recalling the π₯-coordinate of a point on its graph tells us the input value of π₯. And the π¦-coordinate tells us the corresponding output of the function. For example, because the point with coordinates two, negative three lies on the graph, π evaluated at two must be equal to negative three. Negative three is an output of our function. In fact, we can see itβs the largest output. Another way of saying this is itβs the point with largest π¦-coordinate which lies on the graph.

Similarly, we can see that all values of π¦ less than or equal to negative three are possible outputs of the function. Any horizontal line below π¦ is equal to negative three intersects the graph. Therefore, the range of this function is all values less than or equal to negative three. We write this as the left-open, right-closed interval from negative β to negative three, which we can see is option (C). Therefore, the range of π of π₯ is equal to negative three minus the absolute value of two minus π₯ is option (C) the left-open, right-closed interval from negative β to negative three.