### Video Transcript

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

We begin by recalling what we actually mean by the domain and range of a function. We say that the domain of a function is the complete set of possible values for our independent variable, in other words, the set of π₯-values that make the function work and will essentially output real values for π¦. Then, the range is the complete set of all possible resulting values of the dependent variable after we substituted the domain. In other words, the range is the resulting π¦-values we get after substituting in all possible π₯-values.

Now, letβs think about what the function π of π₯ is telling us. We take values of π₯, substitute them into the expression five π₯ plus five, and then make that positive. Then, we subtract nine. There are no values of π₯ which donβt actually make the function work, so the domain of our function is simply all real numbers.

But what about the range? Well, thereβs two ways we can calculate this. Firstly, letβs just think about it algebraically. If we have the absolute value of five π₯ plus five, what do we get when we substitute values of π₯ in? Well, if π₯ is equal to negative one, we get five times negative one plus five, which is zero. And the absolute value of zero is simply zero. For all other real values of π₯, we get a result thatβs greater than zero. And so, we can say that the range of the function the absolute value of five π₯ plus five is greater than or equal to zero.

Weβre going to subtract nine from both sides of this inequality. And we find that the absolute value of five π₯ plus five minus nine must be greater than or equal to negative nine. And so, using set notation for the range, we find itβs greater than or equal to negative nine and less than β.

But there is another way we can find the range. And thatβs to consider the graph of the function. We take the graph of π¦ equals five π₯ plus five. Itβs a single straight line that passes through the π¦-axis at five and the π₯-axis at negative five. We then find the absolute value of all of our outputs, of all of our values of π¦. In other words, any values on the graph that lie below the π₯-axis get reflected in the π₯-axis, as shown. Notice that this does include π¦ equals zero.

By subtracting nine from our function, we translate it down by nine units. The lowest point on our graph has a coordinate of negative five, negative nine. And we said that the range is all possible values of π¦ after weβve substituted our possible values of π₯ in. We see from our graph that π¦ is always greater than or equal to negative nine. And so, we end up with the same range as before. Itβs greater than or equal to negative nine and less than β.