### Video Transcript

Find the horizontal and vertical asymptotes of the function π of π₯ is equal to four minus three divided by π₯.

The question gives us a function π of π₯ and asks us to find all of the horizontal and vertical asymptotes of this function. Letβs start by recalling what we mean by horizontal asymptotes. We say the line π¦ is equal to πΏ is a horizontal asymptote of the function π of π₯ if either of the following are true. Either the limit as π₯ approaches β of π of π₯ is equal to πΏ or the limit as π₯ approaches negative β of π of π₯ is equal to πΏ. This means to find all of the horizontal asymptotes of our function π of π₯, we need to check the limit as π₯ approaches β of π of π₯ and the limit as π₯ approaches negative β of π of π₯.

So letβs start with the limit as π₯ approaches β of π of π₯. In our case, this is the limit as π₯ approaches β of four minus three divided by π₯. First, weβll use the fact that the limit of a difference is equal to the difference of the limits. This gives us the limit as π₯ approaches β of four minus the limit as π₯ approaches β of three divided by π₯. And in our second limit, weβll take out the constant factor of three from our limit. We can now evaluate both of these limits. First, four is a constant. So it doesnβt change as the value of π₯ changes. This means the limit as π₯ approaches β of four is just equal to four.

And in our second limit, we know the limit as π₯ approaches β of the reciprocal function is equal to zero. This means our second limit evaluates to give us zero. And then we have three times zero, which is equal to zero. So weβve shown the limit as π₯ approaches β of four minus three divided by π₯ is equal to four. And therefore, weβve shown π¦ is equal to four is a horizontal asymptote of our function π of π₯. Remember though, we do need to check the limit as π₯ approaches negative β of our function π of π₯ because this could give us a different horizontal asymptote.

So letβs check the limit as π₯ approaches negative β of our function π of π₯. Again, weβll split this into the difference of the limits. This gives us the limit as π₯ approaches negative β of four minus the limit as π₯ approaches negative β of three over π₯. Again, weβll take out our constant of three in our second limit. And now we can evaluate both of these limits. First, the limit as π₯ approaches negative β of the constant four is equal to four. And the limit as π₯ approaches negative β of the reciprocal function is also equal to zero. So the limit as π₯ approaches negative β of π of π₯ is also equal to four. So this function only has one horizontal asymptote which is the line π¦ is equal to four.

We now need to find all of the vertical asymptotes of our function π of π₯. And thereβs a few different ways of doing this. Just like with the horizontal asymptotes, we could work directly with the definition of vertical asymptotes and then apply this to our function π of π₯. However, with vertical asymptotes, itβs often easier to work directly with our function π of π₯. Our function π of π₯ is four minus three divided by π₯. We can consider this as a set of transformations on the function one over π₯. So letβs start with our reciprocal function one over π₯. We know this only has one vertical asymptote, which is the line π₯ is equal to zero.

Next, by looking at our function π of π₯, we can see we multiply this by negative three. Multiplying by negative three is two transformations, so weβll split this into each individual transformation. First, weβll multiply this by negative one. And of course, we know this is the same as reflecting our curve through the π₯-axis. But reflecting our curve through the π₯-axis wonβt change the positions of our vertical asymptotes. It might change whether they go up to β or down to negative β. But it wonβt change where they appear on our π₯-axis. In other words, the graph π¦ is equal to negative one divided by π₯ also has a vertical asymptote when π₯ is equal to zero.

We now want to transform this to the curve π¦ is equal to negative three divided by π₯. Of course, multiplying our entire function by three is a vertical stretch from a factor of three. But again, stretching our function vertically will not change where our vertical asymptotes are. It might change how quickly they go to positive or negative β. But it wonβt change where they appear on our π₯-axis. So our curve π¦ is equal to negative three over π₯ also has a vertical asymptote when π₯ is equal to zero.

The last transformation we need to do to make this curve our function π of π₯ is to translate it up four units. So if we translate our previous curve four units vertically, we get four minus three over π₯, which is our function π of π₯. And of course, translating a curve vertically will not change the positions of its vertical asymptotes. So this curve will also have its only vertical asymptote when π₯ is equal to zero. So weβve now found all of the horizontal and vertical asymptotes of the function given to us in the question.

We could stop here. However, itβs worth talking about using these transformations to find our horizontal asymptotes. To start, we know that our curve π¦ is equal to one over π₯ has the horizontal asymptote π¦ is equal to zero. Letβs now think what a reflection in the π₯-axis would do to our horizontal asymptotes. Well, the horizontal asymptotes would be reflected in the π₯-axis. So we need to multiply our horizontal asymptote by negative one. So the curve π¦ is equal to negative one over π₯ would have its horizontal asymptote the line π¦ is equal to negative one times zero. Of course, in this case, negative one times zero is equal to zero.

Now letβs consider what happens to our horizontal asymptotes when we stretch vertically by a factor of three. This means we need to stretch our horizontal asymptote by a factor of three vertically. This would give us the line π¦ is equal to three times zero. But, of course, in this case, this is just the line π¦ is equal to zero. The last transformation we did is to translate our curve four units vertically. Again, this means that our horizontal asymptotes will be translated four units vertically.

Doing this, we get the line π¦ is equal to zero plus four. And this simplifies to give us the horizontal asymptote π¦ is equal four. And this is the same answer we got before. However, this time, we did it in a more graphical interpretation. Therefore, we were able to find all of the horizontal and vertical asymptotes of our function π of π₯ is equal to four minus three divided by π₯. We showed it had one horizontal asymptote which was the line π¦ is equal to four and one vertical asymptote which was the line π₯ is equal to zero.