### Video Transcript

Find the horizontal and vertical asymptotes of the function π of π₯ is equal to two divided by three π₯ minus five plus seven.

The question gives us a rational function π of π₯. And it wants us to find the horizontal and vertical asymptotes of this function. We recall that the line π¦ equals πΏ is a horizontal asymptote of our function π of π₯. If 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. Similarly, we say that the line π₯ is equal to π is a vertical asymptote of our function π of π₯. If either the limit as π₯ approaches π from the left of π of π₯ is equal to either positive or negative β. Or the limit as π₯ approaches π from the right of π of π₯ is equal to positive or negative β.

Letβs start by using this definition to find the horizontal asymptotes of our function π of π₯. Weβll start by checking the limit as π₯ approaches β of π of π₯. Thatβs the limit as π₯ approaches β of two divided by three π₯ minus five plus seven. The first thing weβll do is use the fact that the limit of a sum is equal to the sum of the limits to rewrite our limit as a sum of two limits. We can now evaluate both of these limits directly.

First, as π₯ approaches β, our numerator two remains constant. However, our denominator of three π₯ minus five is going without bound. So, this limit evaluates to give us zero. In our second limit, as π₯ approaches β, seven remains constant. So, this limit evaluates to give us seven. So, weβve shown the limit as π₯ approaches β of π of π₯ is equal to seven. In other words, π¦ is equal to seven is a horizontal asymptote of our function π of π₯.

Remember, we also need to check the limit as π₯ approaches negative β of π of π₯ since this could give us a different horizontal asymptote. And we could do this using a similar method. However, letβs check what would have happened in our working out if instead weβd taken the limit as π₯ approaches negative β. We can still write our limit as the sum of two limits. We now want to check that both of our limits still evaluate to give us the same values.

The constant seven didnβt change as π₯ changes. So, taking the limit as π₯ approaches negative β does not change the fact that this limit evaluates to give us seven. Similarly, in our first limit, as π₯ approaches negative β, our numerator remains constant. However, our denominator now approaches negative β as π₯ approaches negative β.

However, if our numerator is constant and our denominator is approaching negative β, then our outputs are getting smaller and smaller; theyβre approaching zero. So, again we get the output of seven. This means the only horizontal asymptote of our function π of π₯ is π¦ is equal to seven.

We could now do something similar to find all of the vertical asymptotes of our function π of π₯. However, because π of π₯ is a rational function in this case, we can use a simpler method. We recall for a rational function π of π₯ divided by π of π₯, if π₯ is equal to π is a vertical asymptote of this function, then we must have that π evaluated at π is equal to zero. Itβs worth noting at this point that if π evaluated at π is equal to zero, then the line π₯ is equal to π is not necessarily a vertical asymptote. This just helps us find all of the possible vertical asymptotes.

At this point, we might be worried. Our function π of π₯ is a rational function. However, itβs not in the form π of π₯ divided by π of π₯. So, we might be tempted to write it in this form by cross multiplying. However, this is not necessary, since when we add seven after our function, weβre translating it up seven units. A vertical translation of seven units wonβt change where our vertical asymptotes are.

So, letβs let π of π₯ be π of π₯ translated vertically down seven units. Then, π of π₯ has the same vertical asymptotes as our function π of π₯. Now, our function π of π₯ is in the form π of π₯ divided by π of π₯. So, the only possible vertical asymptote of our function is when π of π₯ is equal to zero. In other words, three π₯ minus five needs to be equal to zero. And we see this is when π₯ is equal to five over three. We now need to check that this is indeed a vertical asymptote of our function.

And because weβre dealing with a rational function in this case, thereβs a simple method. If the polynomial in our numerator π evaluated at π is not equal to zero, then π₯ is equal to π must be a vertical asymptote of our function. However, if we do in fact get that this is equal to zero, then we just remove all of our shared factors of π₯ minus π in our rational function.

In this case, we donβt actually need to worry about this. Our polynomial in our numerator is just the constant two. So, π evaluated at π₯ is equal to five over three is not equal to zero. So, π₯ is equal to five over three must be a vertical asymptote. And remember, this was the only possible vertical asymptote for π of π₯ and π of π₯ and π of π₯ have the same vertical asymptotes.

Therefore, weβve shown the function π of π₯ is equal to two divided by three π₯ minus five plus seven has only one horizontal asymptote when π¦ is equal to seven and it has only one vertical asymptote when π₯ is equal to five over three.