Video: Finding the Asymptotes of a Rational Function

Find the horizontal and vertical asymptotes of 𝑓(π‘₯) = (2π‘₯Β³ + 2)/(π‘₯Β³ βˆ’ 1).

09:33

Video Transcript

Find the horizontal and vertical asymptotes of the function 𝑓 of π‘₯ is equal to two π‘₯ cubed plus two all divided by π‘₯ cubed minus one.

The question is asking us to find the horizontal and the vertical asymptotes of this function. We can recall that a function will have a horizontal asymptote at the line 𝑦 is equal to 𝐿 when either the limit as π‘₯ approaches infinity of the function is equal to 𝐿 or when the limit as π‘₯ approaches negative infinity of the function is equal to 𝐿. It’s worth noting at this point that it is possible to get at most two different horizontal asymptotes, which happens when our limits approach different constants.

For example, if we look at a plot of the graph of 𝑦 is equal to one divided by one plus 𝑒 to the π‘₯, we noticed that it has two different horizontal asymptotes: one at 𝑦 is equal to one and one at 𝑦 is equal to zero. So when we’re finding our horizontal asymptotes, we will need to check both the limit as π‘₯ approaches infinity of the function and the limit as π‘₯ approaches negative infinity of the function. Let’s start with the limit as π‘₯ approaches positive infinity.

We recall that to find the limit of a rational function such as this, we can divide the numerator and the denominator by the highest power of π‘₯ found in the denominator. We can see that the highest power of π‘₯ found in the denominator is π‘₯ cubed. So we divide both the numerator and the denominator of the function in our limit by π‘₯ cubed. We can then split both the fraction in our numerator and the fraction in our denominator into separate fractions. Then, we can cancel all the shared factors of π‘₯ cubed, giving us the limit as π‘₯ approaches infinity of the function two plus two over π‘₯ cubed all divided by one minus one over π‘₯ cubed.

Now we can use the fact that the limit of the quotient between two functions is equal to the quotient of the limit of those two functions to rewrite our limit of the quotient between two functions as the quotient of the limit of two functions. We also know that the limit of the sum of two functions is equal to the sum of the limits of those two functions. So we can use this to rewrite the limit of the sum of two functions in our numerator as the sum of the limit of two functions. We also know that our rule is true for the difference between the limit of two functions. So we can rewrite the limit of the difference between two functions in our denominator as the difference between two limits.

We know that for any constant π‘˜, the limit of the constant π‘˜ as π‘₯ approaches π‘Ž can be evaluated to just be equal to π‘˜. So the limit as π‘₯ approaches infinity of the constant two in our numerator can be evaluated to be equal to two. And the limit as π‘₯ approaches infinity of the constant one in our denominator can be evaluated to just be equal to one. Next, we’re going to use that the limit of a constant π‘˜ multiplied by a function is equal to the constant π‘˜ multiplied by the limits of the function. We can use this to take the constant two out of the limit as π‘₯ approaches infinity of two over π‘₯ cubed in our numerator.

We can now use the fact that the limits of a power of a function is equal to the power of the limits of that function. We can, therefore, write our limit of one over π‘₯ cubed as π‘₯ approaches infinity in both the numerator and the denominator to be equal to the limit of the reciprocal function as π‘₯ approaches infinity cubed, which gives us the following expression to evaluate. Finally, we know that the limit of the reciprocal function as π‘₯ approaches infinity can be evaluated as zero. So both the limits of the reciprocal function in our expression could be evaluated to be zero, giving us two plus two multiplied by zero cubed all divided by one minus zero cubed which we can calculate to be equal to two.

What we have shown is that the limit as π‘₯ approaches infinity of the function two π‘₯ cubed plus two all divided by π‘₯ cubed minus one is evaluated as two. So by using our definition of a horizontal asymptote, we can conclude that there is a horizontal asymptote for the function 𝑓 of π‘₯ at the line 𝑦 is equal to two.

We now remember that we must also check the limit of the function 𝑓 of π‘₯ as π‘₯ approaches negative infinity. We can see that if we want to change our limit to negative infinity, almost all of the rules which we have used would stay the same. We can do the exact same series of calculations which we did before. However, we will end up with the limits of the reciprocal function as π‘₯ approaches negative infinity. But we know that the limit of the reciprocal function as π‘₯ approaches negative infinity is also equal to zero. So we can continue with the same line of calculations which we did before, and we will end up with that there is a horizontal asymptote for the function 𝑓 of π‘₯ on the line 𝑦 is equal to two. Therefore, what we have shown is that the only horizontal asymptote for our function 𝑓 of π‘₯ is the line 𝑦 is equal to two.

So let’s clear some space so we can discuss if there are any vertical asymptotes of the function 𝑓 of π‘₯. We know that there is a vertical asymptote of the function 𝑓 of π‘₯ at the line π‘₯ is equal to π‘Ž if any of the following are true: if the limit of 𝑓 of π‘₯ as π‘₯ approaches π‘Ž from the right is equal to positive or negative infinity, or if the limit of 𝑓 of π‘₯ as π‘₯ approaches π‘Ž from the left is equal to positive or negative infinity, or if the limit of 𝑓 of π‘₯ as π‘₯ approaches π‘Ž is equal to positive or negative infinity. Since our function 𝑓 of π‘₯ is a rational function returns to quotient, we could only have that the limit of 𝑓 of π‘₯ as π‘₯ approaches a point be equal to positive or negative infinity when we divide by zero.

So to find the potential values of our vertical asymptotes, we need to find out when the denominator π‘₯ cubed minus one is equal to zero. We can see that when π‘₯ is equal to one, our denominator will output one cubed minus one, which we know is equal to zero. So, by the factor theorem, we can conclude that π‘₯ minus one is a factor of π‘₯ cubed minus one. Therefore, we can write π‘₯ cubed minus one as the product of π‘₯ minus one and some unknown quadratic. By equating the coefficient of π‘₯ cubed, we can see that we must have that π‘₯ cubed is equal to π‘₯ multiplied by π‘Žπ‘₯ squared, giving us that π‘Ž must be equal to one. If we equate the constants, we can see that negative one must be equal to negative one multiplied by 𝑐, giving us that 𝑐 must be equal to one.

Now, to find the value of 𝑏, we will compare the coefficients of π‘₯ squared on both sides of the equation. On the left, we see that there is no π‘₯ squared term, so we can just write zero π‘₯ squared. On the right-hand side, we can see that our π‘₯ squared terms we will get by multiplying π‘₯ and 𝑏π‘₯ and then adding this to negative one multiplied by π‘₯ squared. This gives us that 𝑏π‘₯ squared minus π‘₯ squared must be equal to zero. And so, 𝑏 is equal to one. This means we can rewrite the denominator of 𝑓 of π‘₯ as π‘₯ minus one multiplied by π‘₯ squared plus π‘₯ plus one.

If we were then to calculate the discriminant of this quadratic, we would get one squared minus four multiplied by one times one which we can calculate to be equal to negative three. Since the discriminant of our quadratic is negative, we can conclude that this quadratic has no real roots. This tells us that the domain of our function 𝑓 is all real values of π‘₯, except when π‘₯ is equal to one. Therefore, the only possible vertical asymptote of the function 𝑓 of π‘₯ will be the line π‘₯ is equal to one.

For a rational function 𝑓 written as a quotient, if 𝑓 of π‘Ž is a nonzero number divided by zero, then π‘₯ equals π‘Ž is always a vertical asymptote of the function 𝑓. In our case, we can see the 𝑓 of π‘₯ is a rational function written as a quotient. So we can evaluate the output of the function 𝑓 of π‘₯ at π‘₯ is equal to one, which is two multiplied by one cubed plus two all divided by one cubed minus one, which we can calculate to be four divided by zero, which is of the form of a nonzero number divided by zero.

Therefore, we can conclude that π‘₯ equals one is a vertical asymptote of the graph of the function of 𝑓 of π‘₯. If we were to plot a graph of our function 𝑓 of π‘₯, we would see that we do get a vertical asymptote at π‘₯ is equal to one. And we get a horizontal asymptote both for the left and the right when 𝑦 is equal to two.

Since we showed earlier that the only horizontal asymptote of the graph of the function of 𝑓 of π‘₯ is the line 𝑦 is equal to two and we just showed that the only vertical asymptote of the graph of the function of 𝑓 of π‘₯ is the line π‘₯ is equal to one. We can conclude that these are the only horizontal and vertical asymptotes of our function.

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