Question Video: Find the Vertical Asymptotes of a Rational Function

Find all the vertical asymptotes for the function 𝑓(π‘₯) = 5/(π‘₯ βˆ’ 3)(3π‘₯Β² + 9π‘₯ βˆ’ 30).

04:50

Video Transcript

Find all the vertical asymptotes for the function 𝑓 of π‘₯ is equal to five divided by π‘₯ minus three multiplied by three π‘₯ squared plus nine π‘₯ minus 30.

The question gives us a rational function 𝑓 of π‘₯. And it wants us to find all of the vertical asymptotes for this function. We might be tempted to start straight from the definition of a vertical asymptote. We recall we call the line π‘₯ is equal to π‘Ž a vertical asymptote of the function 𝑓 of π‘₯ if any of the following are true. Either the limit as π‘₯ approaches π‘Ž of our function 𝑓 of π‘₯ needs to be equal to positive or negative ∞. Or the limit as π‘₯ approaches π‘Ž from the left of 𝑓 of π‘₯ needs to be equal to positive or negative ∞. Or the limit as π‘₯ approaches π‘Ž from the right of 𝑓 of π‘₯ needs to be equal to positive or negative ∞.

We could work directly with this definition for our function 𝑓 of π‘₯. However, remember, we’re given a rational function 𝑓 of π‘₯. So we can use some shortcuts to find our vertical asymptotes. To start, if 𝑓 of π‘₯ is a rational function, let’s say the polynomial 𝑃 of π‘₯ divided by the polynomial 𝑄 of π‘₯. Then if π‘₯ is equal to π‘Ž is a vertical asymptote of our function 𝑓 of π‘₯, we must have our denominator 𝑄 evaluated at π‘Ž is equal to zero. This makes sense because 𝑃 of π‘₯, the numerator, is a polynomial. It’s not gonna be equal to ∞ for any input π‘₯.

But remember, our denominator is also a polynomial. So the only way that this expression is going to give us a vertical asymptote is if our denominator is equal to zero. This gives us a method for finding all of our possible vertical asymptotes. We just need to find all of the values of π‘₯ where our denominator is equal to zero. But remember, this does not guarantee that these will be vertical asymptotes. Instead, we need to remember if our numerator is not equal to zero and our denominator is equal to zero when π‘₯ is equal to π‘Ž. Then π‘₯ is equal to π‘Ž must be a vertical asymptote of our function 𝑓 of π‘₯.

And again, this makes a lot of sense. This time, if we were to take the limit as π‘₯ approaches π‘Ž of our rational function 𝑃 of π‘₯ divided by 𝑄 of π‘₯, our numerator is getting closer and closer to 𝑃 evaluated at π‘Ž, which we’ve said is not equal to zero. However, our denominator is getting closer and closer to zero. And this is not an indeterminate form. So this means it must be a vertical asymptote. And it’s worth pointing out if our numerator does evaluate to give us zero, then it’s inconclusive whether we have a vertical asymptote or not here.

So let’s start by finding all of our possible vertical asymptotes. That’s when our denominator is equal to zero. So we want to find all of the values of π‘₯ where π‘₯ minus three multiplied by three π‘₯ squared plus nine π‘₯ minus 30 is equal to zero. So we want to fully factor the left-hand side of our equation. To start, we notice each term inside of our quadratic shares a factor of three. So we want to take a factor of three from this quadratic. This leaves us with π‘₯ squared plus three π‘₯ minus 10.

And since we’re solving this equal to zero, we don’t need to worry about this factor of three. So now we just need to factorize our quadratic π‘₯ squared plus three π‘₯ minus 10. And we can do this in a few different ways. For example, we can use the quadratic solver on our calculator or we could use the quadratic formula. We could also notice that five multiplied by negative two is equal to negative 10. And five minus two is equal to three. So we can factor this quadratic to give us π‘₯ plus five multiplied by π‘₯ minus two.

So now we fully factored the denominator of our rational function to give us π‘₯ minus three times π‘₯ plus five multiplied by π‘₯ minus two. And, of course, if the product of three factors is equal to zero, then one of these factors must be equal to zero. So we’ve shown the possible vertical asymptotes of our rational function 𝑓 of π‘₯ at the lines π‘₯ is equal to three, π‘₯ is equal to negative five, and π‘₯ is equal to two. But remember, these are not guaranteed vertical asymptotes. We still need to check that these are indeed vertical asymptotes for our function.

To do this, we remember if our numerator is not equal to zero when π‘₯ is equal to π‘Ž and our denominator is equal to zero when π‘₯ is equal to π‘Ž. Then the line π‘₯ is equal to π‘Ž must be a vertical asymptote of our function. And this only works for rational functions. Let’s check our function 𝑓 of π‘₯. We can see the numerator is the constant five, so it’s never equal to zero. So when π‘₯ is equal to three, the numerator is equal to five and the denominator is equal to zero. So π‘₯ is equal to three is a vertical asymptote. In fact, the same is true when π‘₯ is equal to negative five and when π‘₯ is equal to two.

Therefore, all three of our lines are vertical asymptotes of our function 𝑓 of π‘₯. So we were able to show for the function 𝑓 of π‘₯ is equal to five divided by π‘₯ minus three multiplied by three π‘₯ squared plus nine π‘₯ minus 30 has three vertical asymptotes. The line π‘₯ is equal to negative five, the line π‘₯ is equal to two, and the line π‘₯ is equal to three.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.