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GMAT Prep: Help and Review25 chapters | 288 lessons | 15 flashcard sets

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Lesson Transcript

Instructor:
*Peter Kosek*

Peter has taught Mathematics at the college level and has a master's degree in Mathematics.

Rational functions are an extremely useful type of function found in mathematics. They are functions that are fractions whose numerator and denominator are both polynomials. Learn all about them in this lesson!

A **rational function** is a function that is a fraction and has the property that both its numerator and denominator are polynomials. In other words, R(*x*) is a rational function if R(*x*) = p(*x*) / q(*x*) where p(*x*) and q(*x*) are both polynomials. Recall that a **polynomial** is any function of the form f(*x*) = *a*-sub-zero + *a*-sub-1 times *x* + *a*-sub-2 times *x*^2 + . . . + *a*-sub-*n* * *x*^*n*, where *a*-sub-0, a-sub-1, . . ., *a*-sub-*n* are all real numbers and the exponents of each *x* is a non-negative integer.

The definition you just got might be a little overbearing, so let's look at some examples of rational functions:

The function R(*x*) = (*x*^2 + 4*x* - 1) / (3*x*^2 - 9*x* + 2) is a rational function since the numerator, *x*^2 + 4*x* - 1, is a polynomial and the denominator, 3*x*^2 - 9*x* + 2 is also a polynomial.

The function R(*x*) = (-2*x*^5 + 4*x*^2 - 1) / *x*^9 is a rational function since the numerator, -2*x*^5 + 4*x*^2 - 1, is a polynomial and the denominator, *x*^9, is also a polynomial.

The function R(*x*) = 1 / ((*x* - 1)(*x*^2 + 3)) is a rational function since the numerator, 1, is a polynomial (yes, a constant is still a polynomial) and the denominator, (*x* - 1)(*x*^2 + 3), is also a polynomial (it's just in a factored form).

The function R(*x*) = (sqrt(*x*) + *x*^2) / (3*x*^2 - 9*x* + 2) is not a rational function since the numerator, sqrt(*x*) + *x*^2, is not a polynomial since the exponent of *x* is not an integer.

The function R(*x*) = (*x* - 4) / *x*^(-2/3) + 4 is not a rational function since the denominator, *x*^(-2/3) + 4, is not a polynomial since the exponent of *x* is not a non-negative integer.

One of the most unique properties of a rational function is that it may have vertical asymptotes. First off, we should probably define a vertical asymptote. A **vertical asymptote** at a value *x* is when the value of our function approaches either positive or negative infinity when we evaluate our function at values that approach *x* (but are not equal to *x*).

This example may help clarify the idea of a vertical asymptote:

We see there is a vertical asymptote when *x* = 1 since the function is approaching negative infinity as we approach 1 from the left, and the function is approaching positive infinity as we approach 1 from the right.

How do we find the vertical asymptotes (if any exist) if we have been given a rational function? We can use the following theorem:

Theorem: Let R(*x*) be a rational function with no common factors between the numerator and the denominator. Then, the real values of *x* that make our denominator equal to 0 will have vertical asymptotes.

Let's use this theorem to find vertical asymptotes!

Find all vertical asymptotes of the function:

R(*x*) = (-2*x*^3 + 4*x*^2 - 1) / (*x*^2 + *x*)

First, we see that R(*x*) is indeed a rational function with no common factors between the numerator and denominator. That's great because that means we can use the theorem! We have to find what values of *x* make our denominator equal to 0. Since our denominator is *x*^2 + *x*, we'll set it equal to 0 and solve for *x*. Therefore, we have *x*^2 + *x* = 0. Factoring the left hand side, we get *x*(*x* + 1) = 0. Hence, *x* = 0 and -1. Our vertical asymptotes exist at *x* = 0 and *x* = -1.

You might be thinking. . . Wait, we didn't use the numerator! Were we supposed to? Well, we technically did use the numerator since we had to make sure there were no common factors between the numerator and denominator. But, when it comes to actually computing our vertical asymptotes, we only use the denominator!

Let's look at another example.

Find all the vertical asymptotes of the function:

R(*x*) = (*x*^2 + 2) / ((*x* + 3)(*x*^2 + 1))

First, we see that R(*x*) is indeed a rational function (because remember, a factored polynomial is still a polynomial) with no common factors between the numerator and denominator. Once again, that's great news because that means we can use our theorem! We have to find what values of *x* make our denominator equal to 0. Since our denominator is (*x* + 3)(*x*^2 + 1), we'll set it equal to 0 and solve for *x*. Therefore, we have (*x* + 3)(*x*^2 + 1) = 0. Setting each part equal to 0, we get *x* + 3 = 0 and *x*^2 + 1 = 0. Since *x*^2 + 1 = 0 has no real solutions, the only vertical asymptote comes from *x* + 3 = 0. Hence, the only vertical asymptote occurs at *x* = -3.

Remember, a **rational function** is a function that is a fraction where both its numerator and denominator are polynomials. **Vertical asymptotes**, which are when the value of our function approaches either positive or negative infinity when we evaluate our function at values that approach *x* (but are not equal to *x*), may occur in rational functions. As long as there are no common factors between the numerator and denominator, the vertical asymptotes will appear at the *x* values that make our denominator equal to 0.

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GMAT Prep: Help and Review25 chapters | 288 lessons | 15 flashcard sets

- How to Add and Subtract Rational Expressions 8:02
- Practice Adding and Subtracting Rational Expressions 9:12
- How to Multiply and Divide Rational Expressions 8:07
- Multiplying and Dividing Rational Expressions: Practice Problems 4:40
- How to Solve a Rational Equation 7:58
- Rational Equations: Practice Problems 13:15
- Solving Rational Equations with Literal Coefficients 5:26
- Solving Problems Using Rational Equations 5:21
- Finding Constant and Average Rates 6:41
- Solving Problems Using Rates 6:05
- Rational Function: Definition, Equation & Examples 6:28
- Go to Algebra - Rational Expressions: Help and Review

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