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Equivalent Representations of Rational Expressions (College Board AP® Precalculus): Study Guide

Written by: Roger B

Reviewed by: Mark Curtis

Updated on 9 April 2026

Equivalent representations of rational expressions

What is a rational function?

A rational function is a function that can be written as a quotient (fraction) of two polynomial functions
- $r (x) = \frac{p (x)}{q (x)}$
  - where $p (x)$ is the polynomial in the numerator
  - and $q (x)$ is the polynomial in the denominator

What does the factored form tell me about a rational function?

Writing the numerator and denominator of a rational function in factored form makes it easy to see
- Real zeros ( $x$ -intercepts)
  - These come from the zeros of the numerator that are not also zeros of the denominator
  - See the Zeros of Rational Functions study guide
- Vertical asymptotes
  - These occur at values of $x$ where the denominator is zero but the numerator is not zero at that value
  - A vertical asymptote also occurs if the multiplicity of the zero in the denominator is greater than its multiplicity in the numerator
  - See the Vertical Asymptotes of Rational Functions study guide
- Holes
  - These occur at values of $x$ where both the numerator and denominator are zero
  - and the multiplicity of that zero in the numerator is greater than or equal to its multiplicity in the denominator. The common factor "cancels", leaving a hole rather than an asymptote
  - See the Holes of Rational Functions study guide
Factored form also makes it easy to identify the domain of a rational function
- The domain must exclude any $x$ values where the denominator equals zero
  - These correspond to vertical asymptotes or holes
E.g. consider $r (x) = \frac{(x - 3) (x + 1)}{(x + 1) (x - 5)}$
- The numerator has zeros at $x = 3$ and $x = - 1$
  - and the denominator has zeros at $x = - 1$ and $x = 5$
- At $x = - 1$ , both numerator and denominator are zero, with equal multiplicity, so there is a hole at $x = - 1$
  - At $x = 5$ , the denominator is zero but the numerator is not, so there is a vertical asymptote at $x = 5$
  - At $x = 3$ , the numerator is zero but the denominator is not, so there is a zero ( $x$ -intercept) at $x = 3$
- The domain of the function can include real numbers except $x = - 1$ and $x = 5$
  - i.e. except for the points where the denominator is equal to zero

Examiner Tips and Tricks

On the exam, the domain of a rational function is assumed to be all real numbers except for ones which make the denominator equal to zero. Any exceptions to this (for example, a more restricted domain) will be noted explicitly in the question.

What does the standard form of the numerator and denominator reveal?

Having the numerator and denominator in standard form (i.e. expanded polynomial form) makes it easy to see
- the degree, leading terms and leading coefficients of each polynomial
This makes it easy to determine the end behavior of the rational function:
- If the degree of the numerator is less than the degree of the denominator
  - then there is a horizontal asymptote at $y = 0$
- If the degrees are equal
  - then there is a horizontal asymptote at $y = \frac{a_{n}}{b_{n}}$ (the ratio of the leading coefficients)
- If the degree of the numerator is greater than the degree of the denominator
  - then there is no horizontal asymptote
    - The function grows without bound
    - If the numerator's degree exceeds the denominator's by exactly 1, then the graph also has a slant asymptote
- See the End Behavior of Rational Functions study guide

How do I convert between forms for rational functions?

To factor
- Factor the numerator and denominator separately
  - using the same techniques as for polynomials (common factors, quadratic factoring rules)
To expand
- Multiply out the factors in the numerator and denominator separately

Why is it useful to have different forms?

Different forms make different information accessible
- Factored form gives info about zeros, $x$ -intercepts, vertical asymptotes, holes and domain
- Standard form (of numerator and denominator) gives info about end behavior, including horizontal or slant asymptotes
Being able to convert between forms allows you to extract whatever information is needed from a single rational function

Examiner Tips and Tricks

When given a rational function and asked about its graph, the first step is almost always to factor both the numerator and the denominator. This lets you identify

common factors (which give holes)
remaining denominator zeros (which give vertical asymptotes)
and remaining numerator zeros (which give $x$ -intercepts)

Many exam questions about rational functions can be answered just by carefully factoring and comparing zeros.

Worked Example

The function $g$ is given by $g (x) = x^{3} + 2 x^{2} - 15 x$ , and the function $h$ is given by $h (x) = x^{2} + x - 12$ . Let $k$ be the function given by $k (x) = \frac{h (x)}{g (x)}$ . What is the domain of $k$ ?

(A) all real numbers $x$ where $x \neq 0$

(B) all real numbers $x$ where $x \neq - 4$ , $x \neq 3$

(D) all real numbers $x$ where $x \neq - 5$ , $x \neq - 4$ , $x \neq 0$ , $x \neq 3$ , $x \neq 4$

Answer:

The domain of $k$ is all real numbers except where the denominator $g (x) = 0$

Factor $g (x)$

$\begin{array}{rcl} g (x) & = & x^{3} + 2 x^{2} - 15 x \\ = & x (x^{2} + 2 x - 15) \\ = & x (x + 5) (x - 3) \end{array}$

So the zeros of $g$ are $x = 0$ , $x = - 5$ , and $x = 3$ .

The domain of $k$ excludes these three values, regardless of whether they produce vertical asymptotes or holes

Note that the numerator $h (x) = x^{2} + x - 12 = (x + 4) (x - 3)$ has a zero at $x = 3$ in common with the denominator

This means the graph of $k$ has a hole at $x = 3$ (rather than a vertical asymptote)
but $x = 3$ still must be excluded from the domain

Worked Example

Which of the following functions has a zero at $x = 5$ , a vertical asymptote at $x = - 1$ , and a hole at $x = 2$ ?

(A) $h (x) = \frac{x^{2} - 7 x + 10}{x^{2} - x - 2}$

(B) $j (x) = \frac{x^{2} - 3 x + 2}{x^{2} - 7 x + 10}$

(D) $m (x) = \frac{x - 5}{x^{2} - 7 x + 10}$

Answer:

Factor the numerator and denominator of each option:

For option (A)

$\frac{x^{2} - 7 x + 10}{x^{2} - x - 2} = \frac{(x - 5) (x - 2)}{(x - 2) (x + 1)}$

Common factor $(x - 2)$ , so hole at $x = 2$ ✓
Remaining denominator factor $(x + 1)$ , so vertical asymptote at $x = - 1$ ✓
Remaining numerator factor $(x - 5)$ , so zero at $x = 5$ ✓

All three conditions are met.

For option (B)

$\frac{(x - 1) (x - 2)}{(x - 5) (x - 2)}$

This has a hole at $x = 2$
- but the vertical asymptote is at $x = 5$ (not $x = - 1$ )
- and the zero is at $x = 1$ (not $x = 5$ )

For option (C)

$\frac{x - 5}{(x - 2) (x + 1)}$

This has vertical asymptotes at $x = 2$ and $x = - 1$ , but no hole

For option (D)

$\frac{x - 5}{(x - 5) (x - 2)}$

This has a hole at $x = 5$ (not a zero)
- and a vertical asymptote at $x = 2$ (not $x = - 1$ )

(A) $h (x) = \frac{x^{2} - 7 x + 10}{x^{2} - x - 2}$

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Equivalent Representations of Rational Expressions (College Board AP® Precalculus): Study Guide

Equivalent representations of rational expressions

What is a rational function?

What does the factored form tell me about a rational function?

Examiner Tips and Tricks

What does the standard form of the numerator and denominator reveal?

How do I convert between forms for rational functions?

Why is it useful to have different forms?

Examiner Tips and Tricks

Worked Example

Worked Example

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Join the 100,000+ Students that ❤️ Save My Exams

Author: Roger B

Reviewer: Mark Curtis