Table of Contents

Last modified on August 3rd, 2023

The word ‘ratio’ is the root of the term ‘rational’. Thus, a rational function is the ratio of two polynomial functions where the denominator is never 0.

A function of a real variable x, say f(x), is a rational function if it can be represented as:

f(x) = ${\dfrac{p\left( x\right) }{q\left( x\right) }}$, here p(x), q(x) are polynomials in x with q(x) ≠ 0.

- For example, f(x) = ${\dfrac{3x+5}{x^{2}-x-2}}$, is a rational function where the denominator x
^{2}– x – 2 ≠ 0. - Since every constant is a polynomial, a rational function can have a constant numerator. For example, f(x) = ${\dfrac{2}{x + 1}}$.
- The denominator of a rational function is never a constant. For example, ${\dfrac{x + 2}{3}}$ is not a rational function. It is a polynomial or a linear function.
- A function is not rational if the numerator or the denominator is not a polynomial. For example, ${\dfrac{2-\sqrt{x}}{x^{2}+5x+6}}$ is not a rational function as the numerator 2 – ${\sqrt{x}}$ is not a polynomial.
- f(x) = ${\dfrac{cos x}{x +1}}$ is also not a rational function, as cos x is not a polynomial.

Rational functions are applied broadly in science for interpolating and approximating functions in numerical analysis and solving related rates, time, and work problems.

The domain of a rational function f(x) includes all possible real values of x, which is true for the given function. Hence it excludes the values for which the denominator of the rational function is zero.

For example, let us find the domain of f(x) = ${\dfrac{x + 1}{ x^{2}+3x-10}}$.

First, we find the zeros of the denominator x^{2} + 3x – 10

x^{2} + 3x – 10 = 0

⇒ (x – 2) (x + 5) = 0

⇒ x = 2 or x = -5

Hence the rational function f(x) is not defined for x = {2, -5} and thus the domain is {x ∈ ${\mathbb{R}}$ : x ≠ 2, -5}

The range of a rational function y = f(x) is the set of all the values of y for the corresponding values of x.

For example, to find the range of the rational function f(x) = ${\dfrac{x + 1}{ x^{2}+3x-10}}$, we write

y = f(x) = ${\dfrac{x + 1}{ x^{2}+3x-10}}$

⇒ y (x^{2} +3x – 10) – (x – 1) = 0

⇒ (y) x^{2} + (3y – 1) x – (10y +1) = 0 …. (i)

As x is real, the discriminant of (i) is ≥ 0, and thus, y ≠ 0

So, 49 y^{2} – 2y + 1 ≥ 0

⇒ 49 ${\left( y-\dfrac{1}{49}\right) ^{2}}$ + ${\dfrac{48}{49}}$ ≥ 0 ….(ii)

As the lowest possible value of ${\left( y-\dfrac{1}{49}\right) ^{2}}$ is 0, (ii) is true for all values of y.

So, the range of the given function is {y ∈ ${\mathbb{R}}$}

In a rational function, an excluded value is any value of x that makes the function value y undefined. So, these values are excluded from the domain of the function.

For example, the excluded value of the function ${y=\dfrac{3}{x-4}}$ is 4. That is, when x = 4, the value of y is undefined.

An asymptote is a line that the graph of the function approaches but never actually touches. Thus, these values are excluded from the domain of the function. A rational function can have three kinds of asymptotes: vertical (at least one), horizontal (at most one), and oblique (at most one).

Considering the rational function f(x) = ${\dfrac{p\left( x\right) }{q\left( x\right) }}$, we will discuss the different types of asymptotes.

It is a vertical line that the graph of the function approaches but never actually touches. The graph approaches the vertical line x = 0 as the input reaches zero.

A rational function can have one or more vertical asymptotes. To find the vertical asymptote, we set the denominator q(x) = 0.

For example, the function f(x) = ${\dfrac{x}{ x^{2}-1}}$ is undefined when x = ±1

Thus, the vertical asymptotes of the given function are x = {1, -1}

It is a horizontal line that the graph of the function approaches but never touches. A rational function can have a maximum of one horizontal asymptote.

Let the degree of q(x) = d and the degree of p(x) = n. f(x) has a horizontal asymptote only when n ≤ m.

- If n < d, the x-axis or y = 0 is the horizontal asymptote.
- If n = d, the horizontal asymptote is given by ${y=\dfrac{a_{n}}{b_{m}}}$, a
_{n}, b_{m}are the leading coefficients of p(x) and q(x). - If n > d, there is no horizontal asymptote.

For the function f(x) = ${\dfrac{4x}{ x-2}}$, n = d = 1, a_{n }= 4 and b_{d} = 1.

Hence the horizontal asymptote is y = 4.

It is an oblique line that the graph of the function approaches but never touches.

A rational function f(x) has an oblique or slant asymptote when the degree of p(x) is exactly greater than the degree of q(x) by 1.

In this case, the oblique asymptote is y = quotient, where the quotient is obtained when the numerator is divided by the denominator through the long division method.

In the rational function f(x) = ${ \dfrac{x^{2}-5x+4}{x-3}}$, the degree of p(x) = 2, degree of q(x) = 1. Hence the given function has an oblique asymptote.

Now dividing (x^{2} – 5x + 4) by (x – 3) through long division method gives the quotient (x – 2) and the remainder (-2), i.e., f(x) = ${ \dfrac{x^{2}-5x+4}{x-3}}$ = (x-2) – ${\dfrac{2}{ x-3}}$

Thus the oblique asymptote of the function f(x) = ${ \dfrac{x^{2}-5x+4}{x-3}}$ is x – 2s

Suppose a rational function f(x) passes through a coordinate but is undefined at the x value of the same coordinate. Then, that coordinate is called a hole of that given function.

A hole exists at {a, f(a)} only when the denominator and the numerator of a rational function have a common factor (x – a).

For example, let us find the holes of the function f(x) = ${\dfrac{x^{2}-2x-8}{x^{2}+5x+6}}$.

Now, f(x) = ${\dfrac{x^{2}-2x-8}{x^{2}+5x+6}}$ = ${\dfrac{\left( x-4\right) \left( x+2\right) }{\left( x+3\right) \left( x+2\right) }}$

Hence (x + 2) is a common factor of both the numerator and the denominator, and f(x) = ${ \dfrac{x-4}{x+3}}$.

Thus there is a hole at x = (-2), and the coordinate is given by (-2, f(-2)) = (-2, -6)

Like polynomial functions, a rational function has zeros or roots, the points where the graph crosses the x-axis, or f(x) = 0. In a rational function, they are the zeros of the numerator, and they don’t depend on the denominator unless there is a hole.

The end behavior of a function 𝑓(𝑥) describes the function’s behavior when 𝑥→+∞ or 𝑥→−∞. It is equal to the horizontal asymptotes, slant/oblique asymptotes, or the quotient obtained when long dividing the polynomials.

For the parent function ${f\left( x\right) =\dfrac{1}{x}}$, the end behavior is:

- For the horizontal asymptote f(x) → 0 as x → ∞ or -∞
- For the vertical asymptote f(x) → ∞ as x → 0+ and f(x) → -∞ as x → 0-

We will follow the following steps to solve and graph a rational function.

Let, y = f(x) = ${\dfrac{x^{2}-2x-8}{x^{2}+5x+6}}$ be a rational function.

**Step 1: **First, we factor out the common factor (if any) of the numerator and the denominator and write f(x) in the simplest form.

We have already shown that (x + 2) is a common factor of the above function’s numerator and denominator.

Thus, f(x) = ${ \dfrac{x-4}{x+3}}$.

**Step 2: **The x-intercepts or zeros of the function are plotted by putting y = 0, and the y-intercepts are plotted by putting x = 0

So for the function f(x) = ${ \dfrac{x-4}{x+3}}$, the x-intercept is at (4, 0), and the y-intercept is at ${\left( 0,-\dfrac{4}{3}\right)}$.

**Step 3: **We identify and draw the vertical asymptote using a dotted line.

Here, as x = -3 is a zero of f(x), the vertical asymptote is x = -3.

**Step 4: **We draw the horizontal asymptote (if any) using another dotted line.

Here degree of (x – 4) = degree of (x + 3) = 1. Hence the horizontal asymptote of f(x) is at y = ${ \dfrac{1}{1}}$ = 1.

**Step 5:** Then, the holes are plotted (if any).

We have already shown that the hole of f(x) is at (-2, -6)

**Step 6:** Thus, we pick random points to the right and left of both the x-intercepts and the vertical asymptote to find the corresponding values of y.

For the given function ${\dfrac{x^{2}-2x-8}{x^{2}+5x+6}}$, the values are given below:

x | y = f(x) |

-6 | 3.333 |

-5 | 4.5 |

-4 | 8 |

-3 | Vertical Asymptote |

-2 | – 6 |

0 | -1.333 |

1 | –0.75 |

2 | -0.4 |

3 | -0.167 |

4 | 0 (x-intercept) |

5 | 0.125 |

6 | 0.222 |

Finally, these points are plotted and joined by two curves to get the graph of the given rational function.

**Which of the following rational functions is graphed below****A.${\dfrac{1}{x-3}}$ C.${\dfrac{3}{x}}$**

**B.${\dfrac{1}{3x}}$ D.${\dfrac{1}{x+3}}$**

Solution:

Let the required function be f(x)

The graph shows a vertical asymptote at x = 3. So the denominator of f(x) contains the term (x – 3) and has the form:

f(x) = ${\dfrac{p\left( x\right) }{x-3}}$

Here the horizontal asymptote is y = 0, as the curve never touches the x-axis. Hence, the degree of p(x) < the degree of (x – 3) = 1

Thus the required rational function is option A: ${\dfrac{1}{x-3}}$

Let f(x) be a rational function. To find the inverse f-1(x) of a rational function f(x), first f(x) is replaced by y and x, y are interchanged. Then the resulting equation is solved, and we write the value of y in terms of x.

For example, to get the inverse of the function f(x) = ${\dfrac{1+2x}{7+x}}$, we write

y = ${\dfrac{1+2x}{7+x}}$.

Interchanging x and y gives, x = ${\dfrac{1+2y}{7+y}}$ ….(i)

Solving (i) for y, we get y = ${\dfrac{1-7x}{x-2}}$. Thus f^{-1}(x) = ${\dfrac{1-7x}{x-2}}$.

Last modified on August 3rd, 2023