AOS2 Topic 1: Rational Functions

Rational functions play a fundamental role in mathematics and various fields of science and engineering. They are expressions that represent the quotient of two polynomial functions. The concept of rational functions is essential for understanding functions that arise in diverse contexts, from algebraic equations to differential equations.

Definition:

A rational function \( f(x) \) can be expressed as the ratio of two polynomial functions, typically denoted as \( P(x) \) and \( Q(x) \), where \( Q(x) \) is not the zero polynomial. Mathematically, it is represented as:

\[ f(x) = \frac{P(x)}{Q(x)} \]

In this expression:

  • \( P(x) \) is the numerator polynomial, which can be of any degree.
  • \( Q(x) \) is the denominator polynomial, which can also be of any degree, but it cannot be the zero polynomial.

Types of Rational Functions

  • Linear Rational Functions:

    Linear rational functions have both the numerator and denominator polynomials as linear. They are expressed as \( f(x) = \frac{ax + b}{cx + d} \), where \( a, b, c, \) and \( d \) are constants and \( c \neq 0 \). Linear rational functions typically have one vertical asymptote and no horizontal or slant asymptotes.

  • Quadratic Rational Functions:

    Quadratic rational functions have a quadratic numerator polynomial and a linear denominator polynomial. They are expressed as \( f(x) = \frac{ax^2 + bx + c}{dx + e} \), where \( a, b, c, d, \) and \( e \) are constants and \( d \neq 0 \). Quadratic rational functions may have one or two vertical asymptotes and no horizontal or slant asymptotes.

  • Cubic Rational Functions:

    Cubic rational functions have a cubic numerator polynomial and a linear denominator polynomial. They are expressed as \( f(x) = \frac{ax^3 + bx^2 + cx + d}{ex + f} \), where \( a, b, c, d, e, \) and \( f \) are constants and \( e \neq 0 \). Cubic rational functions may have one, two, or three vertical asymptotes and no horizontal or slant asymptotes.

  • Rational Functions with Holes:

    Rational functions with holes have a common factor between the numerator and denominator polynomials, leading to a hole in the graph. They are expressed as \( f(x) = \frac{g(x)}{h(x)} \), where \( g(x) \) and \( h(x) \) are polynomials and \( g(x) \) has a common factor with \( h(x) \). Rational functions with holes have vertical asymptotes at the zeros of the denominator polynomial, but the hole causes a discontinuity at those points.

  • General Rational Functions:

    General rational functions encompass all other types that do not fit into the above categories. They may have higher-degree polynomials in the numerator and/or denominator, resulting in more complex behavior. General rational functions can have various combinations of vertical, horizontal, and slant asymptotes, as well as other features such as intercepts and curves.

    • Domain of a rational function

    All real numbers, with the exception of those that result in a zero denominator, are included in the domain of a rational function.

    How To: Determine the domain of given rational function

  • Put 0 in the denominator.
  • Find the x-values that lead to the denominator equaling 0 by solving the problem.
  • Real numbers make up the domain, with the exception of those from Step 2.

    • Rational functions as a sum of partial fractions:

    Partial fraction decomposition is a method used to express a rational function as a sum of simpler fractions, called partial fractions. This technique is particularly useful for integrating rational functions or solving systems of linear differential equations.

    The process involves breaking down the rational function into partial fractions with denominators that are irreducible factors of the original denominator polynomial. The coefficients of these partial fractions are determined using algebraic techniques such as equating coefficients or evaluating at appropriate values.

    For example, consider the rational function:

    \( f(x) = \frac{2x + 1}{x^2 - 4} \)

    Using partial fraction decomposition, we can express it as:

    \( f(x) = \frac{A}{x - 2} + \frac{B}{x + 2} \)

    Where \( A \) and \( B \) are constants to be determined. Solving for \( A \) and \( B \) allows us to rewrite the original function as a sum of simpler fractions.

    Asymptotes

    Definition:

    Lines that a curve approaches but never touches or crosses are known as asymptotes. Asymptotes in the context of functions show how the function behaves when the independent variable gets closer to particular values. Different kinds of asymptotes exist:

    Vertical Asymptotes:

    These are the points at which the denominator of a rational function becomes zero, or when the function approaches infinity or negative infinity as the independent variable approaches a particular value.

    Horizontal Asymptotes:

    As the independent variable approaches positive or negative infinity, the function's graph approaches horizontal lines known as horizontal asymptotes. When the degree of the polynomials in the denominator and numerator are the same, horizontal asymptotes arise.

    Slant (sometimes called oblique) Asymptotes:

    Slant asymptotes happen when the denominator polynomial's degree is one higher than the numerator polynomial's degree. In this instance, as the independent variable approaches positive or negative infinity, the function becomes closer to a linear function.

    Graphing Rational Functions

    For sketching graphs, it is also useful to write rational functions in the alternative form, that is, with a division performed if possible. For example:

    \( f(x) = \frac{8x^2 - 3x + 2}{x} = \frac{8x^2}{x} - \frac{3x}{x} + \frac{2}{x} = 8x - 3 + \frac{2}{x} \)

    For this example, we can see that \( \frac{2}{x} \rightarrow 0 \) as \( x \rightarrow \pm\infty \), so the graph of \( y = f(x) \) will approach the line \( y = 8x - 3 \) as \( x \rightarrow \pm\infty \).

    We say that the line \( y = 8x - 3 \) is a non-vertical asymptote of the graph. This is a line or curve which the graph approaches as \( x \rightarrow \pm\infty \).

    Important features of a sketch graph are:

    • asymptotes
    • axis intercepts
    • stationary points
    • points of inflection

    Methods for sketching graphs of rational functions include:

    • adding the y-coordinates (ordinates) of two simple graphs
    • taking the reciprocals of the y-coordinates (ordinates) of a simple graph

    Addition of Ordinates

    Key points for addition of ordinates:

    • When the two graphs have the same ordinate, the y-coordinate of the resultant graph will be double this.
    • When the two graphs have opposite ordinates, the y-coordinate of the resultant graph will be zero (an x-axis intercept).
    • When one of the two ordinates is zero, the resulting ordinate is equal to the other ordinate.

    Intersection of Rational Functions with End Behaviour Asymptotes

    To find the location of any points of intersection with the graph of a rational function and its end behavior asymptote, solve a system of two equations consisting of the Reduced Equation \( R(x) \) and the equation of the End Behavior Asymptote, \( EBA(x) \). The End Behavior Asymptote could be either a horizontal asymptote (in the form \( y=c \) ), or a slant asymptote (in the form \( y=mx+b \) ). The Reduced Equation is used to make calculations simpler and to reduce obtaining solutions that are not in the domain of the original rational function.

    Solve \( R(x)=EBA(x) \).

    Each real number solution \( x=c \) in the domain of the rational function is the \( x \)-coordinate of a point of intersection.

    The \( y \)-coordinate of the point of intersection can be calculated by substituting the solution into either original equation. Using \( EBA(c) \) is the more convenient choice.

Example 1

Find the domain of \( f(x) = \frac{x + 3}{x^2 - 9} \)

Solution:

Begin by setting the denominator equal to zero and solving:

\( x^2 - 9 = 0 \)

\( x^2 = 9 \)

\( x = \pm 3 \)

The denominator is equal to zero when \( x = \pm 3 \). The domain of the function is all real numbers except \( x = \pm 3 \).

Analysis:

A graph of this function, as shown in Figure, confirms that the function is not defined when \( x = \pm 3 \).

There is a vertical asymptote at \( x = 3 \) and a hole in the graph at \( x = -3 \).

Example 2

Identifying Vertical Asymptotes

Example:

Find the vertical asymptotes of the graph of \( k(x) = \frac{5x + 2x^2}{2 - x - x^2} \).




Solution

Graph of \( k(x) = \frac{(5 + 2x)^2}{2 - x - x^2} \) with its vertical asymptotes at \( x = -2 \) and \( x = 1 \) and its horizontal asymptote at \( y = -2 \).

First, factor the numerator and denominator.

\( k(x) = \frac{5 + 2x^2}{2 - x - x^2} \)

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

To find the vertical asymptotes, we determine where this function will be undefined by setting the denominator equal to zero:

\( (2 + x)(1 - x) = 0 \)

\( x = -2, x = 1 \)

Neither \( x = -2 \) nor \( x = 1 \) are zeros of the numerator, so the two values indicate two vertical asymptotes. The graph confirms the location of the two vertical asymptotes.

Example 3

Find the domain of the function \( f(x) = \frac{x^2 + 1}{x} \).

Solution

Asymptotes: The vertical asymptote has equation \( x = 0 \), i.e. the y-axis. Dividing through gives:

\[ f(x) = \frac{x^2 + 1}{x} = \frac{x^2}{x} + \frac{1}{x} = x + \frac{1}{x} \]

Note that \( \frac{1}{x} \rightarrow 0 \) as \( x \rightarrow \pm \infty \). Therefore, the graph of \( y = f(x) \) approaches the graph of \( y = x \) as \( x \rightarrow \pm \infty \). The non-vertical asymptote has equation \( y = x \).

Addition of ordinates: The graph of \( y = f(x) \) can be obtained by adding the y-coordinates of the graphs of \( y = x \) and \( y = \frac{1}{x} \).




Intercepts:

There is no y-axis intercept, as the domain of \( f \) is \( \mathbb{R} \setminus \{0\} \). There are no x-axis intercepts, as the equation \( x^2 + \frac{1}{x} = 0 \) has no solutions.

Stationary points:

Given \( f(x) = x + \frac{1}{x} \), we find \( f'(x) = 1 - \frac{1}{x^2} \). Thus, \( f'(x) = 0 \) implies \( x^2 = 1 \), i.e., \( x = \pm 1 \). As \( f(1) = 2 \) and \( f(-1) = -2 \), the stationary points are (1, 2) and (-1, -2).

Points of inflection:

Given \( f''(x) = \frac{2}{x^3} \).

Therefore, \( f''(x) = 0 \) for all \( x \) in the domain of \( f \), and so there are no points of inflection.

Example 4

Sketch the graph of \( y = \frac{{4x^2 + 2}}{{x^2 + 1}} \).

Solution

Axis intercepts

When \( x = 0 \), \( y = 2 \). Since \( 4x^2 + \frac{2}{x^2} + 1 > 0 \) for all \( x \), there are no \( x \)-axis intercepts.

Stationary points

Using the quotient rule:

\( \frac{dy}{dx} = \frac{4x}{(x^2 + 1)^2} \)

\( \frac{d^2y}{dx^2} = \frac{4(1 - 3x^2)}{(x^2 + 1)^3} \)

Thus \( \frac{dy}{dx} = 0 \) implies \( x = 0 \). When \( x = 0 \), \( \frac{d^2y}{dx^2} = 4 > 0 \). Hence there is a local minimum at (0, 2).

Points of inflection

\( \frac{d^2y}{dx^2} = 0 \) implies \( x = \pm \frac{\sqrt{3}}{3} \)

Asymptotes

\( y = 4x^2 + \frac{2}{x^2} + 1 = 4 - \frac{2}{x^2 + 1} \)

The line \( y = 4 \) is a horizontal asymptote, since \( \frac{2}{x^2} + 1 \rightarrow 0 \) as \( x \rightarrow \pm\infty \).



Example 5

For the function \( g(x)=\frac{6x^3-10x}{2x^3+5x^2} \),

    Find Horizontal or Slant Asymptote:

    find any points of intersection between the rational function and its End Behavior Asymptote, solve the system of equations formed by equating the rational function to its slant asymptote.


For the function \( g(x)=\frac{6x^3-10x}{2x^3+5x^2} \),

  • Horizontal Asymptote: Both the numerator and denominator have the same degree (3). Therefore, we can find the horizontal asymptote by taking the ratio of the leading terms. There is a horizontal asymptote at \( y = \frac{6}{2} = 3 \).

  • Points of Intersection: To find if \( g(x) \) intersects with this horizontal asymptote, \( g(x) \) must first of all be reduced. \( R(x)=\frac{6x^2-10}{2x^2+5x} \). Then the equation \( R(x)=3 \) must be solved: \( 6x^2-10 = 3(2x^2+5x) \) → \( 6x^2-10 = 6x^2+15x \) → \( -10 = 15x \) → \( x = -\frac{2}{3} \). Finally, the \( y \)-coordinate of the intersection point is \( EBA\left(-\frac{2}{3}\right) = 3 \), so the point of intersection is \( \left(-\frac{2}{3}, 3\right) \).

    • Example 6

    Let \( y = \frac{x+1}{\sqrt{x - 1}} \).

    1. Find the maximal domain.
    2. Find the coordinates and the nature of any stationary points of the graph.
    3. Find the equation of the vertical asymptote and the behavior of the graph as \( x \to \infinity \).

    Solution:

  • For \(\left( \frac{x + 1}{\sqrt{x - 1}} \right)\) to be defined, we require \(\sqrt{x - 1} > 0\), i.e. \(x > 1\). The maximal domain is \((1, \infty)\).

  • Using the quotient and chain rules: \(\frac{dy}{dx} = \frac{x - 3}{2(x - 1)^{\frac{3}{2}}}\) and \(\frac{d^2y}{dx^2} = \frac{7 - x}{4(x - 1)^{\frac{5}{2}}}\).

    • Thus \(\frac{dy}{dx} = 0\) implies \(x = 3\). When \(x = 3\), \(\frac{d^2y}{dx^2} > 0\). There is a local minimum at \((3, 2\sqrt{2})\).

  • As \(x \to 1\), \(y \to \infinity\). Hence \(x = 1\) is a vertical asymptote. As \(x \to \infinity\), \(y \to \sqrt{x}\).


    • Exercise &&1&& (&&1&& Question)

    Find the domain of \( f(x) = \frac{4x}{5(x-1)(x-5)} \).

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    Exercise &&2&& (&&1&& Question)

    find the vertical asymptotes of the function \( k(x) = \frac{x - 2}{x^2 - 4} \),

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    Exercise &&3&& (&&1&& Question)

    Which type of rational function is this? \( f(x)= \frac{3x+2}{x+1} \).

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    Exercise &&4&& (&&1&& Question)

    As the independent variable approaches positive or negative infinity, the function's graph approaches horizontal lines known as:

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    Exercise &&5&& (&&1&& Question)

    Find the horizontal asymptote of the function \(f(x) = \frac{2x}{x - 3} \).

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    Exercise &&6&& (&&1&& Question)

    Find the point of intersections of \( k(x)=\frac{x^2+4x}{x^3-8} \).

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