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IB DP Maths AI HL Study Notes

2.2.2 Rational Functions

Understanding Rational Functions

Defining Rational Functions

Rational functions are essentially fractions where both the numerator and the denominator are polynomials. The degree (the highest power of x) of these polynomials can vary, providing a wide array of functions with different properties and graphs.

Example 1: Basic Rational Function

Consider R(x) = (x2 - 1) / (x2 + x - 2). Both the numerator and the denominator are polynomials, and thus, R(x) is a rational function. The nuances of its graph and behaviour are dictated by the degrees and factors of the polynomials involved. Understanding function transformations can further help in analysing these graphs.

Asymptotes: Navigating the Infinite

Asymptotes serve as invisible boundaries that the graphs of rational functions may approach but never quite touch. These lines, vertical and horizontal, guide the behaviour of the function as it navigates through the x and y-axes.

Vertical Asymptotes

Vertical asymptotes are vertical lines x = k that the graph approaches but never crosses. They occur where the denominator of the rational function equals zero, assuming this factor is not cancelled out by the numerator.

  • Example 2: For R(x) = 1 / (x - 1), the vertical asymptote is x = 1 because the function becomes undefined at this point.

Horizontal Asymptotes

Horizontal asymptotes guide the behaviour of the graph as x approaches infinity. The position of the horizontal asymptote is determined by comparing the degrees of the numerator and denominator.

  • Example 3: For R(x) = (3x2 + 2) / (x2 - 5), the horizontal asymptote is y = 3 as the degrees of the numerator and denominator are equal, and 3 is the ratio of the leading coefficients.

Holes: The Unseen Points

Holes, or removable discontinuities, are points where the graph is undefined due to a factor that cancels out in the rational function. These points are invisible on the graph but are crucial to understanding the function’s behaviour.

  • Example 4: In R(x) = (x - 2) / (x2 - 4), the factor x - 2 cancels out, leaving R(x) = 1 / (x + 2). However, x = 2 is a hole as it is undefined in the original function.

Understanding inverse functions is also essential when dealing with rational functions and their discontinuities.

IB Maths Tutor Tip: Mastering rational functions involves recognising patterns in graphs and equations. Focus on identifying asymptotes and holes to predict function behaviour and solve real-world problems effectively.

Behaviour of Rational Functions

Sign Analysis

Understanding where a rational function is positive or negative involves analysing the signs of the numerator and denominator in various intervals determined by the zeros and vertical asymptotes.

Increasing/Decreasing Intervals

Identifying where the function is increasing or decreasing can be achieved by analysing the first derivative, which is a fundamental concept in basic differentiation rules, pivotal in calculus.

Example 5: Analysing Behaviour

For R(x) = (x2 - 4) / (x2 - x - 6), as x approaches infinity, R(x) approaches 1, indicating a horizontal asymptote at y = 1. The function is undefined at x = -2 and x = 3, suggesting possible vertical asymptotes or holes, which need further investigation. A deeper analysis of rational functions can provide more insights into their complex behaviour.

Detailed Analysis of a Rational Function

Example 6: Real-world Application

Imagine a scenario where the rational function represents the rate of growth of a population, bacteria, or investment. Understanding the asymptotes, holes, and general behaviour of the function can provide insights into the limitations, potential issues, or future trends of the scenario being modelled.

IB Tutor Advice: During exams, sketch rough graphs of rational functions to visualise asymptotes and holes. This can greatly aid in understanding the function's behaviour and answering questions accurately.

Conclusion

Understanding rational functions, their asymptotes, holes, and general behaviour is crucial in various mathematical and real-world applications. These concepts lay the foundation for further studies in algebra, calculus, and mathematical modelling, enabling students to solve complex problems and create mathematical models in diverse scenarios. This knowledge is not only pivotal for your IB Mathematics journey but also applicable in various fields such as finance, engineering, and physics, where rational functions are used to model and solve real-world problems.

FAQ

Vertical asymptotes in rational functions can represent boundaries or limitations in real-world scenarios. For instance, in a business model, if a cost function has a vertical asymptote at x = 100, it might imply that the model is not valid or practical when producing more than 100 items. It could indicate infinite costs or other impractical scenarios when crossing that boundary, thus highlighting limitations or critical points in practical applications.

Graphically, holes in rational functions are points that are excluded from the domain and are typically represented as an open circle on the graph. In practical terms, a hole might signify an undefined or impractical scenario within a particular model. For example, in a financial model, a hole might indicate a point where an investment strategy is not defined due to external factors, such as regulatory restrictions or market limitations, providing a visual cue of an exception in the model.

No, a rational function cannot have more than one horizontal asymptote. The behaviour of the function as x approaches positive or negative infinity is determined by the degrees of the polynomials in the numerator and denominator. Depending on these degrees, the function may have one horizontal asymptote or none, but never more than one. The position and existence of the horizontal asymptote are guided by the rules discussed in the study notes.

The degrees of the numerator and denominator play a pivotal role in determining the end behaviour and horizontal asymptotes of a rational function. If the degree of the numerator is less than the denominator, the x-axis (y = 0) is the horizontal asymptote. If both degrees are equal, the horizontal asymptote is found by dividing the leading coefficients. If the numerator’s degree is greater, there is no horizontal asymptote, and the function may approach infinity or negative infinity as x approaches infinity.

The domain of a rational function includes all real numbers except those that make the denominator zero (excluding holes). Vertical asymptotes and holes define exclusions in the domain. The range of a rational function, particularly concerning asymptotes and holes, is influenced by the behaviour of the graph as it approaches these features. For instance, if there is a horizontal asymptote, the function may approach a particular y-value but never reach it, influencing the range. Understanding the domain and range is crucial for interpreting the applicability and limitations of rational functions in various contexts.

Practice Questions

Determine the vertical and horizontal asymptotes of the rational function R(x) = (2x^2 + x - 3) / (x^2 - 5x + 6).

To find the vertical asymptotes, we set the denominator equal to zero and solve for x. Factoring the denominator, we get (x - 2)(x - 3) = 0, which gives us x = 2 and x = 3 as the vertical asymptotes. For the horizontal asymptote, we compare the degrees of the numerator and denominator. Since they are equal, we take the ratio of the leading coefficients, which is 2/1 or 2. Therefore, the horizontal asymptote is y = 2.

Analyse the behaviour of the rational function R(x) = (x^2 - 4) / (x^2 - x - 6) as x approaches infinity and negative infinity. Also, identify any holes in the graph.

As x approaches infinity, the highest degree term dominates the behaviour of the function. Thus, R(x) approaches 1, indicating a horizontal asymptote at y = 1. Similarly, as x approaches negative infinity, R(x) also approaches 1. To identify holes, we factorise the numerator and denominator. The numerator becomes (x + 2)(x - 2) and the denominator becomes (x + 2)(x - 3). The factor x + 2 cancels out, leaving R(x) = 1 / (x - 3). However, x = -2 is a hole because it is undefined in the original function.

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