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Functions - Properties of Rational Functions

Grade 12IB

Review the key concepts, formulae, and examples before starting your quiz.

๐Ÿ”‘Concepts

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Definition and Domain: A rational function is defined as f(x)=P(x)Q(x)f(x) = \frac{P(x)}{Q(x)} where P(x)P(x) and Q(x)Q(x) are polynomials. The domain consists of all real numbers except those where Q(x)=0Q(x) = 0. Visually, these exclusions create breaks or gaps in the graph, representing values where the function is undefined.

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Vertical Asymptotes (VA): These occur at values of xx where the denominator Q(x)=0Q(x) = 0 but the numerator P(x)โ‰ 0P(x) \neq 0 (after simplification). On a graph, the curve will get closer and closer to these vertical dashed lines, moving toward +โˆž+\infty or โˆ’โˆž-\infty, but will never touch or cross them.

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Horizontal Asymptotes (HA): These indicate the end behavior of the function as xโ†’ยฑโˆžx \to \pm\infty. If the degree of the numerator nn is less than the degree of the denominator mm, the HA is y=0y = 0. If n=mn = m, the HA is the ratio of the leading coefficients. Visually, the graph approaches this horizontal line at the far left and right edges.

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Oblique (Slant) Asymptotes: When the degree of the numerator is exactly one higher than the degree of the denominator, the graph follows a linear path y=ax+by = ax + b as xx becomes very large or very small. This line is found using polynomial long division. The graph will approach this diagonal line instead of a horizontal one.

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Holes (Removable Discontinuities): A hole occurs at x=cx = c if (xโˆ’c)(x - c) is a common factor in both the numerator and denominator. Graphically, the curve appears continuous, but there is a single point missing, represented by an open circle โˆ˜\circ at that specific coordinate.

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Intercepts: The yy-intercept is found by evaluating f(0)f(0), showing where the graph crosses the vertical axis. The xx-intercepts (zeros) are the roots of the numerator P(x)=0P(x) = 0 (provided they are not also roots of the denominator), showing where the graph crosses the horizontal axis.

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Sign Diagrams and Behavior: To understand the shape of the graph, we use critical values (zeros and VAs) to test intervals. This tells us if the graph is above (f(x)>0f(x) > 0) or below (f(x)<0f(x) < 0) the xx-axis, helping to determine how the curve approaches its asymptotes.

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Transformations of 1x\frac{1}{x}: Rational functions in the form f(x)=axโˆ’h+kf(x) = \frac{a}{x-h} + k are shifts of the parent function y=1xy = \frac{1}{x}. The value hh represents a horizontal shift (and the VA), kk represents a vertical shift (and the HA), and aa represents a vertical stretch or reflection.

๐Ÿ“Formulae

General Rational Form: f(x)=anxn+anโˆ’1xnโˆ’1+โ‹ฏ+a0bmxm+bmโˆ’1xmโˆ’1+โ‹ฏ+b0f(x) = \frac{a_n x^n + a_{n-1} x^{n-1} + \dots + a_0}{b_m x^m + b_{m-1} x^{m-1} + \dots + b_0}

Horizontal Asymptote (if n<mn < m): y=0y = 0

Horizontal Asymptote (if n=mn = m): y=anbmy = \frac{a_n}{b_m}

Vertical Asymptote: Solve Q(x)=0Q(x) = 0 for simplified f(x)f(x)

Oblique Asymptote: y=mx+cy = mx + c (Quotient of P(x)Q(x)\frac{P(x)}{Q(x)})

Transformation Form: f(x)=kxโˆ’h+vf(x) = \frac{k}{x-h} + v

๐Ÿ’กExamples

Problem 1:

For the function f(x)=2x2โˆ’8x2โˆ’xโˆ’2f(x) = \frac{2x^2 - 8}{x^2 - x - 2}, find the coordinates of any holes and the equations of all asymptotes.

Solution:

  1. Factorize: f(x)=2(x2โˆ’4)(xโˆ’2)(x+1)=2(xโˆ’2)(x+2)(xโˆ’2)(x+1)f(x) = \frac{2(x^2 - 4)}{(x - 2)(x + 1)} = \frac{2(x - 2)(x + 2)}{(x - 2)(x + 1)}. \ 2. Hole: The factor (xโˆ’2)(x - 2) is common. A hole exists at x=2x = 2. Simplify the function to f(x)=2(x+2)x+1f(x) = \frac{2(x + 2)}{x + 1} and plug in x=2x = 2: y=2(2+2)2+1=83y = \frac{2(2 + 2)}{2 + 1} = \frac{8}{3}. Hole is at (2,83)(2, \frac{8}{3}). \ 3. Vertical Asymptote: Remaining denominator x+1=0โ‡’x=โˆ’1x + 1 = 0 \Rightarrow x = -1. \ 4. Horizontal Asymptote: Degrees are equal (n=2,m=2n=2, m=2). y=21=2y = \frac{2}{1} = 2.

Explanation:

We first factor the expression to see if any terms cancel. Canceling terms indicate 'holes' rather than asymptotes. The remaining zeros of the denominator define the vertical asymptotes, and the ratio of leading coefficients defines the horizontal asymptote because the degrees are equal.

Problem 2:

Find the oblique asymptote of the function g(x)=x2+3x+5xโˆ’1g(x) = \frac{x^2 + 3x + 5}{x - 1}.

Solution:

  1. Check Degrees: Degree of numerator (2) is one higher than denominator (1), so an oblique asymptote exists. \ 2. Polynomial Division: Divide x2+3x+5x^2 + 3x + 5 by xโˆ’1x - 1. \ x2รทx=xx^2 \div x = x. \ x(xโˆ’1)=x2โˆ’xx(x - 1) = x^2 - x. \ (x2+3x)โˆ’(x2โˆ’x)=4x(x^2 + 3x) - (x^2 - x) = 4x. \ 4xรทx=44x \div x = 4. \ 4(xโˆ’1)=4xโˆ’44(x - 1) = 4x - 4. \ 5โˆ’(โˆ’4)=95 - (-4) = 9. \ 3. Result: g(x)=x+4+9xโˆ’1g(x) = x + 4 + \frac{9}{x - 1}. \ 4. Equation: The oblique asymptote is y=x+4y = x + 4.

Explanation:

To find an oblique asymptote, perform long division (or synthetic division). As xx approaches infinity, the remainder term 9xโˆ’1\frac{9}{x-1} approaches zero, leaving the linear quotient as the path the function follows.