Lesson 5: Rational Functions PDF
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National University
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This document appears to be a set of lesson notes on rational functions and equations. It includes definitions, examples, word problems, and practice exercises. The problems cover topics like constructing functions, finding domain, range, asymptotes, and solving rational equations.
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^ - x ^ - x RATIONAL KA BA? ^ Instructions Instructions - x Five review questions about todayβs lesson will be flashed on the screen. Two choices will be given per question: UTAK...
^ - x ^ - x RATIONAL KA BA? ^ Instructions Instructions - x Five review questions about todayβs lesson will be flashed on the screen. Two choices will be given per question: UTAK vs. PUSO. Studentβs task is to choose between the two and explain briefly their chosen answer. Students who are first to raise their hand and get the correct answer will earn 1 slay chip each. ^ Instructions Number One: - x ππ Is this a function or an equation: π π = ? πβπ FUNCTION EQUATION ^ Instructions Number Two: - x What is the LCM of 5 and 15? 5 15 ^ Instructions Number Three: - x πβπ What value of x will make the expression undefined? π+π -2 2 ^ Instructions Number Four: - x Which are the correct factors of ππ β ππ β π? (π β π)(π + π) (π β π)(π + π) ^ Instructions Number Five: - x ππβπ What is the ratio of the leading coefficients in ? ππβπ π π π π ^ - x Lesson 5: Rational Functions Rational Functions and its Asymptotes; and Rational Equations ^ Lesson Objectives - x At the end of this lesson, you will be able to: Distinguish rational functions and rational equations; Determine the domain and asymptote/s of rational functions; Solve rational equations with or without extraneous solutions using algebraic techniques; and Represent real-life situations using rational functions and equations. ^ - x 01 Rational Function Definition, Real-Life Representation, and Domain ^ Definition of Terms - x Rational Function β A rational function π π₯ , is a function of the form π(π) π π = ; where π π₯ and π(π₯) are polynomial π(π) functions and π π β π. β The denominator of a rational function cannot be zero. Any value of the variable that would make the denominator zero is not permissible. ^ Examples ofofRational Definition Functions Terms/Concepts - x π+π π πβπ π π = π π = π π = π πβπ π π βπ π β βπ πβ π πβ π πβ π ^ Word Problem #1.1 - x During the first semester of the school year the officersβelect of the Student Government decided to divide their budget evenly to the different committees. If their budget is β±40,000, construct a function f which would Since the budget is give the amount of money each of the x divided evenly to number of committees would receive. the number of committees, the Number of Committees (x) π 4 6 8 function would be: πππππ Amount Allocated π π = π β±ππ, πππ β±ππ, πππ β±π, πππ. ππ β±π, πππ for Each Committee where π₯ β 0. ^ Word Problem #1.2 - x Suppose a sponsor wants to supplement the budget allotted for each committee by donating additional β±750.00 per committee. If h(x) represents the new amount allotted per committee, construct a function representing the relationship. Number of Committees (x) π 4 6 8 Amount Allocated for β±ππ, πππ β±ππ, πππ β±π, πππ. ππ β±π, πππ Each Committee + β±πππ + β±πππ + β±πππ + β±πππ πππππ The representation of the function is π π = + πππ. π ^ Word Problem #2 - x A car is to travel a distance of 65 kilometers. Express the velocity (v) as a function of travel time (t) in hours. [Note: πππ π‘ππππ = π£ππππππ‘π¦ β π‘πππ] Time (hours) 1 2 3 5 10 Velocity ΰ΄₯ ππ ππ. π ππ. π ππ π. π (km/hr) ππ Thus, the function π π = can represent v as a function of t. π ^ Definition Domain of Terms/Concepts and Range of Rational Functions - x Domain Range The domain of a The range of a rational π π rational function is function is the set of all π(π) the set of all real outputs (y-values) that it numbers so that π π β produces. π. ^ Domain Definition of Terms/Concepts - x π+π π πβπ π π = π π = π π = π πβπ π π βπ π β βπ πβ π πβ π πβ π Domain = π₯ β β | π₯ β 1 Domain = π₯ β β | π₯ β 0 Domain = π₯ β β | π₯ β β2, 2 The domain is the set of all The domain is the set of all The domain is the set of all real real numbers except 1. real numbers except 0. numbers except -2 and 2. ^ Definition ofRange Terms/Concepts - x To get the range of rational functions: π+π π π = 1. Replace π π₯ with π¦. πβπ 2. Solve the equation for π₯. 3. Set the denominator of the π+π π= resulting equation β 0 and solve for πβπ π¦. π πβπ =π+π 4. Set of all real numbers other than ππ β π = π + π the values of π¦ mentioned in the ππ β π = π + π last step is the range. π(π β π) = π + π Range = π¦ β β | π¦ β 1 π+π The range of the function is the set of all real π= ; πβ π πβπ numbers except 1. ^ - x 02 Asymptotes of Rational Functions Vertical and Horizontal Asymptotes, and Holes ^ Definition of Terms - x Asymptote An asymptote is a line (or curve) which acts as a boundary for the graph of a function. ^ Definition Asymptotes of Rational Terms/Concepts Functions - x Vertical Horizontal Asymptote Asymptote A simplified rational The horizontal asymptote is function has vertical determined by looking at asymptote(s) at the the degrees of the zeroes of the numerator and denominator of a rational denominator. function. ^ Vertical Asymptote - x π(π₯) Given the rational function π π₯ = , if π(π₯) and π(π₯) π(π₯) have no common factors, then π(π₯) has vertical asymptote(s) when π π₯ = 0. Vertical Asymptotes occur when the denominator of the simplified rational function is equal to 0. Note: simplified rational function has cancelled any factors common to both the numerator and denominator. ^ Vertical Definition of Asymptote Terms/Concepts - x Example #1: π+π π π = πβπ Check if the function is already simplified. Once simplified, set the denominator equal to zero. πβπ=π π=π Vertical Asymptote: π₯ = 1 ^ Vertical Definition of Asymptote Terms/Concepts - x Example #2: πβπ πβπ π π = π = π βπ (π β π)(π + π) πβπ π π π = π = π βπ π+π π+π=π π = βπ Vertical Asymptote: π₯ = β2 ^ Definition of Terms - x Hole(s) Holes occur in the graph of a rational function whenever the numerator and denominator have common factors. The holes occur at the x-value(s) that make the common factors equal to 0. ^ Definition ofHole Terms/Concepts - x From Example #2 of V.A.: πβπ πβπ π π = π = π βπ (π β π)(π + π) πβπ π π π = π = π βπ π+π πβπ=π π=π There is a hole at x = 2; (2, undefined) ^ - x Are there any connections between the domain, vertical asymptote(s), and hole(s) of rational functions? OBSERVATION CHECK! ^ Horizontal Asymptote - x π(π₯) Given the rational function π π₯ = , the horizontal π(π₯) asymptote can be determined by looking at the degrees of π(π₯) and π π₯. If the degree of p(x) is less than the degree of q(x), then the horizontal asymptote is π = π. If the degree of p(x) is equal to the degree of q(x), then the ππππ πππ πππππππππππ ππ π(π) horizontal asymptote is π =. ππππ πππ πππππππππππ ππ π(π) If the degree of p(x) is greater than the degree of q(x), then there is no horizontal asymptote. ^ Horizontal Definition Asymptote of Terms/Concepts - x π+π ππ β π πβπ π π = π π = π π = π πβπ π π βπ Horizontal Asymptote: No Horizontal Horizontal Asymptote: 1 π¦= =1 Asymptote π¦=0 1 ^ Vertical and Horizontal Asymptote with Hole - x πβπ π π = π π βπ Vertical Asymptote: π₯ = β2 Horizontal Asymptote: π¦=0 Hole at π₯ = 2 (2, π’ππππππππ) ^ TryTerms/Concepts Definition of this! - x π₯ 2 + 9π₯ + 20 (π₯ + 5)(π₯ + 4) π π₯ = 2 = π₯ β π₯ β 20 (π₯ β 5)(π₯ + 4) Vertical Asymptote: π = π Horizontal Asymptote: π = π Hole at π = β4 ^ - x 02 Rational Equations Definition, Solving for the Variable, and Real-Life Representation ^ Definition of Terms - x Rational Equation A rational equation is an equation that involves one or more rational expressions. A rational expression is a fraction in which the numerator and denominator are both polynomials. ^ Rational Equation - x The general form of the rational equation looks like this: π·(π) πΉ(π) = πΈ(π) πΊ(π) Where: P(x) and Q(x) are polynomials in x (the variable). R(x) and S(x) are other polynomials in x. ^ Examples ofofRational Definition Equations Terms/Concepts - x π π π π + = π= βπ π π π π π π π = + πβπ πβπ π ^ How to Solve? - x Solving Rational Equations 1. Find the LCM of the denominators (LCD). 2. Clear denominators by multiplying both sides of the equation by the LCM. 3. Solve the resulting polynomial equation. 4. Check the solutions. ^ DefinitionExample #1 of Terms/Concepts - x Solve for the value of x: Solution: π π π π₯ + 1 = π₯ + = 5 4 2 π π π 4π₯ 5 10π₯ + = 20 20 20 4π₯+5 10π₯ LCD: 20 20 = 20 20 20 4π₯ + 5 = 10π₯ π 4π₯ β 10π₯ = β5 Therefore, the value of x is. π π π= π ^ DefinitionExample #2 of Terms/Concepts - x Solve for the value of x: Solution: 6 π π₯ = β1 π₯ π= βπ π₯2 6 π₯ π = β π₯ π₯ π₯ π₯2 6 π₯ π₯ = β π₯ LCD: π π₯ π₯ π₯ 2 π₯ =6βπ₯ π₯2 + π₯ β 6 = 0 Therefore, the value of x are π₯β2 π₯+3 =0 β π and π. π = π πππ π = βπ ^ Definition of Terms - x Extraneous Solution An extraneous solution to a rational equation is an algebraic solution that would cause any of the expressions in the original equation to be undefined. ^ DefinitionExample #3 of Terms/Concepts - x Solve for the value of x: Solution: π π π = + πβπ πβπ π π π π 6 6π₯ π₯(π₯β1) = + = + πβπ πβπ π (6)(π₯β1) (6)(π₯β1) (6)(π₯β1) 6 6π₯ π₯ π₯β1 [ 6 π₯β1 ] = + [ 6 π₯β1 ] 6 π₯β1 6 π₯β1 6 π₯β1 6 = 6π₯ + π₯(π₯ β 1) LCD: π (π β π) 6 = 6π₯ + π₯ 2 β π₯ π₯ 2 + 5π₯ β 6 = 0 Therefore, the only solution for the (π₯ β 1)(π₯ + 6) = 0 equation is βπ. π = π πππ π = βπ EXTRANEOU S ^ Word Problem #1 - x Jason's car uses 20 gallons of gas to travel 400 miles. If Jason currently has 7 gallons of gas in his car, how much gas is needed to travel 200 miles? Let x = number of additional gasoline needed to travel 200miles Represent the equation thru Ratio and Proportionality: 20 πππππππ π₯+7 = 400 πππππ 200 πππππ ^ Word Problem #1 - x Solution: 20 πππππππ π₯+7 = 400 πππππ 200 πππππ 20 π₯+7 = LCD: πππ 400 200 20 2(π₯+7) = 400 400 20 2(π₯+7) 400 400 = 400 Since x = 3, and x+7 is needed to 400 20 = 2(π₯ + 7) travel the 200 miles, therefore, Jason 20 = 2π₯ + 14 needs (3)+7 = 10 gallons of gasoline to 2π₯ = 20 β 14 travel 200 miles. 2π₯ = 6 π=π ^ Word Problem #2 - x A water tank can be filled by two pipes, A and B. Pipe A can fill the tank in 3 hours, while pipe B can fill it in 4 hours. If both pipes are opened simultaneously, how long will it take to fill the tank? Let t represent the time it takes to fill the tank when both pipes are open. 1 The rate of pipe A is tank per hour 3 1 The rate of pipe B is β tank per hour. 4 ^ Word Problem #2 - x Solution: Let this be represented by equation: 1 1 1 =3+4 LCD: πππ π‘ 1 π‘πππ 1 π‘πππ 1 π‘πππ = + 12 4π‘ 3π‘ π‘ 3βπ 4βπ = + 12π‘ 12π‘ 12π‘ 1 tank = Pipe A + Pipe B 12 4π‘ 3π‘ 12π‘ = + 12π‘ 12π‘ 12π‘ 12π‘ π Therefore, it will take π hours to 12 = 4π‘ + 3π‘ π fill the tank if both pipes are open 12 = 7π‘ simultaneously. ππ π π= π ππ π π πππππ ^ - x What are the distinct features of rational expression, rational function, and rational equation? OBSERVATION CHECK! ^ - x PRACTICE EXERCISE ^ Rational Functions - x A. Construct the function for the word problem below: Joy has 10 cups of flour to be used in baking cakes, she wanted to split it evenly among the containers that she will use so that she can adjust the measurements of other ingredients. Construct a function f which would give the number of cups of flour each of the number of containers x will have. B. Find the domain, range, vertical asymptote, and horizontal asymptote of the function: 2π₯ + 5 π π₯ = π₯β1 ^ Rational Equations - x C. Solve for what is being asked: 5π₯ 3π₯+4 1. Solve for x: = π₯β2 π₯β2 2. Working alone, it takes Tony 11 hours to complete a restoration project on a truck. Steve can perform the same task in 110 hours. How long would it take if they worked together? ^ - x Discovery Problem Letβs Discover! ^ Inverse Function - x Find the inverse of the function: π π = ππ + π Note: To find the inverse of a function algebraically, interchange x and y and solve for y. ^ - x Got any questions?