5.1 We will begin this section with the definition of a rational expression. We

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1 Basic Properties and Reducing to Lowest Terms 5.1 We will begin tis section wit te definition of a rational epression. We will ten state te two basic properties associated wit rational epressions and go on to apply one of te properties to reduce rational epressions to lowest terms. Recall from Capter R tat a rational number is any number tat can be epressed as te ratio of two integers: Rational numbers = a b a and b are integers, b 0 A rational epression is defined similarly as any epression tat can be written as te ratio of two polynomials: Rational epressions = P Q P and Q are polynominals, Q? 0 Some eamples of rational epressions are a b 1 b a Basic Properties For rational epressions, multiplying te numerator and denominator by te same nonzero epression may cange te form of te rational epression, but it will always produce an epression equivalent to te original one. Te same is true wen dividing te numerator and denominator by te same nonzero quantity. PROPERTY Properties of Rational Epressions If P, Q, and K are polynomials wit Q? 0 and K? 0, ten P = PK Q QK and P Q = P K/Q/K Reducing to Lowest Terms Te fraction 6_ 8 can be written in lowest terms as 3_ 4. Te process is sown ere: 6 = = 3 4 Reducing 6_ 3_ 8 to 4 involves dividing te numerator and denominator by 2, te factor tey ave in common. Before dividing out te common factor 2, we must notice tat te common factor is 2! (Tis may not be obvious because we are very familiar wit te numbers 6 and 8 and terefore do not ave to put muc tougt into finding wat number divides bot of tem.) We reduce rational epressions to lowest terms by first factoring te numerator and denominator and ten dividing bot numerator and denominator by any factors tey ave in common.

2 2 CHAPTER 5 Rational Epressions and Rational Functions Reduce 2 9 Eample 1 to lowest terms. 3 Factoring, we ave = ( + 3)( 3) 3 Te numerator and denominator ave te factor 3 in common. Dividing te numerator and denominator by 3, we ave ( + 3)( 3) = + 3 = For te problem in Eample 1, tere is an implied restriction on te variable : It cannot be 3. If were 3, te epression ( 2 9) 0_ would become ( 3) 0, an epression tat we cannot associate wit a real number. For all problems involving rational epressions, we restrict te variable to only tose values tat result in a nonzero denominator. Wen we state te relationsip = + 3 we are assuming tat it is true for all values of ecept = 3. Here are some oter eamples of reducing rational epressions to lowest terms. Eample 2 Eample 3 Eample 4 Reduce y 2 5y 6 y 2 to lowest terms. 1 y 2 5y 6 y 2 = (y 6)(y + 1) 1 (y 1)(y + 1) = y 6 y 1 2a Reduce a 2 to lowest terms. 12a + 8 2a a 2 12a + 8 = 2(a 3 8) 4(a 2 3a + 2) = 2(a 2)(a2 + 2a + 4) 4(a 2)(a 1) = a2 + 2a + 4 2(a 1) Reduce a 3a 2 to lowest terms. a 3 + 3a a 3a 2 = ( 3) + a( 3) a 3 + 3a () 3() = ( 3)( + a) ()( 3) = + a Te answer to Eample 4 cannot be reduced furter. It is a fairly common mistake to attempt to divide out an or an a in tis last epression. Remember, we can di-

3 5.1 Basic Properties and Reducing to Lower Terms 3 vide out only te factors common to te numerator and denominator of a rational epression. Te net eample involves wat we call a trick. Te trick is to reverse te order of te terms in a difference by factoring 21 from eac term in eiter te numerator or te denominator. Te net eamples illustrate ow tis is done. Eample 5 Reduce to lowest terms: a b b a Te relationsip between a b and b a is tat tey are opposites. We can sow tis fact by factoring 21 from eac term in te numerator: a b b a = 1( a + b) b a = 1(b a) b a Factor 1 from eac term in te numerator. Reverse te order of te terms in te numerator. = 1 Divide out common factor b a Eample 6 Reduce to lowest terms: We begin by factoring te numerator: 2 25 = ( 5)( + 5) 5 5 Te factors 5 and 5 are similar but are not eactly te same. We can reverse te order of eiter by factoring 21 from it. Tat is: 5 = 1( 5 + ) = 1( 5). ( 5)( + 5) = ( 5)( + 5) 5 1( 5) = = ( + 5) Rational Functions Te function sown in Eample 8 is called a rational function because te rigt side, 785 t, is a rational epression (te numerator, 785, is a polynomial of degree 0). We can etend our knowledge of rational epressions to functions wit te following definition: DEFINITION rational function A rational function is any function tat can be written in te form f() = P() Q() were P() and Q() are polynomials and Q() 0.

4 4 CHAPTER 5 Rational Epressions and Rational Functions Eample 7 f(22), and f(2). For te rational function f() = 4, find f(0), f(24), f(4), 2 To find tese function values, we substitute te given value of into te rational epression, and ten simplify if possible. f(0) = = 4 2 = 2 f( 2) = 2 4 = = 3 2 f( 4) = 4 4 = = 4 3 f(2) = 2 4 = Undefined f(4) = = 0 2 = 0 Because te rational function in Eample 7 is not defined wen is 2, te domain of tat function does not include 2. We ave more to say about te domain of a rational function net. Te Domain of a Rational Function If te domain of a rational function is not specified, it is assumed to be all real numbers for wic te function is defined. Tat is, te domain of te rational function f() = P() Q() is all for wic Q() is nonzero. Eample 8 Find te domain for eac function. a. f() = 4 2 b. g() = c. () = 2 9 a. Te domain for f() = 4 is { 2}. 2 b. Te domain for g() = is { 1}. + 1 c. Te domain for () = 2 is { 3, 3}. 9 Notice tat, for tese functions, f(2), g(21), (23), and (3) are all undefined, and tat is wy te domains are written as sown.

5 5.1 Basic Properties and Reducing to Lower Terms 5 Difference Quotients Te diagram in Figure 1 is an important diagram from calculus. Altoug it may look complicated, te point of it is simple: Te slope of te line passing troug te points P and Q is given by te formula Slope of line troug PQ = m = f() f(a) f(a) y P y = f() Run = a f() Q Rise m = Run Rise = f() f(a) a a FIGURE 1 f() f(a) Te epression is called a difference quotient. Wen f() is a polynomial, it will be a rational epression. Eample 9 If f () , find f() f(a). f () f (a) = (3 5) (3a 5) = 3 3a = 3() = 3 If f() = 2 f() f(a) Eample 10 4, find and simplify. Because f() and f(a) 5 a 2 2 4, we ave f() f(a) = (2 4) (a 2 4) = 2 4 a = 2 a 2 ( + a)() = = + a Factor and divide out common factor.

6 6 CHAPTER 5 Rational Epressions and Rational Functions Te diagram in Figure 2 is similar to te one in Figure 1. Te main difference is in ow we label te points. From Figure 2, we can see anoter difference quotient tat gives us te slope of te line troug te points P and Q. Slope of line troug PQ = m = f( + ) f() f() y P y = f() Run f( + ) Q Rise m = Run Rise = f( + ) f() + Eamples 7 and 8 use te same functions used in Eamples 5 and 6, but tis time te new difference quotient is used. Eample 11 If f() = 3 5, find f( + ) f()/. Te epression f ( 1 ) is given by f( + ) = 3 ( + ) 5 Using tis result gives us = f( + ) f() = ( ) (3 5) = 3 = 3 If f() = 2 f( + ) f() Eample 12 4, find. Te epression f( 1 ) is given by f ( + ) = ( + ) 2 4 Using tis result gives us = f( + ) f() = ( ) ( 2 4) = = (2 + ) = 2 + FIGURE 2

7 5.1 Basic Properties and Reducing to Lower Terms 7 Getting Ready for Class After reading troug te preceding section, respond in your own words and in complete sentences. A. Wat is a rational epression? B. Eplain ow to determine if a rational epression is in lowest terms. C. Wen is a rational epression undefined? D. Eplain te process we use to reduce a rational epression or a fraction to lowest terms.

8 Problem Set If g() = + 3, find g(0), g(23), g(3), g(21), and g(1), if possible. 1 1 t + 1 t If g() = 2, find g(0), g(22), g(2), g(21), and g(1), if possible. 3. If (t) = t 3, find (0), (23), (3), (21), and (1), if possible. 4. If (t) = t 2, find (0), (22), (2), (21), and (1), if possible. State te domain for eac rational function. 5. f() = 3 6. f() = g() = g() = (t) = t 4 3 t (t) = t 5 16 t 2 25 Reduce eac rational epression to lowest terms y y 13. a4 81 a a 12 a2 a 2 + 8a y y 2 5y y 2y 22 y 6y 4y 2 2 y + 100a + 10a a 50a 2a ( 3)2 ( + 2) ( + 2) 2 ( 3) 20. ( 4)3 ( + 3) ( + 3) 2 ( 4) am 4an 23. 3n 3m ad2 24. ad d ab a + b 1 ab + a + b cd 4c 9d + 6 6d 2 13d a3 + b 3 a 2 b b 2 a2 a 3 b a a ay + 2y 4a a 2 + 6a 2 2 3a + 2 6a y 2 9 y y y 3 + 8

9 Problem Set Refer to Eamples 5 and 6 in tis section, and reduce te following to lowest terms y y y y a 2 9a 2 6a a 2 a 2 2a + 1 Simplify eac epression. (3 5) (3a 5) (2 + 3) (2a + 3) 53. (2 4) (a 2 4) 54. (2 1) (a 2 1) For te functions below, evaluate a. f() f(a) b. f( + ) f() 55. f() = f() = f() = f() = f() = f() = f() = f() = f() = f() = f() = f() = Applying te Concepts 67. Diet Te following rational function is te one we mentioned in te introduction to tis capter. Te quantity W() is te weigt (in pounds) of te person after weeks of dieting. Use te function to fill in te table. Ten compare your results wit te grap in te capter introduction. W() = 80(2 + 15) + 6 Weeks Weigt (lb) W()

10 10 CHAPTER 5 Rational Epressions and Rational Functions 68. Drag Racing Te following rational function gives te speed V(), in miles per our, of a dragster at eac second during a quarter-mile race. Use te function to fill in te table. V() = Time (sec) Speed (mi/r) V() Getting Ready for te Net Section Multiply or divide, as indicated y y y y 5 Factor y 78. a 2 5a y a + b + ya + yb Etending te Concepts 81. Te graps of two rational functions are given in Figures 5 and 6. Use te graps to find te following. a. f(2) b. f( 1) c. f(0) d. g(3) e. g(6) f. g( 1) g. f(g(6)). g(f( 2)) y = f() = 4 y y = g() = 6 y (1, 4) (2, 2) (4, 1) ( 4, 1) 2 ( 2, 2) 3 ( 1, 4) (1, 6) (2, 3) (3, 2) (6, 1) ( 6, 1) 2 ( 3, 2) 3 ( 2, 3) 4 5 ( 1, 6) 6 FIGURE 5 FIGURE 6