The graphs of Rational Functions

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1 Lecture 4 5A: The its of Rtionl Functions s x nd s x + The grphs of Rtionl Functions The grphs of rtionl functions hve severl differences compred to power functions. One of the differences is the behvior of left nd right ends of the grph. The left nd right ends of the grph of rtionl function my hve y vlues tht increse to positive infinity or decrese to negtive infinity like power functions but they my lso pproch horizontl symptotes. Rtionl functions re represented by mny more types of grphs thn power function we lerned to grph in the lst section. This mens tht the mteril in this section will be broken down into more specific cses nd more theorems will be needed to nlyze the behvior of the grphs. We will strt by looking t the behvior of the end points, then consider wht effect the ddition of verticl symptotes hs on the grph nd then dd roots to the grph. With ll this informtion we will then be ble to grph rtionl functions. This section will focus on rtionl functions with polynomil function in the numertor nd the denomintor. The terms of the polynomils with hve powers of x tht re positive integers. The terms will be written in decresing powers of x. f(x) = n xn ± n 1 x n 1 ± n x n ± b m x m ± b m 1 x m 1 ± b m x m ±...+ b 0 where n z + nd n R f (x) = 3x 6x x 1 Exmples f (x) = 5x3 x 4 x 3x Note: This section will only del with rtionl functions tht hve no common fctors. A lter section will consider wht issues my exist if the rtionl function hs common fctors. 4-5A Lecture! Pge 1 of 1! 018 Eitel

2 Nottion: The its of rtionl functions s x nd s x + Left End! Right End If we wnt to know bout the behvior! If we wnt to know bout the behvior of the left end of the grph we sk:! of the right end of the grph we sk: Wht re the y vlues pproching! Wht re the y vlues pproching s the x vlues pproch negtive infinity?! s the x vlues pproch positive infinity? The nottion used to sk this question is! g(x) h(x)! The nottion used to sk this question is x g(x) h(x) Wht prt of the polynomils in rtionl expression determine the behvior of the left nd right ends of the grph? We hve seen tht the leding term of power function ws the only term of the polynomil tht determined the behvior of the right nd left ends of the grph. This is lso true for rtionl functions but we must use the leding terms of the polynomil in both the numertor nd denomintor to determine the behvior of the right nd left ends of the grph of rtionl function. Given the rtionl function f(x) where f(x) = g(x) h(x) = n x n ± n 1 x n 1 ± n x n b m x m ± b m 1 x m 1 ± b m x m b 0 for n,m Z + nd n,b n R the rtio of the leding terms of the numertor nd denomintor is n xn b m x m This rtio lone cn be used to determine the behvior of the right nd left ends of the grph of rtionl function. The first step in determining the behvior of the function is to write the rtio of the leding terms nd then use the quotient rule to reduce the rtio to simplest terms. Exmple 1! Exmple 9x 3 5x f (x) = 6x 5 x 3 5x the rtio of the leding terms of the numertor nd demomintor is 9x 3 6x 5 which 3 x! f (x) = 1x5 5x 3 + 8x 6x 3 3x 7x the rtio of the leding terms of the numertor nd demomintor is 1x 4 6x 3 which x 4-5A Lecture! Pge of 1! 018 Eitel

3 There re severl different behviors tht occur depending on the rtio of the leding terms of the numertor nd denomintor Limits of Rtionl Functions Given the rtionl function f (x) where f (x) = g(x) h(x) = n xn ± n 1 x n 1 ± n x n b m x m ± b m 1 x m 1 ± b m x m +...b 0 for n,m Z + nd n,b n R Cse 1 If f (x) is rtionl function nd the exponent of x in the leding term of the denomintor is greter thn the exponent of x in the leding term of the numertor then the the rtio of the leding terms of the numertor nd denomintor will reduce to If n Z + bx n nd,b R, b 0 then bx n = 0 nd x bx n = 0 In English: If f (x) is rtionl function nd the exponent of the highest power in the denomintor is greter thn the highest power in the numertor both ends of the grph pproch the horizontl symptote y = 0 which is the x xis. Both ends of the grph get closer nd closer to the x xis but do not touch the xis. Exmple 1! Exmple 8x 3 x + 5x 3 6x 5 + 6x 4 x + x 6x 3 + 7x 5 3x 6 x 4 + 5x 3 4x 4 3x x 3! 4 3x = 0 nd 4 3x = 0 x 3 = 0 nd x 3 = 0 The left nd right ends of the grph both pproch the x xis The left nd right ends of the grph both pproch the x xis 4-5A Lecture! Pge 3 of 1! 018 Eitel

4 The it cn be found by looking t the reduced form of the rtio of the leding terms. It cn lso be found by simply inspecting the rtio nd using the english sentence tht describes the it. In English: If f (x) is rtionl function nd the exponent of the highest power in the denomintor is greter thn the highest power in the numertor both ends of the grph pproch the horizontl symptote y = 0 which is the x xis. Exmple 1!! Exmple 1 English pproch!! n x n reduce b n x m pproch x 3 +1 x 5 3x + 4 x x 5 3x + 4 the exponent of the highest power in the denomintor is greter then the highest power in the denomintor so the y vlues t both ends of the grph pproch 0 x x 3 +1 x 5 3x + 4 = 0 x 3 +1 x 5 3x + 4 = 0! 3x 3x = 0 nd The left nd right ends of the grph both pproch the x xis nd x 3 +1 x 5 3x + 4 = 0 x 3 +1 x 5 3x + 4 = 0 3x = 0 4-5A Lecture! Pge 4 of 1! 018 Eitel

5 Cse Given the rtionl function f (x) where f (x) = g(x) h(x) = n xn ± n 1 x n 1 ± n x n b m x m ± b m 1 x m 1 ± b m x m +...b 0 for n,m Z + nd n,b n R If f (x) is rtionl function nd the exponent of x in the leding term of the denomintor equls the exponent of x in the leding term of the numertor then the rtio of the leding terms of the numertor nd denomintor will reduce to b If,b R, b 0 then f(x) = b nd x f(x) = b As x (the left end of the grph) or s x + (the right end of the grph ) the y vlues on both ends of the grph pproch b The line y = b is horizontl symptote t both ends of the grph. In English: If f (x) is rtionl function nd the exponent of the highest power in the numertor is equl to the highest power in the denomintor then both ends of the grph pproch horizontl symptote of y =. Both ends of the grph get closer nd b closer to the line y = b but do not touch the line. Exmple 1! Exmple 1x 4 x 3 + 5x 3 6x 4 + 6x 4 x + x 1x 6 9x + 4x 3x 6 + 4x x 3 = nd = The left nd right ends of the grph both pproch the line y =! 4 4 = 4 nd The left nd right ends of the grph both pproch the line y = 4 4 = 4 4-5A Lecture! Pge 5 of 1! 018 Eitel

6 The it cn be found by looking t the reduced form of the rtio of the leding terms. It cn lso be found by simply inspecting the rtio nd using the english sentence tht describes the it. In English: If the exponent of the highest power in the numertor is equl to the highest power in the denomintor then both ends of the grph pproch horizontl symptote of y = b. Exmple 1!! Exmple 1 English pproch!! reduce n xn b n x m pproch 6x 3 + 5x x 3 3x + 4 6x 3 + 5x x 3 3x the exponent of the highest power in the denomintor is equl to the highest power in the denomintor so the y vlues t both ends of the grph pproch 6 / = 3 x 6x 3 + 5x x 3 3x + 4 = 6 = 3 6x 3 + 5x x 3 3x + 4 = 6 = 3! 6 6 = 3 nd 6 = 3 The left nd right ends of the grph both pproch the line y = 5 nd 6x 3 + 5x x 3 3x = 3 6x 3 + 5x x 3 3x = 3 4-5A Lecture! Pge 6 of 1! 018 Eitel

7 Cse 3 If the exponent of x in the leding term of the numertor is greter thn the exponent of x in the leding term of the denomintor then the the rtio of the leding terms of the numertor nd denomintor will reduce to b xn b xn = If n Z + + or nd nd,b R then b xn = + or depending on the sign of b xn when you subsitute negitive vlue for x into b xn depending on the sign of b xn when you subsitute positive vlue for x into b xn As x (the left end of the grph) the y vlues increse to positive infinity y + or decrese to negtive infinity y depending on the sign of b xn for positive vlues of x As x + (the right end of the grph) the y vlues increse to positive infinity y + or decrese to negtive infinity y depending on the sign of b xn for negtive vlues of Exmple 1! Exmple 4x 3 x 3 + 5x 3 x 3x + x 6x 5 + 5xx 4 x x 3 6x + 7 x 3x if x is negitive if x is positive ( #) = (+ #) = +! if x is negitive if x is positive 3( #) = + 3(+ #) = + x = nd The left end of the grph pproches x = + The right end of the grph pproches + 3x = + nd 3x = + The left end of the grph pproches + The right end of the grph pproches + 4-5A Lecture! Pge 7 of 1! 018 Eitel

8 Exmple 3! Exmple 4 8x 6 4x 3 x x 3 + 3x + 8 4x 6 x 3 + x 3 x + 3x 3 4x3 x4 if x is negitive if x is positive if x is negitive if x is positive 4( #) 3 = + 4(+#) 3 = ( #) 4 = (+#) 4 = 4x3 = + nd 4x3 = x4 = nd x4 = The left end of the grph pproches + The right end of the grph pproches! The left end of the grph pproches The right end of the grph pproches If the rtio of the leding terms of the rtionl function b xn then the behvior of the ends of the grph is the sme s the behvior for the ends of the grph for power functions. hs even exponent Model: y = 3x y + y + hs odd exponent Model: y = x 3 y + hs positive coefficient y Model: y = 3x Model: y = x 3 y + hs negtive coefficient y y y 4-5A Lecture! Pge 8 of 1! 018 Eitel

9 Expnding fctored forms to find the leding terms It is esy to find the rtio of the leding terms of rtionl function if the polynomils in the numertor nd denomintors re written s polynomils with the terms listed in decresing powers of x. The rtio of the leding terms of f (x) = 3x3 x + 4x + 3 x 5 + 5x 4 x 7x is 3x 3 x 5 In mny cses the rtionl expression will be in fctored form like the function shown below. f (x) = (3x 4)3 (x + 3) 6x (x 1) If the rtionl function is written with the polynomils listed in fctored form you will need to expnd the fctors to find the leding terms. Do I relly need to FOIL the fctors? Tht would be lrge job nd tke lot to time. The good news is tht ll you relly only need to find the leding term for the expression in the numertor nd denomintors. We cn find the leding terms without FOILing the entire list of fctors. Find the rtio of the leding terms of the rtionl function! Exmple 1 (x 1) 3 (x + 3) f (x) = (x 5) (x 1) 3 rewrite ech fctor with only the leding term for ech fctor = (x)3 (x) (x) (x) 3 multiply those terms = 8x3 x 4x x 3 = 8x4 4x 5 reduce the expression to find the leding term in reduced form x 4-5A Lecture! Pge 9 of 1! 018 Eitel

10 ! Exmple Find the rtio of the leding terms of the rtionl function (3x 4) 3 (x + 3) 4 6x (x 1) rewrite ech fctor with only the leding term for ech fctor = (3x)3 (x) 4 9x (x) multiply those terms = 7x3 x 4 9x x = 7x7 9x 4 reduce the expression to find the leding term in reduced form 3x 3! Exmple 3 Find the rtio of the leding terms of the rtionl function 4x 3 (x 1) (x + 5) 3 (x + 3) 4 rewrite ech fctor with only the leding term for ech fctor = 4x3 (x) (x) 3 (x) 4 multiply those terms = 4x3 x 8x 3 x 4 = 4x5 8x 7 reduce the expression to find the leding term in reduced form 1 x 4-5A Lecture! Pge 10 of 1! 018 Eitel

11 ! Exmple 4 Find the rtio of the leding terms of the rtionl function 8x(x + 5)(x 1) (x 1) (x + 7) rewrite ech fctor with only the leding term for ech fctor 8x (x)(x) = (x) (x) multiply those terms 8x x x = 4x x = 8x4 4x 4 reduce the expression to find the leding term in reduced form! Exmple 5 Find the rtio of the leding terms of the rtionl function (x + 3) (x 1) 3 x (x ) rewrite ech fctor with only the leding term for ech fctor = (x) (x) 3 x (x) multiply those terms = x x 3 x x = x5 x 3 reduce the expression to find the leding term in reduced form x 4-5A Lecture! Pge 11 of 1! 018 Eitel

12 Limits s x pproches infinity for rtionl functions expressed in fctored form. Exmple 1 x (x 1) (x + ) 3 (x 1) rtio of the leding terms is x (x) (x) 3 (x) = x3 x 3 = x x = 0 nd x = 0 The left end of the grph pproches 0 The right end of the grph pproches 0 Exmple x (x 1) 3 (x + 5) (x 1) 3 rtio of the leding terms is x (x) 3 (x) (x) 3 = 8x5 4x 5 = = nd = The left end of the grph pproches The right end of the grph pproches 4-5A Lecture! Pge 1 of 1! 018 Eitel

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