Chpter 3: Polynomil nd Rtionl Functions Section 3. Power Functions & Polynomil Functions... 94 Section 3. Qudrtic Functions... 0 Section 3.3 Grphs of Polynomil Functions... 09 Section 3.4 Rtionl Functions... 8 Section 3.5 Inverses nd Rdicl Functions... 3 Section 3. Power Functions & Polynomil Functions A squre is cut out of crdord, with ech side hving some length L. If we wnted to write function for the re of the squre, with L s the input, nd the re s output, you my recll tht re cn e found y multiplying the length times the width. Since our shpe is squre, the length & the width re the sme, giving the formul: A ( L) L L L Likewise, if we wnted function for the volume of cue with ech side hving some length L, you my recll tht volume cn e found y multiplying length y width y height, which re ll equl for cue, giving the formul: 3 V ( L) L L L L These two functions re emples of power functions; functions tht re some power of the vrile. Definition A power function is function tht cn e represented in the form p Where the se is the vrile nd the eponent, p, is numer. Emple Which of our toolkit functions re power functions? The constnt nd identity functions re power functions, since they cn e written s 0 nd respectively. The qudrtic nd cuic functions re oth power functions with whole numer powers: 3 nd. The rtionl functions re oth power functions with negtive whole numer powers since they cn e written s nd. The squre nd cue root functions re oth power functions with frctionl powers 3 since they cn e written s or. This chpter is prt of Preclculus: An Investigtion of Functions Lippmn & Rsmussen 0. This mteril is licensed under Cretive Commons CC-BY-SA license.
3. Power nd Polynomil Functions 95 Try it Now. Wht point(s) do the toolkit power functions hve in common? Chrcteristics of Power Functions Shown to the right re the grphs of 4 6,, nd, ll even whole numer powers. Notice tht ll these grphs hve firly similr shpe, very similr to the qudrtic toolkit, ut s the power increses the grphs fltten somewht ner the origin, nd grow fster s the input increses. 6 4 To descrie the ehvior s numers ecome lrger nd lrger, we use the ide of infinity. The symol for positive infinity is, nd for negtive infinity. When we sy tht pproches infinity, which cn e symoliclly written s, we re descriing ehvior we re sying tht is getting lrge in the positive direction. With the even power function, s the input ecomes lrge in either the positive or negtive directions, the output vlues ecome very lrge positive numers. Equivlently, we could descrie this y sying tht s pproches positive or negtive infinity, the f() vlues pproch positive infinity. In symolic form, we could write: s ±, f (). Shown here re the grphs of 3 5 7,, nd, ll odd whole numer powers. Notice ll these grphs look similr to the cuic toolkit, ut gin s the power increses the grphs fltten ner the origin nd grow fster s the input increses. For these odd power functions, s pproches negtive infinity, f() pproches negtive infinity. As pproches positive infinity, f() pproches positive infinity. In symolic form we write: s, f () nd s, f (). 7 5 3 Definition The ehvior of the grph of function s the input tkes on lrge negtive vlues ( ) nd lrge positive vlues ( ) s is referred to s the long run ehvior of the function.
96 Chpter 3 Emple Descrie the long run ehvior of the grph of 8. 8 Since hs whole, even power, we would epect this function to ehve somewht like the qudrtic function. As the input gets lrge positive or negtive, we would epect the output to grow in the positive direction. In symolic form, s ±, f (). Emple 3 Descrie the long run ehvior of the grph of 9 Since this function hs whole odd power, we would epect it to ehve somewht like the cuic function. The negtive in front of the function will cuse verticl reflection, so s the inputs grow lrge positive, the outputs will grow lrge in the negtive direction, nd s the inputs grow lrge negtive, the outputs will grow lrge in the positive direction. In symolic form, for the long run ehvior we would write: s, f () nd s, f (). You my use words or symols to descrie the long run ehvior of these functions. Try it Now. Descrie in words nd symols the long run ehvior of 4 Tretment of the rtionl nd rdicl forms of power functions will e sved for lter. Polynomils An oil pipeline ursts in the Gulf of Meico, cusing n oil slick roughly in circulr shpe. The slick is currently 4 miles in rdius, ut tht rdius is incresing y 8 miles ech week. If we wnted to write formul for the re covered y the oil slick, we could do so y composing two functions together. The first is formul for the rdius, r, of the spill, which depends on the numer of weeks, w, tht hve pssed. Hopefully you recognized tht this reltionship is liner: r( w) 4 + 8w We cn comine this with the formul for the re, A, of circle: A( r) πr Composing these functions gives formul for the re in terms of weeks: A ( w) A( r( w)) A(4 + 8w) π (4 + 8w)
3. Power nd Polynomil Functions 97 Multiplying this out gives the formul A( w) 576π + 384πw + 64πw This formul is n emple of polynomil. A polynomil is simply the sum of terms consisting of trnsformed power functions with positive whole numer powers. Definitions A polynomil is function of the form f + + ( ) + + 0 n n Ech of the i constnts re clled coefficients nd cn e positive, negtive, whole numers, decimls, or frctions. A term of the polynomil is ny one piece of the sum, ny trnsformed power function i i. Ech individul term is The degree of the polynomil is the highest power of the vrile tht occurs in the polynomil. The leding term is the term contining the highest power of the vrile; the term with the highest degree. The leding coefficient is the coefficient on the leding term. Becuse of the definition of the leding term we often rerrnge polynomils so tht the powers re descending nd the prts re esier to determine. n +... + + n + 0 Emple 4 Identify the degree, leding term, nd leding coefficient of these polynomils: 3 3 + 4 5 3 g( t) 5t t + 7t h ( p) 6 p p 3 For the function f(), the degree is 3, the highest power on. The leding term is the 3 term contining tht power, 4. The leding coefficient is the coefficient of tht term, -4. For g(t), the degree is 5, the leding term is For h(p), the degree is 3, the leding term is 5 5t, nd the leding coefficient is 5. 3 p, so the leding coefficient is -.
98 Chpter 3 Definition For ny polynomil, the long run ehvior of the polynomil will mtch the long run ehvior of the leding term. Emple 5 Wht cn we determine out the long run ehvior nd degree of the eqution for the polynomil grphed here? Since the grph grows lrge nd positive s the inputs grow lrge nd positive, we descrie the long run ehvior symoliclly y writing: s, f (), nd s, f (). In words we could sy tht s vlues pproch infinity, the function vlues pproch infinity, nd s vlues pproch negtive infinity the function vlues pproch negtive infinity. We cn tell this grph hs the shpe of n odd degree power function which hs not een reflected, so the degree of the polynomil creting this grph must e odd. Try it Now 3. Given the function 0.( )( + )( 5) use your lger skills write the function in polynomil form nd determine the leding term, degree, nd long run ehvior of the function. Short Run Behvior Chrcteristics of the grph such s verticl nd horizontl intercepts nd the plces the grph chnges direction re prt of the short run ehvior of the polynomil. Like with ll functions, the verticl intercept is where the grph crosses the verticl is, nd occurs when the input vlue is zero. Since polynomil is function, there cn only e one verticl intercept, which occurs t 0, or the point ( 0, 0 ). The horizontl intercepts occur t the input vlues tht correspond with n output vlue of zero. It is possile to hve more thn one horizontl intercept.
3. Power nd Polynomil Functions 99 Emple 6 Given the polynomil function ( )( + )( 4), given in fctored form for your convenience, determine the verticl nd horizontl intercepts. The verticl intercept occurs when the input is zero. f ( 0) (0 )(0 + )(0 4) 8. The grph crosses the verticl is t the point (0, 8) The horizontl intercepts occur when the output is zero. 0 ( )( + )( 4) when, -, or 4 The grph crosses the horizontl is t the points (, 0), (-, 0), nd (4, 0) Notice tht the polynomil in the previous emple, which would e degree three if multiplied out, hd three horizontl intercepts nd two turning points - plces where the grph chnges direction. We will mke generl sttement here without justifiction t this time the resons will ecome cler lter in this chpter. Definition A polynomil of degree n will hve: At most n horizontl intercepts. An odd degree polynomil will lwys hve t lest one. At most n- turning points Emple 7 Wht cn we conclude out the grph of the polynomil shown here? Bsed on the long run ehvior, with the grph ecoming lrge positive on oth ends of the grph, we cn determine tht this is the grph of n even degree polynomil. The grph hs horizontl intercepts, suggesting degree of or greter, nd 3 turning points, suggesting degree of 4 or greter. Bsed on this, it would e resonle to conclude tht the degree is even nd t lest 4, so it is proly fourth degree polynomil.
00 Chpter 3 Try it Now 4. Given the function 0.( )( + )( 5) determine the short run ehvior. IMPORTANT FEATURES OF THIS SECTION: Power Functions Polynomils Coefficients Leding coefficient Term Leding Term Degree of polynomil Long run ehvior Short run ehvior Try it Now Answers. (0, 0) nd (, ) re common to ll power functions. As pproches positive nd negtive infinity, f() pproches negtive infinity: s ±, f () ecuse of the verticl flip. 3 3. The leding term is 0., so it is degree 3 polynomil, s pproches infinity (or gets very lrge in the positive direction) f() pproches infinity, nd s pproches negtive infinity (or gets very lrge in the negtive direction) f() pproches negtive infinity. (Bsiclly the long run ehvior is the sme s the cuic function) 4. Horizontl intercepts re (, 0) (-, 0) nd (5, 0), the verticl intercept is (0, ) nd there re turns in the grph.
3. Qudrtic Functions 0 Section 3. Qudrtic Functions In this section, we will eplore the fmily of nd degree polynomils, the qudrtic functions. While they shre mny chrcteristics of polynomils in generl, the clcultions involved in working with qudrtics is typiclly little simpler, which mkes them good plce to strt our eplortion of short run ehvior. In ddition, qudrtics commonly rise from prolems involving re nd projectile motion, providing some interesting pplictions. Emple A ckyrd frmer wnts to enclose rectngulr spce for new grden. She hs purchsed 80 feet of wire fencing to enclose 3 sides, nd will put the 4 th side ginst the ckyrd fence. Find formul for the re of the fence if the sides of fencing perpendiculr to the eisting fence hve length L. In scenrio like this involving geometry, it is often helpful to drw picture. It might lso e helpful to introduce temporry vrile, W, to represent the side of fencing prllel to the 4 th side or ckyrd fence. Grden W L Since we know we only hve 80 feet of fence ville, we know tht L + W + L 80, or more simply, L + W 80 This llows us to represent the width, W, in terms of L: W 80 Now we re redy to write n eqution for the re the fence encloses. We know the re of rectngle is length multiplied y width, so A LW L( 80 L) A( L) 80L L This formul represents the re of the fence in terms of the vrile length L. L Bckyrd Short run Behvior: Verte We now eplore the interesting fetures of the grphs of qudrtics. In ddition to intercepts, qudrtics hve n interesting feture where they chnge direction, clled the verte. You proly noticed tht ll qudrtics re relted to trnsformtions of the sic qudrtic function.
0 Chpter 3 Emple Write n eqution for the qudrtic grphed elow s trnsformtion of, then epnd the formul nd simplify terms to write the eqution in stndrd polynomil form. We cn see the grph is the sic qudrtic shifted to the left nd down 3, giving formul in the form g ( ) ( + ) 3. By plugging in cler point such s (0,-) we cn solve for the stretch fctor: (0 + ) 3 4 Written s trnsformtion, the eqution for this formul is g ( ) ( + ) 3. To write this in stndrd polynomil form, we cn epnd the formul nd simplify terms: g( ) ( + ) 3 g( ) ( + )( + ) 3 g( ) ( + 4 + 4) 3 g( ) + + 3 g( ) + Notice tht the horizontl nd verticl shifts of the sic qudrtic determine the loction of the verte of the prol; the verte is unffected y stretches nd compressions.
3. Qudrtic Functions 03 Try it Now. Below, coordinte grid hs een superimposed over the qudrtic pth of sketll. Find n eqution for the pth of the ll. Does he mke the sket? Definition The stndrd form of qudrtic is + + c The trnsformtion form of qudrtic is ( h) + k The verte of the qudrtic is locted t (h, k) Becuse the verte cn lso e seen in this formt it is often clled verte form s well In the previous emple, we sw tht it is possile to rewrite qudrtic in trnsformed form into stndrd form y epnding the formul. It would e useful to reverse this process, since the trnsformtion form revels the verte. Epnding out the generl trnsformtion form of qudrtic gives: ( h) + k ( h)( h) + k ( h + h ) + k h + h + k This should e equl to the stndrd form of the qudrtic: h + h + k + + c The second degree terms re lredy equl. For the liner terms to e equl, the coefficients must e equl: h, so h This provides us method to determine the horizontl shift of the qudrtic from the stndrd form. We could likewise set the constnt terms equl to find: h + k c, so k c h c c 4 c 4 In prctice, though, it is usully esier to rememer tht k is the output vlue of the function when the input is h, so k f (h). From http://log.mrmeyer.com/?p4778, Dn Meyer, CC-BY
04 Chpter 3 Definition For qudrtic given in stndrd form, the verte (h, k) is locted t: h, k f f ( h) Emple 3 Find the verte of the qudrtic 6 + 7. Rewrite the qudrtic into trnsformtion form (verte form). The horizontl component of the verte will e t The verticl component of the verte will e t h 3 3 f 6 () 6 4 3 3 6 + 7 Rewriting into trnsformtion form, the stretch fctor will e the sme s the in the originl qudrtic. Using the verte to determine the shifts, 3 + 5 5 Try it Now. Given the eqution g( ) 3 + 6 write the eqution in Stndrd Form nd then in Trnsformtion/Verte form. In ddition to enling us to more esily grph qudrtic written in stndrd form, finding the verte serves nother importnt purpose it llows us to determine the mimum or minimum vlue of the function, depending on which wy the grph opens. Emple 4 Returning to our ckyrd frmer from the eginning of the section, wht dimensions should she mke her grden to mimize the enclosed re? Erlier we determined the re she could enclose with 80 feet of fencing on three sides ws given y the eqution A( L) 80L L. Notice tht qudrtic hs een verticlly reflected, since the coefficient on the squred term is negtive, so grph will open downwrds, nd the verte will e mimum vlue for the re. In finding the verte, we tke cre since the eqution is not written in stndrd polynomil form with decresing powers. But we know tht is the coefficient on the squred term, so -, 80, nd c 0.
3. Qudrtic Functions 05 Finding the verte: 80 h 0, k A(0) 80(0) (0) 800 ( ) The mimum vlue of the function is n re of 800 squre feet, which occurs when L 0 feet. When the shorter sides re 0 feet, tht leves 40 feet of fencing for the longer side. To mimize the re, she should enclose the grden so the two shorter sides hve length 0 feet, nd the longer side prllel to the eisting fence hs length 40 feet. Short run Behvior: Intercepts As with ny function, we cn find the verticl intercepts of qudrtic y evluting the function t n input of zero, nd we cn find the horizontl intercepts y solving for when the output will e zero. Notice tht depending upon the loction of the grph, we might hve zero, one, or two horizontl intercepts. zero horizontl intercepts one horizontl intercept two horizontl intercepts Emple 5 Find the verticl nd horizontl intercepts of the qudrtic 3 + 5 We cn find the verticl intercept y evluting the function t n input of zero: f (0) 3(0) + 5(0) Verticl intercept t (0,-) For the horizontl intercepts, we solve for when the output will e zero 0 3 + 5 In this cse, the qudrtic cn e fctored, providing the simplest method for solution 0 (3 )( + ) 0 3 0 + or Horizontl intercepts t,0 nd (-,0) 3 3 Notice tht in the stndrd form of qudrtic, the constnt term c revels the verticl intercept of the grph.
06 Chpter 3 Emple 6 Find the horizontl intercepts of the qudrtic + 4 4 Agin we will solve for when the output will e zero 0 + 4 4 Since the qudrtic is not fctorle in this cse, we solve for the intercepts y first rewriting the qudrtic into trnsformtion form. 4 h k f ( ) ( ) + 4( ) 4 6 () ( + ) 6 Now we cn solve for when the output will e zero 0 ( + ) 6 6 ( + ) 3 ( + ) + ± 3 ± 3 The grph hs horizontl intercepts t ( 3,0) nd ( + 3,0) Try it Now 3. In Try it Now prolem we found the stndrd & trnsformtion form for the eqution g( ) 3 + 6. Now find the Verticl & Horizontl intercepts (if ny). Since this process is done commonly enough tht sometimes people find it esier to solve the prolem once in generl then rememer the formul for the result, rther thn repeting the process. Bsed on our previous work we showed tht ny qudrtic in stndrd form cn e written into trnsformtion form s: + + c 4 Solving for the horizontl intercepts using this generl eqution gives:
3. Qudrtic Functions 07 c 4 0 + + strt to solve for y moving the constnts to the other side 4 + c divide oth sides y 4 + c find common denomintor to comine frctions 4 4 4 + c comine the frctions on the left side of the eqution 4 4 + c tke the squre root of oth sides c 4 4 + ± sutrct / from oth sides c ± 4 comining the frctions c 4 ± Notice tht this cn yield two different nswers for Definition For qudrtic given in stndrd form, the qudrtic formul gives the horizontl intercepts of the grph of the qudrtic. c 4 ± Emple 7 A ll is thrown upwrds from the top of 40 foot high uilding t speed of 80 feet per second. The ll s height ove ground cn e modeled y the eqution 40 80 6 ) ( + + t t t h. Wht is the mimum height of the ll? When does the ll hit the ground? To find the mimum height of the ll, we would need to know the verte of the qudrtic. 5 3 80 6) ( 80 h, 40 40 5 80 5 6 5 + + h k The ll reches mimum height of 40 feet fter.5 seconds
08 Chpter 3 To find when the ll hits the ground, we need to determine when the height is zero when h(t) 0. While we could do this using the trnsformtion form of the qudrtic, we cn lso use the qudrtic formul: 80 ± t 80 4( 6)(40) 80 ± 8960 ( 6) 3 Since the squre root does not evlute to whole numer, we cn use clcultor to pproimte the vlues of the solutions: 80 8960 80 + 8960 t 5.458 or t 0. 458 3 3 The second nswer is outside the resonle domin of our model, so we conclude the ll will hit the ground fter out 5.458 seconds. Try it Now 4. For these two equtions determine if the verte will e mimum vlue or minimum vlue.. g ( ) 8 + + 7. g ( ) 3(3 ) + IMPORTANT FEATURES OF THIS SECTION Qudrtic functions Stndrd form Trnsformtion form/verte form Verte s mimum / Verte s minimum Short run ehvior Verte / Horizontl & Verticl intercepts Qudrtic formul Try it Now Answers. The pth psses through the origin with verte t (-4, 7). 7 h( ) ( + 4) + 7. To mke the shot, h(-7.5) would 6 need to e out 4. h( 7.5).64 ; he doesn t mke it.. g ( ) 6 + 3 in Stndrd form; g( ) ( 3) + 4 in Trnsformtion form 3. Verticl intercept t (0, 3), NO horizontl intercepts. 4.. Verte is minimum vlue. Verte is mimum vlue
3.3 Grphs of Polynomil Functions 09 Section 3.3 Grphs of Polynomil Functions In the previous section we eplored the short run ehvior of qudrtics, specil cse of polynomils. In this section we will eplore the short run ehvior of polynomils in generl. Short run Behvior: Intercepts As with ny function, the verticl intercept cn e found y evluting the function t n input of zero. Since this is evlution, it is reltively esy to do it for ny degree polynomil. To find horizontl intercepts, we need to solve for when the output will e zero. For generl polynomils, this cn e chllenging prospect. While qudrtics cn e solved using the reltively simple qudrtic formul, the corresponding formuls for cuic nd 4 th degree polynomils re not simple enough to rememer, nd formuls do not eist for generl higher degree polynomils. Consequently, we will limit ourselves to three cses: ) The polynomil cn e fctored using known methods: gretest common fctor nd trinomil fctoring. ) The polynomil is given in fctored form 3) Technology is used to determine the intercepts Emple : Find the horizontl intercepts of + 6 4 3. We cn ttempt to fctor this polynomil to find solutions for f() 0 6 4 3 + 0 Fctoring out the gretest common fctor 4 3 + Fctoring the inside s qudrtic ( ) 0 ( )( ) 0 ( ) Then rek prt to find solutions 0 0 or 0 or This gives us 5 horizontl intercepts. ( ) ± ± 0 Emple : Find the verticl nd horizontl intercepts of g ( t) ( t ) (t + 3) The verticl intercept cn e found y evluting g(0). g (0) (0 ) ((0) + 3)
0 Chpter 3 The horizontl intercepts cn e found y solving g(t) 0 ( t ) (t + 3) 0 Since this is lredy fctored, we cn rek it prt: ( t ) 0 (t + 3) 0 t 0 or 3 t t Emple 3 Find the horizontl intercepts of 3 h ( t) t + 4t + t 6 Since this polynomil is not in fctored form, hs no common fctors, nd does not pper to e fctorle using techniques we know, we cn turn to technology to find the intercepts. Grphing this function, it ppers there re horizontl intercepts t -3, -, nd Try it Now. Find the verticl nd horizontl intercepts of the function f 4 ( t) t 4t Grphicl Behvior t Intercepts If we grph the function 3 ( + 3)( ) ( + ), notice tht the ehvior t ech of the horizontl intercepts is different. At the horizontl intercept -3, coming from the ( + 3) fctor of the polynomil, the grph psses directly through the horizontl intercept. The fctor is liner (hs power of ), so the ehvior ner the intercept is like tht of line - it psses directly through the intercept. We cll this single zero, since the zero is formed from single fctor of the function. At the horizontl intercept, coming from the ( ) fctor of the polynomil, the grph touches the is t the intercept nd chnges direction. The fctor is qudrtic (degree ), so the ehvior ner the intercept is like tht of qudrtic it ounces off of
3.3 Grphs of Polynomil Functions the horizontl is t the intercept. Since ( ) ( )( ), the fctor is repeted twice, so we cll this doule zero. 3 At the horizontl intercept -, coming from the ( +) fctor of the polynomil, the grph psses through the is t the intercept, ut flttens out it first. This fctor is cuic (degree 3), so the ehvior ner the intercept is like tht of cuic, with the sme 3 S type shpe ner the intercept tht the toolkit hs. We cll this triple zero. By utilizing these ehviors, we cn sketch resonle grph of fctored polynomil function without needing technology. Definition p If polynomil contins fctor of the form ( h), the ehvior ner the horizontl intercept h is determined y the power on the fctor. p p p 3 Single zero Doule zero Triple zero For higher even powers 4,6,8 etc the grph will still ounce off of the grph ut the grph will pper fltter with incresing even power s it pproches nd leves the is. For higher odd powers, 5,7,9 etc the grph will still pss through the grph ut the grph will pper fltter with incresing odd power s it pproches nd leves the is. Emple 4 Sketch grph of ( + 3) ( 5) This grph hs two horizontl intercepts. At -3, the fctor is squred, indicting the grph will ounce t this horizontl intercept. At 5, the fctor is not squred, indicting the grph will pss through the is t this intercept. Additionlly, we cn see the leding term, if this polynomil were multiplied out, would 3 e, so the long-run ehvior is tht of verticlly reflected cuic, with the outputs decresing s the inputs get lrge positive, nd the inputs incresing s the inputs get lrge negtive.
Chpter 3 To sketch this we consider the following: As the function f () so we know the grph strts in the nd qudrnt nd is decresing towrd the horizontl is. At (-3, 0) the grph ounces off of the horizontl is nd so the function must strt incresing. At (0, 90) the grph crosses the verticl is t the verticl intercept Somewhere fter this point the grph must turn ck down / or strt decresing towrd the horizontl is since the grph psses through the net intercept t (5,0) As the function f () so we know the grph continues to decrese nd we cn stop drwing the grph in the 4 th qudrnt. Using technology we see tht the resulting grph will look like: Solving Polynomil Inequlities One ppliction of our ility to find intercepts nd sketch grph of polynomils is the ility to solve polynomil inequlities. It is very common question to sk when function will e positive nd negtive. We cn solve polynomil inequlities y either utilizing the grph, or y using test vlues. Emple 5 Solve ( + 3)( + ) ( 4) > 0 As with ll inequlities, we strt y solving the equlity ( + 3)( + ) ( 4) 0, which hs solutions t -3, -, nd 4. We know the function cn only chnge from positive to negtive t these vlues, so these divide the inputs into 4 intervls.
3.3 Grphs of Polynomil Functions 3 We could choose test vlue in ech intervl nd evlute the function ( + 3)( + ) ( 4) t ech test vlue to determine if the function is positive or negtive in tht intervl Intervl Test in intervl f( test vlue) >0 or <0? < -3-4 7 > 0-3 < < - - -6 < 0 - <.< 4 0 - < 0 > 4 5 88 > 0 On numer line this would look like: positive 0 negtive 0 negtive 0 positive From our test vlues, we cn determine this function is positive when < -3 or > 4, or in intervl nottion, (, 3) (4, ) We could hve lso determined on which intervls the function ws positive y sketching grph of the function. We illustrte tht technique in the net emple Emple 6 Find the domin of the function v( t) 6 5t t A squre root only is defined when the quntity we re tking the squre root of is zero or greter. Thus, the domin of this function will e when 6 5t t 0. Agin we strt y solving the equlity 6 5t t 0. While we could use the qudrtic formul, this eqution fctors nicely to ( 6 + t )( t) 0, giving horizontl intercepts t nd t -6. Sketching grph of this qudrtic will llow us to determine when it is positive: From the grph we cn see this function is positive for inputs etween the intercepts. So 6 5t t 0 for 6 t, nd this will e the domin of the v(t) function.
4 Chpter 3 Try it Now 3. Given the function g( ) 6 use the methods tht we hve lerned so fr to find the verticl & horizontl intercepts, determine where the function is negtive nd positive, descrie the long run ehvior nd sketch the grph without technology. Writing Equtions using Intercepts Since polynomil function written in fctored form will hve horizontl intercept where ech fctor is equl to zero, we cn form n eqution tht will pss through set of horizontl intercepts y introducing corresponding set of fctors. Definition If polynomil hs horizontl intercepts t,,, n, then the polynomil cn e written in the fctored form p p p ( ) ( ) ( n n ) where the powers p i on ech fctor cn e determined y the ehvior of the grph t the corresponding intercept, nd the stretch fctor cn e determined given vlue of the function other thn the horizontl intercept. Emple 7 Write n eqution for the polynomil grphed here This grph hs three horizontl intercepts: -3,, nd 5. At -3 nd 5 the grph psses through the is, suggesting the corresponding fctors of the polynomil will e liner. At the grph ounces t the intercept, suggesting the corresponding fctor of the polynomil will e nd degree or qudrtic. Together, this gives us: ( + 3)( ) ( 5) To determine the stretch fctor, we cn utilize nother point on the grph. Here, the verticl intercept ppers to e (0,-), so we cn plug in those vlues to solve for
3.3 Grphs of Polynomil Functions 5 (0 + 3)(0 ) (0 5) 60 30 The grphed polynomil would hve eqution ( + 3)( ) ( 5) 30 Try it Now 3. Given the grph, determine nd write the eqution for the grph in fctored form. Estimting Etrem With qudrtics, we were le to lgericlly find the mimum or minimum vlue of the function y finding the verte. For generl polynomils, finding these turning points is not possile without more dvnced techniques from clculus. Even then, finding where etrem occur cn still e lgericlly chllenging. For now, we will estimte the loctions of turning points using technology to generte grph. Emple 8 An open-top o is to e constructed y cutting out squres from ech corner of 4cm y 0cm sheet of plstic then folding up the sides. Find the size of squres tht should e cut out to mimize the volume enclosed y the o. We will strt this prolem y drwing picture, leling the width of the cut-out squres with vrile, w. w w
6 Chpter 3 Notice tht fter squre is cut out from ech end, it leves (4-w) cm y (0-w) cm for the se of the o, nd the o will e w cm tll. This gives the volume: 3 V ( w) (4 w)(0 w) w 80w 68w + 4w Using technology to sketch grph llows us to estimte the mimum vlue for the volume, restricted to resonle vlues for w vlues from 0 to 7. From this grph, we cn estimte the mimum vlue is round 340, nd occurs when the squres re out.75cm squre. To improve this estimte, we could use fetures of our technology if ville, or simply chnge our window to zoom in on our grph. From this zoomed-in view, we cn refine our estimte for the m volume to out 339, when the squres re.7cm squre. Try it Now 4. Use technology to find the Mimum nd Minimum vlues on the intervl [-, 4] of 3 the eqution 0.( ) ( + ) ( 4).
3.3 Grphs of Polynomil Functions 7 IMPORTANT FEATURES OF THIS SECTION Short Run Behvior Intercepts (Horizontl & Verticl) Methods to find Horizontl intercepts Fctoring Methods Fctored Forms Technology Grphicl Behvior t intercepts Single, Doule nd Triple zeros (or power, & 3 ehviors) Solving polynomil inequlities using test vlues & grphing techniques Writing equtions using intercepts Estimting etrem Try it Now Answers. Verticl intercept (0, 0) Horizontl intercepts (0, 0), (-, 0), (, 0). Verticl intercept (0, 0) Horizontl intercepts (-, 0), (0, 0), (3, 0) The function is negtive from (, -) nd (0, 3) The function is positive from (-, 0) nd (3, ) 3 The leding term is so s, g() nd s, g () 3 3. ( ) ( + ) ( 4) 8 4. Approimtely, (0, -6.5) minimum nd pproimtely (3.5, 7) mimum.
Section 3.4 Rtionl Functions In the lst few sections, we hve uilt polynomils sed on the positive whole numer power functions. In this section we eplore the functions sed on power functions with negtive integer powers, the rtionl functions. Emple You pln to drive 00 miles. Find formul for the time the trip will tke s function of the speed you drive. You my recll tht multiplying speed y time will give you distnce. If we let t represent the drive time in hours, nd v represent the velocity (speed or rte) t which we drive, then vt distnce. Since our distnce is fied t 00 miles, vt 00. Solving this reltionship for the time gives us the function we desired: 00 t ( v) 00v v While this type of reltionship cn e written using the negtive eponent, it is more common to see it written s frction. This prticulr emple is one of n inversely proportionl reltionship where one quntity is constnt divided y the other quntity. Notice tht this is trnsformtion of the reciprocl toolkit function. Severl nturl phenomen, such s grvittionl force nd volume of sound, ehve in mnner inversely proportionl to the squre of the second quntity. For emple, the k volume, V, of sound herd t distnce d from the source would e relted y V d for some constnt vlue k. These functions re trnsformtions of the reciprocl squred toolkit function We hve seen the grphs of the sic reciprocl function nd the squred reciprocl function from our study of toolkit functions. These grphs hve severl importnt fetures. This chpter is prt of Preclculus: An Investigtion of Functions Lippmn & Rsmussen 0. This mteril is licensed under Cretive Commons CC-BY-SA license.
3.4 Rtionl Functions 9 Let s egin y looking t the reciprocl function,. As you well know, dividing y zero is not llowed nd therefore zero is not in the Domin, nd so the function is undefined t n input of zero. Short run ehvior: As the input ecomes very smll or s the input vlues pproch zero from the left side, the function vlues ecome very lrge in negtive direction, or pproch negtive infinity. We write: s 0, f (). As we pproch 0 from the right side, the input vlues re still very smll, ut the function vlues ecome very lrge or pproch positive infinity. + We write: s 0 f (). This ehvior cretes verticl symptote. An symptote is line tht the grph pproches. In this cse the grph is pproching the verticl line 0 s the input ecomes close to zero. Long run ehvior: As the vlues of pproch infinity, the function vlues pproch 0. As the vlues of pproch negtive infinity, the function vlues pproch 0. Symoliclly, s ± 0 Bsed on this long run ehvior nd the grph we cn see tht the function pproches 0 ut never ctully reches 0, it just levels off s the inputs ecome lrge. This ehvior cretes horizontl symptote. In this cse the grph is pproching the horizontl line f ( ) 0 s the input ecomes very lrge in the negtive nd positive direction. Definition A verticl symptote of grph is verticl line where the grph tends towrds positive or negtive infinity s the inputs pproch. As, f () ±. A horizontl symptote of grph is horizontl line where the grph pproches the line s the inputs get lrge. As ±,.
0 Chpter 3 Try it Now:. Use symolic nottion to descrie the long run ehvior nd short run ehvior for the reciprocl squred function. Emple Sketch grph of the reciprocl function shifted two units to the left nd up three units. Identify the horizontl nd verticl symptotes of the grph, if ny. Trnsforming the grph left nd up 3 would result in the eqution + 3, or equivlently y giving the terms common denomintor, + 3 + 7 + Shifting the toolkit function would give us this grph. Notice tht this eqution is undefined t -, nd the grph lso is showing verticl symptote t -. As,, nd s +, As the inputs grow lrge, the grph ppers to e leveling off t f ( ) 3, indicting horizontl symptote t f ( ) 3. As ±, 3. Notice tht horizontl nd verticl symptotes shifted long with the function. Try it Now. Sketch the grph nd find the horizontl nd verticl symptotes of the reciprocl squred function tht hs een shifted right 3 units nd down 4 units. In the previous emple, we shifted the function in wy tht resulted in function of the 3 + 7 form. This is n emple of generl rtionl function. +
3.4 Rtionl Functions Definition A rtionl function is function tht cn e written s the rtio of two polynomils, p() nd q(). p p( ) 0 + + + + p q q( ) + + + + 0 q Emple 3 A lrge miing tnk currently contins 00 gllons of wter, into which 5 pounds of sugr hve een mied. A tp will open pouring 0 gllons per minute of wter into the tnk t the sme time sugr is poured into the tnk t rte of pound per minute. Find the concentrtion (pounds per gllon) of sugr in the tnk fter t minutes. Notice tht the wter in the tnk is chnging linerly, s is the mount of sugr in the tnk. We cn write n eqution independently for ech: wter 00 + 0t sugr 5 + t The concentrtion, C, will e the rtio of pounds of sugr to gllons of wter 5 + t C( t) 00 + 0t Finding Asymptotes nd Intercepts Given rtionl eqution, s prt of discovering the short run ehvior we re interested in finding ny verticl nd horizontl symptotes, s well s finding ny verticl or horizontl intercepts s we hve in the pst. To find verticl symptotes, we notice tht the verticl symptotes occurred when the denomintor of the function ws undefined. With few eceptions, verticl symptote will occur whenever the denomintor is undefined. Emple 4 5 + Find the verticl symptotes of the function k( ) To find the verticl symptotes, we determine where this function will e undefined y setting the denomintor equl to zero: 0 ( + )( ) 0,
Chpter 3 This indictes two verticl symptotes, which look t grph confirms. The eception to this rule occurs when oth the numertor nd denomintor of rtionl function re zero. Emple 5 Find the verticl symptotes of the function k( ) 4 To find the verticl symptotes, we determine where this function will e undefined y setting the denomintor equl to zero: 4 0 4, However, the numertor of this function is lso equl to zero when. Becuse of this, the function will still e undefined t, since 0 0 is still undefined, ut the grph will not hve verticl symptote t. The grph of this function will hve the verticl symptote t -, ut t the grph will hve hole; single point where the grph is not defined, indicted y n open circle. Definition The verticl symptotes of rtionl function will occur where the denomintor of the function is equl to zero nd the numertor is not zero. A hole will occur in rtionl function if n input cuses oth the numertor nd denomintor to oth e zero.
3.4 Rtionl Functions 3 To find horizontl symptotes, we re interested in the ehvior of the function s the input grows lrge, so we consider long run ehvior of the numertor nd denomintor seprtely. Recll tht polynomil s long run ehvior will mirror tht of the leding term. Likewise, rtionl function s long run ehvior will mirror tht of the rtio of the leding terms of the numertor nd denomintor functions. There re three distinct outcomes when this nlysis is done: Cse : The degree of the denomintor > degree of the numertor 3 + Emple: + 4 5 3 3 In this cse, the long run ehvior is. This tells us tht s the inputs grow 3 lrge, this function will ehve similrly to the function. As the inputs grow lrge, the outputs will pproch zero, resulting in horizontl symptote t f ( ) 0. As ±, 0 Cse : The degree of the denomintor < degree of the numertor 3 + Emple: 5 3 In this cse, the long run ehvior is 3. This tells us tht s the inputs grow lrge, this function will ehve similrly to the function 3. As the inputs grow lrge, the outputs will grow nd not level off, so this grph hs no horizontl symptote. Insted, the grph will pproch the slnted line 3. As ±, f () ±, respectively. Ultimtely, if the numertor is lrger thn the denomintor, the long run ehvior of the grph will mimic the ehvior of the reduced long run ehvior frction. As nother 5 3 emple if we hd the function with long run ehvior + 3 5 3 4 3, the long run ehvior of the grph would look similr to tht of n even polynomil nd s ±, f (). Cse 3: The degree of the denomintor degree of the numertor 3 + Emple: + 4 5
4 Chpter 3 3 In this cse, the long run ehvior is f ) 3. This tells us tht s the inputs ( grow lrge, this function will ehve the similrly to the function 3, which is horizontl line. As ±, 3, resulting in horizontl symptote t f ( ) 3. Definition The horizontl symptote of rtionl function cn e determined y looking t the degrees of the numertor nd denomintor. Degree of denomintor > degree of numertor: Horizontl symptote t f ( ) 0 Degree of denomintor < degree of numertor: No horizontl symptote Degree of denomintor degree of numertor: Horizontl symptote t rtio of leding coefficients. Emple 6 In the sugr concentrtion prolem from erlier, we creted the eqution 5 + t C( t). 00 + 0t Find the horizontl symptote nd interpret it in contet of the scenrio. Both the numertor nd denomintor re liner (degree ), so since the degrees re equl, there will e horizontl symptote t the rtio of the leding coefficients. In the numertor, the leding term is t, with coefficient. In the denomintor, the leding term is 0t, with coefficient 0. The horizontl symptote will e t the rtio of these vlues: As ±, f ( ). 0. This function will hve horizontl symptote t 0 This tells us tht s the input gets lrge, the output vlues will pproch /0. In contet, this mens tht s more time goes y, the concentrtion of sugr in the tnk will pproch one tenth of pound of sugr per gllon of wter or /0 pounds per gllon. Emple 7 Find the horizontl nd verticl symptotes of the function ( )( + 3) ( )( + )( 5) The function will hve verticl symptotes when the denomintor is zero cusing the function to e undefined. The denomintor will e zero t, -, nd 5, indicting verticl symptotes t these vlues.
3.4 Rtionl Functions 5 The numertor is degree, while the denomintor is degree 3. Since the degree of the denomintor is greter thn the degree of the numertor, the denomintor will grow fster thn the numertor, cusing the outputs to tend towrds zero s the inputs get lrge, nd so s ±, 0. This function will hve horizontl symptote t f ( ) 0. Try it Now 3. Find the verticl nd horizontl symptotes of the function ( )( + ) ( )( + 3) Intercepts As with ll functions, rtionl function will hve verticl intercept when the input is zero, if the function is defined t zero. It is possile for rtionl function to not hve verticl intercept if the function is undefined t zero. Likewise, rtionl function will hve horizontl intercepts t the inputs tht cuse the output to e zero. It is possile there re no horizontl intercepts. Since frction is only equl to zero when the numertor is zero, horizontl intercepts will occur when the numertor of the rtionl function is equl to zero. Emple 8 Find the intercepts of ( )( + 3) ( )( + )( 5) We cn find the verticl intercept y evluting the function t zero (0 )(0 + 3) 6 3 f ( 0) (0 )(0 + )(0 5) 0 5 The horizontl intercepts will occur when the function is equl to zero: ( )( + 3) 0 This is equivlent to when the numertor is zero ( )( + )( 5) 0 ( )( + 3), 3 Try it Now 4. Given the reciprocl squred function tht is shifted right 3 units nd down 4 units. Write this s rtionl function nd find the horizontl nd verticl intercepts nd the horizontl nd verticl symptotes.
6 Chpter 3 From the previous emple, you proly noticed tht the numertor of rtionl function revels the horizontl intercepts of the grph, while the denomintor revels the verticl symptotes of the grph. As with polynomils, fctors of the numertor my hve powers. Hppily, the effect on the shpe of the grph t those intercepts is the sme s we sw with polynomils. When fctors of the denomintor hve power, the ehvior t tht intercept will mirror one of the two toolkit reciprocl functions. We get this ehvior when the degree of the fctor in the denomintor is odd. The distinguishing chrcteristic is tht on one side of the verticl symptote the grph increses, nd on the other side the grph decreses. We get this ehvior when the degree of the fctor in the denomintor is even. The distinguishing chrcteristic is tht on oth sides of the verticl symptote the grph either increses or decreses. For emple, the grph of ( + ) ( 3) is shown here. ( + 3) ( ) At the horizontl intercept - corresponding to the ( +) fctor of the numertor, the grph ounces t the intercept, consistent with the qudrtic nture of the fctor. At the horizontl intercept 3 corresponding to the ( 3) fctor of the numertor, the grph psses through the is s we d epect from liner fctor. At the verticl symptote -3 corresponding to the ( + 3) fctor of the denomintor, the grph increses on oth sides of the symptote, consistent with the ehvior of the toolkit.
3.4 Rtionl Functions 7 At the verticl symptote corresponding to the ( ) fctor of the denomintor, the grph increses on the left side of the symptote nd decreses s the inputs pproch the symptote from the right side, consistent with the ehvior of the toolkit. Emple 9 ( + )( 3) Sketch grph of ( + ) ( ) We cn strt our sketch y finding intercepts nd symptotes. Evluting the function t zero gives the verticl intercept: (0 + )(0 3) f (0) 3 (0 + ) (0 ) Looking t when the numertor of the function is zero, we cn determine the grph will hve horizontl intercepts t - nd 3. At ech, the ehvior will e liner, with the grph pssing through the intercept. Looking t when the denomintor of the function is zero, we cn determine the grph will hve verticl symptotes t - nd. Finlly, the degree of denomintor is lrger thn the degree of the numertor, telling us this grph hs horizontl symptote t y 0. To sketch the grph, we might strt y plotting the three intercepts. Since the grph hs no horizontl intercepts etween the verticl symptotes, nd the verticl intercept is positive, we know the function must remin positive etween the symptotes, letting us fill in the middle portion of the grph. Since the fctor ssocited with the verticl symptote t - ws squred, we know the grph will hve the sme ehvior on oth sides of the symptote. Since the grph increses s the inputs pproch the symptote on the right, the grph will increse s the inputs pproch the symptote on the left s well. For the verticl symptote t, the fctor ws not squred, so the grph will hve opposite ehvior on either side of the symptote. After pssing through the horizontl intercepts, the grph will then level off towrds n output of zero, s indicted y the horizontl symptote.
8 Chpter 3 Try it Now ( + ) ( ) 5. Given the function, use the chrcteristics of polynomils ( ) ( 3) nd rtionl functions to descrie the ehvior nd sketch the function. Since rtionl function written in fctored form will hve horizontl intercept where ech fctor of the numertor is equl to zero, we cn form numertor tht will pss through set of horizontl intercepts y introducing corresponding set of fctors. Likewise since the function will hve verticl symptote where ech fctor of the denomintor is equl to zero, we cn form denomintor tht will ehiit the verticl symptotes y introducing corresponding set of fctors. Definition Writing rtionl functions from intercepts nd symptotes If rtionl function hs horizontl intercepts t,,,, nd verticl symptotes t m v, v,, v then the function cn e written in the form p p pn ( ) ( ) ( n ) q q qn ( v) ( v ) ( vm ) where the powers p i or q i on ech fctor cn e determined y the ehvior of the grph t the corresponding intercept or symptote, nd the stretch fctor cn e determined given vlue of the function other thn the horizontl intercept, or y the horizontl symptote if it is nonzero. n Emple 0 Write n eqution for the rtionl function grphed here. The grph ppers to hve horizontl intercepts t - nd 3. At oth, the grph psses through the intercept, suggesting liner fctors. The grph hs two verticl symptotes. The one t - seems to ehiit the sic ehvior similr to, with the grph incresing on one side nd decresing on the other. The symptote t is ehiiting ehvior similr to decresing on oth sides of the symptote., with the grph
3.4 Rtionl Functions 9 Utilizing this informtion indictes n eqution of the form ( + )( 3) ( + )( ) To find the stretch fctor, we cn use nother cler point on the grph, such s the verticl intercept (0,-) (0 + )(0 3) (0 + )(0 ) 6 4 8 4 6 3 4( + )( 3) This gives us finl eqution of 3( + )( ) IMPORTANT FEATURES OF THIS SECTION Inversely proportionl; Reciprocl toolkit function Inversely proportionl to the squre; Reciprocl squred toolkit function Horizontl Asymptotes Verticl Asymptotes Rtionl Functions Finding intercepts, symptotes, nd holes. Given eqution sketch the grph Identifying the function from grph Try it Now Answers. Long run ehvior, s ±, 0 Short run ehvior, s 0, f () (there re no horizontl or verticl intercepts). The function nd the symptotes re shifted 3 units right nd 4 units down. As 3, f () nd s ±, 4
30 Chpter 3 3. Verticl symptotes t nd -3; horizontl symptote t y 4 4. For the trnsformed reciprocl squred function, we find the rtionl form. 4( 3) 4( 6 + 9) 4 + 4 35 4 ( 3) ( 3) ( 3)( 3) 6 + 9 Since the numertor is the sme degree s the denomintor we know tht s ±, 4. 4 is the horizontl symptote. Net, we set the denomintor equl to zero to find the verticl symptote t 3, ecuse s 3, f (). We set the numertor equl to 0 nd find the horizontl intercepts re t 35 (.5,0) nd (3.5,0), then we evlute t 0 nd the verticl intercept is t 0, 9 5. Horizontl symptote t y /. Verticl symptotes re t, nd 3. Verticl intercept t (0, 4/3), Horizontl intercepts (, 0) nd (-, 0) (-, 0) is doule zero nd the grph ounces off the is t this point. (, 0) is single zero nd crosses the is t this point.
3.5 Inverses nd Rdicl Functions 3 Section 3.5 Inverses nd Rdicl Functions In this section, we will eplore the inverses of polynomil nd rtionl functions, nd in prticulr the rdicl functions tht rise from finding the inverses of qudrtic functions. Emple A prolic trough wter runoff collector is uilt s shown elow. Find the surfce re of the wter in the trough s function of the depth of the wter. 3ft in 8 in Since it will e helpful to hve n eqution for the prolic cross sectionl shpe, we will impose coordinte system t the cross section, with mesured horizontlly nd y mesured verticlly, with the origin t the verte of the prol. y From this we find n eqution for the prolic shpe. Since we plced the origin t the verte of the prol, we know the eqution will hve form y ( ). Our eqution will need to pss through the point (6,8), from which we cn solve for the stretch fctor : 8 6 8 36 Our prolic cross section hs eqution y ( ) Since we re interested in the surfce re of the wter, we re interested in determining the width t the top of the wter s function of the wter depth. This is the inverse of the function we just determined. However notice tht the originl function is not oneto-one, nd indeed given ny output there re two inputs tht produce the sme output, one positive nd one negtive.
3 Chpter 3 To find n inverse, we cn restrict our originl function to limited domin on which it is one-to-one. In this cse, it mkes sense to restrict ourselves to positive vlues. On this domin, we cn find n inverse y solving for the input vrile: y y ± y This is not function s written. Since we re limiting ourselves to positive vlues, we eliminte the negtive solution, giving us the inverse function we re looking for ( y) y Since mesures from the center out, the entire width of the wter t the top will e. Since the trough is 3 feet (36 inches) long, the surfce re will then e 36(), or in terms of y: Are 7 7 y The previous emple illustrted two importnt things: ) When finding the inverse of qudrtic, we hve to limit ourselves to domin on which the function is one-to-one. ) The inverse of qudrtic function is squre root function. Both re toolkit functions nd different types of power functions. Functions involving roots re often clled rdicl functions. Emple Find the inverse of ( ) 3 4 + From the trnsformtion form of the eqution, we cn see the verte is t (,-3), nd tht it ehves like sic qudrtic. Since the grph will e decresing on one side of the verte, nd incresing on the other side, we cn restrict this function to domin on which it will e one-to-one y limiting the domin to. To find the inverse, we strt y writing the function in stndrd polynomil form, replcing the f() with simple vrile y. Since this is qudrtic eqution, we know tht to solve it for we will wnt to rrnge the eqution so tht it is equl to zero, which we cn do y sutrcting y from oth sides of the eqution. y 4 + 0 4 + y In this formt there is no esy wy to lgericlly put on one side & everything else on the other, ut we cn recll tht given sic qudrtic in stndrd form + + c we cn solve for y using the qudrtic formul