PRECALCULUS ALGEBRA MAC 1140
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Section 115 Binomial Theorem Binomial Coef cients Dennmon c e quot J 1 01 J Where n 1 new recursweer nl1 n71 n72 3 2 1 event hence Tm ahenw ThTs lsthE eehhmeh Remark The above hmaheh is read as h taken at a me ur n huusef The number h ewerem Ttems Ferexampie Tryeu have 52 mrrerem earns you wanna draw 5 a hand 52l 52 rrum them Then you could have c5 5m 2375040 msuhct possible Bumbl atl s Vuu can use thTs memud m calcula a e your chance at WThhThg a lottery What s your chance in become a mllllunalre DVEml hf Binomial Theorem n 1 l 1 gt17 1 n n oaba baba 0 J n 1 aquot1 1 aquotquotb 1 251 1 abH 1 15quot 0 1 J mil 1 7 217151 n J Home The eperaueh h the paremhesTs must be regarded as summaheh EVEN We Want b m apahu a 75quot Exerclse1 1152aPTlEvaluate the binary coef cient g gt o 8 Exercise 2 1152aPTEvaluate the binary coef cient g 00 05 0120 01 010 Exercise 3 1151aPTFind the coef cient 0f 375 in the expansion of 211 11 5 0 g 10 11 0 M432 1391 5 0 7m 11 5 ow o 1 Exercise 4 1151aPT Find the coef cient of m4 in the expansion of 3 m J 101 6 o m3 0 10 6 o m3 it 6 o 39T 11 1 0 m3 Section 36 Complex Zeros of REAL Polynomials Must understand 34 and 35 first Recollection of 34 and 35 Section 34 Eyery real polynomlal l e polynomlal Wltn real uefflclents can he completely tactoreo lnto ractors that are eltnerllneartactors ufdegree l orlrreooclhle he tactoreo lnto REAL llnear tactors l e lts olscrlmlnant ls NEGAT v Section 35 Eyery real ooaoratlc tonctlon whose olscrlmlnant ls NEGAGWE lE lrreooclhle has a palr of conlogate zeros TWO complex zeros that are conlogates of each other he complex zeros of a REAL polyrlomlal come frum those lrreouclhle quadratlc facturs And they show up lrl palrs Fur each palr the Wu complex numbers are corllugate of the other 2 if a REAL polyrlomlal has a complex zero then the corllugate uf tnls zero must also he a zero ottllls polyrlomlal A unsequence of l Exercisei 3936t1aPTSelect the choice with ALL the remaining zeros of a poly nomial with real coe ioients and degree 5 having mos of s 4 e 3139 O l 25i O l l Bi 2 5i 0 none of these 0743l 25i 3 A REAL polynomlal of degree n most naye n zeros lt complex nomhers are allowed Thus lt could he tactoreo lnto n llneartactors lfyuu see a llnearractor llxe sz or Xrarbl where z ls a complex nomher NOT real then the tactoreo torm must naye anutherllnearfactur xrz or xahl A corollary of l and 2 Attention Pay attention to the signs Exercise 2 6t1bPTSelect the polynomial with real coef cients of degree 5 having zeros of 5 71i 39 Exponential Functions Exponential Functions Afunction of the form fx a where a is a positive real number such that a gt0 and a 1 The domain off is the set of all real numbers Why is a limited to positive numbers Whyis 1 excluded from being a possible base When a gt 1 Domain 7 0000 Range 0 00 Increasing 7 00 00 There are no xintercepts and the yintercept is The giaph alvmys contains the points 01 1 a an The xaxis yo is ahon39zontal asymptote When 0 lt a lt 1 Domain 7 0000 R 3 0 00 Decreasing 7 00 00 There are no xintercepts and the y intercept is 1 The graph aimys contains the points 01 and 121 The xaxis y0 is ahorizontal asymptote Two Special Bases When a 10 this is called the common base n a e this is called the natural base 0 e is named a er Leonard Euler e is a transcendental irrational number like 7 o The de nition of e is n 2 lim 13 a 2 718281827 new Section 31 Quadratic Functions Graphs l Quadratrc mnctron compieted square form contarns rnore rnrotnan tne otherform yaxrh2ka 0 i Vertex h k Examplei 311cPTFind the quadratic equation whose graph has vertex at 71 a and y intercept 74 0 y a2a12 2 0 y 21 12 2 76141 2 If y g22 1 0 y6x7122 Example 2 12dFTFind the vertex of the graph 0171 1E3223fa U o 2 3 o 32 o 33 0 3v 2 0 3 2 2 Axis of Symmetry rstne verticai nnetnrougn thevenex xh 3 Tne paraboia opens a Up r agt0 vertex rstne rnrnrrnurn pornt xr 2 2 la 2 4 n r 2 3 A 5 b V Down if alt0 vertex is the maximum point REMARK To graph a quadratic function DO 13 and 4 if necessary Exercise 3 312 PTJ 1 2 a lt 0 has 0 O O O O P a maximum value of 2 a maximum value of 4 a minimum value of at a minimum value of 2 a maximum value of 2 Intercepts a b V yintercept set x0 and solve fory only one xintercept The number of xintercepts may be determined by considering the vertex and the direction the parabola opens To find the value of xintercepts set y0 and solve for x V Quadratic function polynomial form yeax bxca 0 1 i 1 Vertex wH 2a 2a Exemusea S12aPTThe graph of Le quadmtic function m 1 7 m 9 has 9 vertex at 36773 O vertex at 73 36 O vertex at 03 9 vertex at 730 9 vertex at 30 75 2 Axts at Symmetry me same as umptaed squaredfurm e xn2 a a Open mrectmn me same as befure REMARK To graph a quadratic function no 1a and 4 If necessary Examples 31 3bPTSe1ect the equation 0139 the following graph D11124r3571 0 J1 221 1 Exercise7 393L3cPTSelect the equation of the following graph Note c is a constant 9 y 74x251c 9 y417751c 9 y74139275139c 9 141751 c 4 meme 5 a yrintercept me same as bEfDrE n xrintercept by dmg nut n and N We can EasHy rewrite me puiynumiai furm win we cumpieieu square furm Exerclsea 3L1aPTThe gmph ofthe quadratic function 1 O 0 zintercepts 3910has 0 None of these 0 2 Iintercepts 0 1 Iiutercept b 7411 xrintercept ur by ndmg uutme discriminant Ab2 7 4a gtn TWO xrintemepts CROSSES Ab2 7411 u ONE xrimercem TOUCHES BUT DO NOT CROSS Which pmnt an we Xrast dues me graph much Abz 74a ltn NO xrintercepuDOES NOT TOUCH NOR CROSS Exerclses 311bPTTlie graph of the quadratic function has 0 n zintercepiis D 2 Antercepts D 1 zinterccpt Section 31 6 THINGS YOU SHOULD KNOW You can nd out the vertex of a quadratic function by using the formula given in class7 ie Or by completing the square Completing the square is equivalent to nding the vertex using the formula HOWEVER7 in most cases7 completing the square can save you much more time Here7 we show some most common forms7 for which you can complete the square swiftly If none of the common forms ts the expression you see7 use the formula instead fxz22z1x12h71k0 fzz272z1z712h1k0 More fz z2i4x4xi22 h2k0 fz z2i6z9zi32 h3k0 fz z2i8z 16 9amp4 h Izhk 0 Variants fz 28x15281671x42717 h747k1 fz 7x26x15 7x276x15 7x276x9159 7z73224 h 3k 24 Extra Credit Hint look at the last example to gure out how to deal with the leading coef cient a if it is not 1 1 f 2x2 4x 9 2 f 7z2 4x 7 5 3 fz 72952 1295 10 Section 35 Complex Numbers EXPONENTS AND RADICALS REQUIRED Definition 1 z39 1 1 i is the building block of complex numbers It handles the difficulty we usually have to take a square root of a negative real number i could be considered the same as x in the expressions However idoes have the following properties 0 139 z393z39 z392 11ori4 z3922 12 1 Remark The powers ofl repeat with EVERY 4th power Example 1 351aPT Write 290 i in a bi form 0 2 f o 2 i o 2 i o 2 o 2 395 Complex number standard form abi where a and b are real numbers a is called the real part B is called the imaginary part Treating i as x the algebraic operations for real numbers carry over I Equality abi cdi cgt ac and bd 0 Addition abicdi cgt acbdi O Subtration abicdi cgt acbdi O Multiplication abicdiacadibcibdacbdadbci The following operations are different from real numbers I Conjugate if zabi then the conjugate ofz is abi denoted by E Conjugate is to negate the imaginary part 2 Z 2a 2 E 2bz39 22 a2 b2 this trick is important in doing complex division Actually we can a understand ms as Extra Credit snuw tnanne abuve 6 ruenmres are me can We nterpra me ruenmy i a2 52 by usmg the mnerenee ufsquares runnura7 rpmranuwwrmen 1 W W 2 z a 5 Exampl 35 in the form 2 hi 1 O 2 O O smvrng ouaurau Equauuns wrtn NEGATWE Drsenrnrnants REEaH Ab2 74a gtn TWO xrmtertepts Abz 7441 n ONE xrmtemept 552741 ltn no xrmtemept r e nu rear smutmns Huwever aner We have tnaL runnura mtmduced by the quadrant rbtebz 74a 2a urnprex numbers We Knuw me quadrant functer wmn rear uemmems m ms ease wm nave a PAH er CONJUGATE urnprex numbers as We smutmns THEV ARE NOT XJNTERCEPTS THOUGH Remark Huw Intake square rent at a negatwe number 1716 47 IBM Examples 352aPT Select the real and complex zeros of f 212 6x 5 O 32 12i32 0 none of these 0 32 12i 32 O 32 3 fx1x fX1X El RahanalFuncums Funchanlnsumte Mathematics c mutants Hume This is probably die simplest ofrational functions 1 f x i 36 Here is how mis functionlooks on a giapii wiui an mm of10 10 and ay extentof1010 First notice die x and yaxes They are dam in red The mimicquot fx 1 x is dam in green I i A 39 A funquot m m 39 As we x i ii LILL39 Il L 982007 f4 fx1x will s only see this section ofme graph holds what is termed an a ymptote and computers along with rap 39e calculators otteu have a dif culttime drawing function eat asymptotes output alu 39 output value Here is a picture showing that Large osmve lnpul yields mall poslhve oulpul r will negative value Here is a picture showing this idea Large negalwe lnpul yields mall negalwe Small negalwe l npul yleldS large oulpul This makes eomplete sense ifyou think aboutlt for a moment Consider alarge 982007 f4 fX 1 X Page 3 of 4 positive input value say one million i e 1000000 The output of fX 1X would be one millionth i e 11000000 or 0000001 This and other representative examples are shown in the following table Input value or X Output value or y 1000000 0000001 or 11000000 0000001 or 11000000 1000000 4000000 0000001 or 11000000 0000001 or 11000000 1000000 Let us look at this function as it leaves the graph at the top and bottom You should notice that the green function line approaches but does not touch the y aXis If you graphed the function on a set of X y axes that went up to positive one million and down to negative one million the function line would still not touch the yaXis though it would get very close Just think about the X y values in the table above At say an output value or y value of 1000000 the input would not be 0 And of course the input value or X value must be 0 for the graph to touch the yaXis This type of behavior about the yaXis is called asymptotic behavior And in this case the yaXis would be called a vertical asymptote of the function That is the function approaches the yaXis ever closer and closer but never touches it Notice that the XaXis functions as a horizontal asymptote for this function That is as the function line stretches out to the left or right it gets closer and closer to the XaXis but it never touches it So for the function fX 1X the yaXis is a vertical asymptote and the XaXis is a horizontal asymptote In the following diagram of this function the asymptotes are drawn as white lines httpidmindnetNzonammtsfunctionTnstituterationalFunctionsoneOv 982007 fx1x A A J polynomial although both are quite simple polynomials Be sure that you Page 4 Bf 4 and m u go on to the other rational function information Back Rational Functions Function Institute Mathematics Contents Index Email Home 982007 Section 114 Mathematical Induction Statement Sequence rennuia paiametenzeu bythe inuex n of st equence Example atements 11121221231331234 04 25 gt 2 2 2 i 2 Here S 2 2101 i is tne statement era general index n i Note in tne above Example tne ntn statement is Essennaiiy eiaiming a rermuia rer tne rinite sum er a paitieuiar antnmetie seuuenee 1234 gt Whuse rirst tem is i anu wnese eemmen uirrerenee is i How do we prove that the an of the statements in the sequence are true i Verity tne Validity er eaen statement it is usually easy te verify tne rirst a eeupie er statements auttnere are infinitely many statements To prove tne Validity er tne rirlh statement DlRECTLV But in mest cases tnis is pruhibitively naiu Principles of Mathematical Induction o Showthefirststatemert Sims true Easic step e snew tnat lF Sh is true for a paitieuiar xi THEN SM tne neltt statement must aise bEtmE Thatis shim Sh 2S lnductlve step wlth any given Index n Exerc 11 41mmth 112 and at such hat 1 In A a In or an n 004224301327 06258329 0 None of these 0 12 5613 Exercise 2 1142 sPTTo prove by induction that Q 7 5 11 2n 1071 n2 is true for all positive integers n we assume 9 7 5 11 21 101 1272 is true for some positive integer k and show that 97 l 5 l 11 2139 l A1014 l I39c12 where A is 013 2 09 21 010 ri Ol2 2k 0 None of these Exercise 3 1142bPT T0 prove by induction that 10 7 4 13 3n 152371 3712 is true for all positive integers n we assume 10 7 4 13 316 23k 3k2 is true for some positive integer k and Show that 10 T4 13 3k 13 3k l 1 A where A is 0 23k 1 312 1 01 2301 1 3112 J l 015 230 1 31 1v 015 23k 1 30 1 1 0152311 31k 1 Exercise 4 1143aPTjT0 prove by induction that 712 5n 2 is divisible by 2 is true for all positive integers n we assume k2 5k 2 is divisible by 2 is true for some positive integer k and we show that A is divisible by 2 Where A is okg Sk 21 019239 i 5k1 2 0 None of these OIk12 5k1 2 Ok12 5ik1 21 Exercise 5 1143bPTT0 prove by induction that n2 7n 4 is divisible by 2 is true for all positive integers n we assume k2 7k 4 is divisible by 2 is true for some positive integer k and we Show that k2 7k 4 A is divisible by 2 where A is 0 None of these 0 209 2 o 2U 1 O 2k3 o 2lk 3 Fulynumial Lens Divisiun Page 1 Df 10 Algebra Solver Sm m Ranam mm Polynomial Long Division An Exmle In his sectinn yuu will learn haw a rewrite a ratinnal functinn such as 3112124173 12 31 3 in the farm 2 2 3117223531037 3 31 7 11 The expressinn 31 e 11 is called the autism he expressinn 12 31 3 is called the divilnr and the term 281 30 is called the x ndzr What 5 special abuut the way he expressinn abuve is written The remainder ZShSU has degree 1 and is thus less than the degree nf he divisnr 12 31 3 It is always pussible a rewrite a ratinnal functinn in this manner DIVISION ALGORITHI If Ar and dz 0 are pelynemials and the degree Bf dx is less than Dr equal a the degree Bf httpwww snsmath cumailgebrafacnrfacUlfacUl html 8282007 Fulynumia Lung Divisieh Page 2 Df 10 Ar then there exist unique pnlynumials qx and rx se that m NH rz 41 W and sn that the degree nf rx is less than the degree nf d x In the special case where rtr0 we say that dx 39vidu even1y intu Ar Haw dn ynu dn this Let s lnuk at nur example 31121Z413 1 31 3 in mere detail Write the expressinn in a farm reminiscent nf lung divisinn 1 27 12 31 3 31 21 41 3 First divide the leading term 3139 Df the numeratnr pnlynnmial by the leading term 12 uf the div39 Dr and write the answer 3r Dn the tap line 31 1 27 12 313 31 21 41 3 an multiply this term 3r by the divisur 12 31 3 and write the answer 311Z 31 3 31 912 91 under the numeratnr pnlynnmial lining up terms nf Equal dEgree 31 1 27 12 31 3 31 21 41 3 311 912 91 Next subtract the last line frum the line abnve it 31 1Z313 31 212 41 3 311 912 91 1112 51 3 httpwwwsesmathcuma1gebrafacterfacolfacolhtml 8282007 Pulynumia Lung Diyisiun Page 3 uf 10 New repeat the prucedure Divide the leading term 71112 nf the pulynumial en the last line by the leading term 12 uf the diyisur tu ubtain all and add this term tn the 3x en the tap line 31 1Z313 31 7212 41 e3 311 912 91 71112 751 7 Then multiply back 71112 31 3 1112 7 331 a 33 and write the answer under the last line pulynumial lining up terms uf equal degree 31 71 1Z313l31 7212 41 3 31 711 1 27 1Z313 31 e1 41 e3 311 912 91 27 111 751 73 71112 7331 733 231 30 nu are dune In the next 51810 Van wnuld divide 281 by 2 nut yielding a Dulynnmial expressinn The remainder is the last line 28x60 and he quntient is he expressinn an the very up 3r11 Cnnsmuentlyr 311 e 212 41 e 231 30 317 11 7 1Z313 12313 Hm tn check yuuz answer The easiest way tu check yuur answer algebraically is tu multiply butn sides by the isur diy 31 e 212 41 e 3 31 e 111Z 31 3 231 30 httpwww susmath cuma1gebrafacturfacUlfacol ntml 8282007 Pulynumia Lung Divisiun Page 4 cf 10 then tn multiply nut 31 s 2124zs 3 3z 9z29zs 1112s 33 3328z30 and then tn simplify the right side 1 2 z 2 31 an 4zs3 31 an 4zs3 Indeed bath sides are eq 1 Other ways f checking include graphing bath sides ua e if ynu have a graphing calculatur Dr plugging in a few numbers en buth sides this is nut always 100 funlpruuf Anathar Exllwls Letls use pulynumial lung divisiun tn rewrite z s 1 z 2 39 Write the expressinn in a farm reminiscent uf lung divisiun 17 z 2 1 s1 First divide the leading term 1 cf the numeratnr pulynumial by the leading term 1 cf the divisur and write the answer 12 en the tap line 12 17 z 2 1 s1 Nuw multiply this term 12 by the divisur rtz and write the answer 12z 2 11 212 under the numeratur pulynunial carefully lining up terms uf equal degree zz 12 1 1 212 Next subtract the last line frm the line abuve it httpwww susnath cuma1gebrafacturfacUlfacol html 8282007 Fulynumia Lung Divisien Page 5 ef 10 2 an 1 2 71 212 7212 71 a New repeat the prncedure Divide the leading term 7212 Df he pulynumial en the last line by the leading term 1 ef the divisnr a ubtain 72x and add his term a z 1 line an the up 12 721 1 z 2 z 71 1 212 27 721 71 Then multiply back 7241 2 7212 r 41 and write the answer under the last line pelynemial lining up terms ef equal degree 1 721 1 2 z 71 1 212 7212 71 7212 74 Yeu have a repeat the prncedure Dne mere time Divide httpwww susmath cuma1gebrafacnrfacUlfac l nuul 8282007 Pelynemia Lung Divisieh 12 21 4 e2 1 1 1 212 7M 1 4 Multiply back 1 21 4 1 2 1 1 1 212 212 1 212 41 41 1 41 8 and subtract 12 21 4 1 2 1 1 1 212 212 1 212 41 41 1 41 8 9 e are dune In the next step we wuuld divide x pulynumial expressinn The remainder is the last me e9 quetieht is the expressinn en the very tep 12 e 2 e e 1 2 e m z 214z2 79 by 1 het yielding a ef degree 0 CDnsEquent 1y Page 6 ef 10 and the An Exmle Len Polynomial Divisinn and Factorial Let s use pelynemial lung divisieh te rewrite httpwww sesmath cuma1gebrafacterfacUlfacol html 8282007 Fulynunial Lung Diyisiun Page 7 uf 10 751231715 123 39 Write the expressinn in a farm reminiscent uf lung divisiun 27 12 3 1 esz 31 715 rm 1 uf tne nuneratur pulynunial by the leading term 12 39ne First divide the leadin h g te cf the diyissr and write the answer 1 an the up 1 1 27 12 3 1 esz 31 715 NEW multiply this term 1 by the diyisur z2 3 and write the answer 2 1 11 3 1 31 under the numeratnr pslynsnial carefully lining up terns uf equal degree 1 zz3 z 7512 31 715 1 31 Next subtract the last line frm the line abuye it 1 zz3 z 7512 31 715 1 31 esz 715 New repeat the prucedure Divide the leading term 7512 Df the pulynunial en the last line by the tn 1 leading term 12 uf the diyisur tu ubtain 5 and add this term tu en the tap line z 75 12 3 z 7512 31 715 1 31 751 715 Then multiply back 7512 3 7512 e 15 and write the answer under the last line pulynunial lining up terns uf equal degree httpwww susnath cuma1gebrafacturfacUlfacol htnl 8282007 Fulynumia Lens Divisiun Page 8 Df 10 z 5 zz3 z 7512 31 715 I 75z2 715 z 31 512 715 Subtract the last line frum the line abnve it 1 75 zz3 z 7512 31 715 1 31 7512 715 7512 715 0 Ynu are dune In his case the remainder is 0 se 12 3 divides evenly intu 11751231715 CDnsEquent 1y eszz3zels 0 5 z Haulers z e 5 Multiplying bath sides by he divisur 2 3 yields 1 7 512 31 e 15 z e 5cZ 3 In his case we have incurred the pulynnmial z e 512 31 e 15 ie we have written it as a prnduct uf Wu easier lnwer degree pnlynnmials Exercise 1 Use lens pulynumial divisiun a rewrite httpwww susmath cuma1gebrafacnrfacUlfac l html 8282007 Fulynumia Lens Divisieh Page 9 Df 10 Exercise 2 Use lens pulynumial divisieh a rewrite Mr Exercise 3 Use lung pnlynumial divisiun a rewrite 12 e 71 4 z e 512 e l m Ennis 4 Use lens pulynumial divisieh a rewrite Anna Exercise 5 Use lung pnlynumial divisiun a rewrite z 7 12 1 Back Next Algebra Trigunumetry Cemlex Variables Calculus Differential E uatiehs Matrix Algebra g S 0 S MATHematics hams DD yuu heed mare help Please past ynur Questiun Dn eur S 0 S Mathematics CyberBuard httpwww susmath cuma1gebrafacterfacUlfacol html 8282007 Section 37 Rational Functions Must understand 33 Come back to real numbers Rational Function 1 x Where Fgtlt 3 gtlt are Wu F lv mlals 406 2 Domain The demam offgtlt defined above lSiHE setofall real X such that qx 0 5e ifthe qgtlt does not have a zero then the demam is all real numbers Exercisel 71nPTClmoso the domain of the rational function fr so 3 U 3 3 U 3 DC 0073 U 730 U 03 U 390 00 no 3 4amp0 U 0130 0045 L 31u13 u 300 Exerc39 2 se 37i1bPTChoosc the domain of Lhe rational function shown below f ham 3 U 400 79073 u 73 0 u 14 U 400 734 00 Do 70073u 7311U400 1 3 Recall Inverse Variation fx7 x Note that the lines never cross the axes W they get closer and closerto gtlt 0 and y 0 but x and y never equal zero See the supplementary material formore details Vertical Asymptote and Behavior near the Zeros of the Denominator The function blows up shooting vertically up or down hearthe zeros orthe denominator x lust as inverse variation function does For example if fx P and r is a zero of q x X Cl i Xr but nevertouches it Such vertical line is called vertical asymptote lll le More formally the line ofxr is a vertical asymptote orthe graph of x it fxgtoo 85 X gt 7 Remark A rational function could have many vertical asymptotes ohe zero of om correspond to one vertical asymptote Exercise 3 3937 2aPTSelecc the choice having ALL the vertical asymptotes of Lhe function fz 2 0 y21 121392 2 122 1392tlx l y2 Exercised 3TAlaPTSnluct the graph of y 21 I z x 1 m4 1 7 2er Horizontal Asymptote p0 rm the degree at pgtlt s m me degree at em 5 n Tnen wnen mSn me Let fx grapn mm W have a hunzunta asympmte o r mltn n e fx 5 pruper men yn s the hunzunta asymptute o r nm men y s the hunzunta asymmme wnere an s the Emma nean me swgn A s the Emma enemment er qx uemment er pgtlt nean the sgn o otnenmse NO hunzunta asympmtes Remark A ratmna functer nas at must une hunzunta asymptute Exerc s 37l2bPTSelect the choice having ALL the horizontal asymptotes of Hz 1 7 1 No horizontal asymptote y 11 y 1 1 41 0r 4 391 1 y 0 Oblique Asym ptote Let fx F8 the degree gr pgtlt is m the degree ufqgtlt is n Tnen Wnen m n1 q 6 men there is an ubllque asyrnpmle Which is the QUOTlENT cumputed by lung dmsmn Otherwise NO ubllque asymptute 3733T T1Fiud lrhu ohliquv asymptote 01 2L1 y 2 7 2 No oblique asymptote Extra Credit Use the behavmr gr inverse vanaudn funmun near xu m Explain the behavmr gr 3 general ratmnal functer near the zerds gr us denummatur o Wnen mltn yE lsthe hurlzuntal asyrnpmle a o Wnen mn y lsthe hurlzuntalasymptute o Wnen rnnmne gunmenldr Fm lsthE ubhque asymptute Hlnt Put Fm lnm W W tne farm pm uatn2nrgtc Wnat s tne degree uf guuuenux Wnat tne W W 1x ends behavmr gr 7 W 0 Otherwise nu hurlzuntal asymptute nurubhque asymptute Section 43 Logarithmic Functions ALERT Based on 41 42 Logarithms 1 De nmon y 1ognx read as y is iuganmm in we base a erx a x a man a gt 011 1 Piuggmg me rst in me secund we have paw 1n anumer wum intuitiveiy lognb gives me puwer a needs in be raise 1e 1e umam x 0 Basic prupemes 10g U because a logna because a log aquotm because a ahS39m x 1 by aefmmon sub e but mosnmpomrm o Cummun iug logx logn x o Naturai iug 1m 1og x 1 Where e 2 7182818284 59045 23536 Example1 4319PTIf 15 41hen Ologa 4 Ologa4 Olog4a O 10g4m O loga4 1 010g45n Example 2 I131cPTlog 5 0 0 5 0 up 0 En ma 0 Examples 0711 i o a 1 On function Loga hmic Func n and Exponen We Knuw that ylognx xa39v se w fxlognx men w 5 me mverse er fx A e fquotx gx ax That s m saymatme uga mmmmnctmn and Then we reeau rrum Serum 4 me re atmn between a functer and us mverse o The Dumamuf y foam sme Range er ylognx 0m 5 me Dumam at y 10g o The Range at y and ms mverse lognoz Exercised 43 10PTThe domain of at 11731 7 1 is l O 00 I O E 00 39l O i oo i l 0 00 o The graph of y ax and y logax are symmetric about the line yx for any agt1a 1 y 1nx and y ex are the inverse function of each other Properties The graph of fx loga x a gt1 looks similar to fx1nx o Strictly increasing ie if x1 gtxZ then a gt a o y a 00 as x a 00 o y gt 700 as x gt 0 Le x0 is a vertical asymptote o The xintercept is l The graph contains the points 10 and a l o y 1nx and y ex are the inverse function of each other 1 x y loglZx and y are the inverse function of each other lt x logmx 4 39Y12X Properties The graph of fx loga x0 lt alt 12 looks similar to fx10gnzx o Strictly decreasing ie if x1 gtxz then a lt a o y gt 00 as x gt 0 Le y0 is a vertical asymptote 0 y gt 700 as x a 00 The xintercept is 1 The graph contains the points 10 and a1 1 x O yloglZx and y3 are the inverse function ofeach other Exercise 5 432aMSPTSelect ALL the correct equations for the given graph i B y loga ar0 lt a lt1 D N one of these E y loga a gt 1 Section 43 Logarithmic Function Inverse of Exponential Functions Recall an exponential is de ned as fxax agt0 ail The inverse function is defined implicitly as xayagt0a l Logarithmic Functions The logarithmic function to the base a where a gt 0 and 5 1 is denoted by ylogdx read as 31 is the logarithm to the base a ofx and is de ned by y loga x if and onlyif x ay Two Special Bases 39 a 10 7 log is also WIitten as log 7 Common Log 39 a e 7 logy is also WIitten as In 7 Natural Log When a gt 1 Domain 000 Range OO 00 Increasing 0 00 There are no yintercepts andthe xintercept is 1 e ys contains the points 10 and a1 The yaxis x0 is avenical asymptote When 0 lt a lt 1 Domain 000 Range 00 00 Decreasing 0 00 There are no yintercepts andthe xintercept is 1 e ys contains the points 10 and a1 The yaxis x0 is avenical asymptote Domain of Logarithmic Functions Logarithmic mctions will become the third item on our list of functions we need to check 7 For a logarithmic function to be de ned the argument must be bigger than zero 7 Why Transformations Vertical Shift c gt O 0 y fx c Shift f up 6 units Add 6 to the y coordinates y fx c Shift f down 6 units Subtract c from the y coordinates Horizontal Shift c gt O y fx c Shift f left 6 units Subtract c from the x coordinates y fx 0 Shift f right 6 units Add 6 to the x coordinates Vertical Stretch or Compression y kf13 If k gt 1 it is a vertical stretch and the graph is stretched vertically away from the x axis If 0 lt k lt 1 it is a vertical compression and the graph is compressed vertically toward the x axis Multiply the y coordinate by k Horizontal Stretch or Compression y fUm If k gt 1 it is a horizontal compression and the graph is compressed horizon tally toward the y axis If 0 lt k lt 1 it is a horizontal stretch and the graph is stretched horizontally away from the y axis Multiply the x coordinate by 1k Reflection about x axis y f13 Multiply the y coordinate by 1 Reflect about the x axis Reflection about y axis y f 13 Multiply the x coordinate by 1 Reflect about the y axis Unifying Vertical and Horizontal Shifts a j outside the original function ex pression f13 corresponds to up positive y direction corresponds to down negative y direction 0 j inside the original function expres sion f13 corresponds to left neg ative x direction corresponds to right positive x direction o Aren t they confusing YES and NO Here we introduce a view point to unify vertical shifts and horizonal shifts We call them general shifts For a function regard it as an equa tion y fx the graph of y yofx xoxo gtOyo gt O shifts the original one right by 5130 positive a direction up by yo positive y direction Extra Credit 0 How should we shift the graph ofy f13 to obtain that of y I yo f13 5130IZO gt Oy0 gt O o Is the method in this section con sistent with the first two sections Why Check 0 Does the same approach analogously applies to reflections How Present your own theory unify reflection about x axis and y axis Hint do some thing to the sign outside f13 Section 105 Matrix Algebra Equality Two m by n matrices A and B are said to be equal written as A B provided that each entry aU in A is equal to the corresponding entry by in B AdditionSubtraction To add or subtract two matrices the matrices must be the same dimension and you addsubtract the corresponding entries Zero Matrix A matrix whose entries are all zero Scalar Multiplication When multiplying a matrix by a real number often referred to as a scalar multiply each entry of the matrix by the scalar Matrix Multiplication Starting with an n x rand an rx m matrix Multiply each row i of the rst matrix by each column j of the second matrix according to the following rule 1 2 a 392 mi mwzc quotWCW cw The result goes in theimjM position of the new matrix Section 33 Polynomial Functions and Models De nition A golmomial nctt on is a function ofthe fonn fx anx HIM a1xl 11 a1a are real numbers where await coef cients d n is anonnegative integer The degree ofthe polynomial is the highest power ofthe variable Real Roots Real Zeros and X intercepts These are all names for the same thing 7 Find the xintercepts roots zeros of x xZ 2x 3 x 1 3 7 Imagine going backwards if you have x2 3 as the roots zeros intercepts of a polynomial then list the factoIs x2 and x 3 x3 Rule for Roots If a real number r is a root ofa functionf 7 x ris afactoroff 7 and in mction notation fr 0 Multiplicity Ifa factor X r is repeated en r is called a multiple zero of the function fx XlZX33 7 xl is repeated twice thus 1 has a multiplicity of two 7 X3 is repeated three times thus 3 has a multiplicng of three Multiplicity and Graphs Ifr is a zero of even multiglicigy the graph touches but does not cross the Xaxis at r Ifr is a zero of odd multiglicigg the graph crosses the Xaxis at r Section 34 The Real Zeros of a Polynomial Function Division Algorithm for Polynomials amp gow or x qltxgltxgt rm gm gm fis the dividend g is the divisor q is the quotient r is the remainder Polynomial Long Division Step 1 Fill in any missing terms in the dividend or divisor Step 2 Divide the rst term of the dividend by the rst term of the divisor 7 This Will indicate Where to begin Step 3 Follow the typical steps in long division multiply and subtract 7 Repeat steps 2 and 3 until you get a emainder or zero Remainder Theorem Letfbe a polynomial function lffx is divided byx c then the remainder isfc Example If fx 3x34 is divided byx 3 then the remainder is 7 its 3334 85 Factor Theorem lffc 0 thenx c is a factor of fx lfx c is a factor offx thenfc 0 fx74x3 5962 8 c72 fez 60 x 0 With c 2 is X c a factor off 0 Therefore X 2 x2 is not a factor of f Number and Bound of Zeros A polynomial cannot have more zeros than its degree If fxX anrlxn391 alx a0 then whichever of the following two numbers is the smallest provides the bound for the real zeros of a polynomial M1 Maxl aolla1 an71 MZ 1Max a0 a1 anrl Rational Zeros Theorem 0 Let f be a polynomial function of degree 1 or higher of the form fxanxquot aHx39 1 alx1 aoarl 0 a0 0 Where each coefficient is an integer prq in lowest terms is a rational zero of f then 19 must be a factor of a0 and q must be a factor of an Factoring Polynomials 0 Every polynomial function with real coef cients can be uniquely factored into a product of linear factors andor irreducible quadratic factors 7 A polynomial function with real coefficients of odd degree has at least one real zero Intermediate Value Theorem 0 Let f denote a continuous function If a lt b and if at and fb are of opposite sign then f has at least one zero between a and b F Section 31 Power Functions SEC 25 NEEDED G raphs 1 Power function of degree n basic form not transformed y fx ax a 0 n must be an integer positive whole number When n is EVEN I I r l I l I i I l I I I I I I l I I l I 1 yfxgt0 forallx The graph is symmetric about the yaxis 2 3 The graph always contains 00 11 11 4 The graph TOUCHes the x axis only at the origin 00 When n is ODD when agt0 y fx gt 0 for all xgt0 y fx lt 0 for all xlt0 when alt0 y fx lt 0 for all xgt0 y fx gt 0 for all xlt0 2 The graph is symmetric about the origin 3 The graph always contains 00 11 11 4 The graph CROSSes the xaxis only at the origin 00 Transformations of Power Functions SEE SEC 25 General forms ofthe power functions of degree n y fx ax7h ka 0 Doesn t it look very like a quadratic function What s different here Procedures for graphing For EVEN power the same method for graphing a quadratic function carries over However in general for both EVEN and ODD power we should use the following procedures Graph the basic form y l a l x 2 Re ect it about XaXis if alt0 SEE SEC 25 DO STEP 2 BEFORE YOU SHIFT THE GRAPH 3 Shi it vertically and horizontally SEE SEC 25 Example 1 32 1bPTSelect the equation of Lhe following graph o J5393 15 o y r15 o yw1 o y E15 Exemusez 321aPTSelect the graph ofy 7139s 7 2 Section 93 Hyperbola Pay Close Attention to the Difference from Ellipse Application Definition Given any POSITIVE constant 2a and two xed puints F1 F9 fed the set Luf all points P such that ldtP F jI MP F2 2a is called a hymnbola Note dL l Bl denutos the 1 mice h uln 4 to B MAJ 13 13952 I yA yglg dPF1 EIIPFgl Ea 1 dip F11 diPFgl i252 my r1r 2 The line containing the fuci 1711 is called thre trmlsvorsc axis The midpoint of the line segment joining the foci is called the armor of the lijTerbola The line through the center and perpendicular to the transverse axis is called the cm jugatc axis The hyptrhola consists of two separate Cul V39E39S called branches The 2 points where the hyperbole intersects the transverse axis are the Vern395 139 Iquot of the hyperlmla Two Fundamental Forms of Hyperbolas Pay attention to the differences and similarities 1 The transverse aXlS Isthe xaxls andme center 5 me ungm Aquot mumm 3 m hnmlmln r Hanna39s mij is Luv IALW m center is at m ongm Tim on m a Thu ernccs are u Iup 4 n Ls mm of m lisrzmw Ivunrwu up wruccs I Jctvnnines hemlnpv Ur n 101eran 5 as39mplole t in HALF um ihhmrr hrth mu m lem r b1 a1 2 Thetransverse aXIS Is the yaxls and me center 5 me ungm Y 4 equation uf the hvperbola whom namvaxse am Is the yraxjs The sum is at me ongm The fuel are n Uri and F2 u ey U a The BzLiCBs are v are and I There am We OELIQUE asympmms y i where a lt c u a is HALF of me distance between ehe elncas b detmmmEs ma slap of ma hwmhole s aspnptotas c is HALF af elm distance hemeen my fem Then 8 b2 e2 ATTN a n and furme nypemma have mvrerem re auun as a n and rurme empse Extra Credit HEW can yuu EU a mm m gwen me Equatmn at an empse7 Hm can yuu em a 7mm u gwen me Equatmn ur a hyperbma DD 3 n and denute me same an5 M me empse w a ewpse am u may have umerent meamngs Example1 9413PTSclcct the graph of o Exemlsez 94 1bPTSelect the equation of the following graph 1 09 16 g 09 1671 gg DIS 971 1 016 J Exercised 9 4 2nPTSeleuL he equmm or the hyperbola Wm center at 0 0 tow at 730 and venex at 70 u 1 Exercised 942bPTSclcct the foci of the hyperbola given by 1 7 3L 1 O 170 O 1140 o 157 O 014 0 None of these Exercises 9 42rPTFer ma aqympmrm of m hypmhnla gwm by g 7 H oyi I O 0 II II H w W W a m s 8 H H Growth and Decay Uninhibited Growth of Cells Nz Noe k gt o ND is the initial number of cells k isthe growth rate Uninhibited Radioactive ecay AzA0eh k gt o AEl is the initial amount of radioactive material k is a negative number representing the decay rate Section 101 Systems of Linear Equations in Two Variables Geometric Meaning of Two Linear Equations in Two Variables 1 Reminder of linear equation in two variables A linear equation containing two variables x and y is an equation of the form AxByC where A B and C The graph of such an equation is a LINE in xyspace 2 A system of two linear equations in two variables Thus the system of two linear equations containing the variables x and y is a pair of LlNEs in xyspace Each x y pair that satisfies the system of two equations must satisfy both equations ie the pair x y must be on both lines a ff Jquot quot v x r a Warming lines symm m Pamllal lungs syslnm c B lntiuem Ilnes Kirsten las mezsulutian has no milic has Imilnn aly Fumsulunms 3 Possible solutions of linear systems I Exactly ONE solution UNIQUE solution The solution is exactly the point where the two lines which the two equations represent intersect O NO solution This is the second case where two lines are parallel to each other There is no point that could be on both lines because parallel lines never intersect O lNFlNlTELY MANY solutions This is the third case where the two line overlaps Essentially this is to say that the two equations in the system are the same ATTN IN NO CASE can a linear system has exactly two solutions A system of equations is said to be consistent if the system has AT LEAST ONE solution If a system does not have a solution the system is said to be inconsistent Solving System of Equations by Elimination Exercise 1 Section 101 Systems of Linear Equations Two Equations Containing Two Variables The Idea Behind Solving Systems of Two Equations When we solve a single equation for a single variable 3x 8 we are solving for the values ofx which make the equation true When we solve a system oflinear equations we want the values of X and y which make the equations true x y 2 x7 y 0 The Graphical Interpretation For a single equation and single variable the solution is found on a number line For a system of linear equations with two variables the equations are lines and the solution is found in the Cartesian coordinate plane as an ordered pair 7 How many solutions could We have De nitions and the Number of Solutions If the lines intersect then the system of equations has one solution given by the point of intersection The system is consistent and the equations are independent If the lines are parallel then the system of equations has no solution because the lines never intersect The system is inconsistent De nitions and the Number of Solutions If the lines are coincident then the system of equations has in nitely many solutions represented by the totality of points on the line The system is consistent and the equations are dependent Method 1 Substitution Step 1 Solve one of the equations for one variable pick the easiest one Step 2 Substitute the solution into the other equation Step 3 Solve for the variable that is le Step 4 Substitute the solution into one of the equations and solve for the other variable Method 2 Elimination I Step 1 Line up the two equations so that the X s y s and constants are in their own column I Step 2 Identify a variable to eliminate by addition I Step 3 Multiply both sides of the equations by a number ifnecessary I Step 4 Add the equations and solve for the remaining variable 7 Why can we do this I Step 5 Substitute the answer into an equation and solve for the other variable Obtaining an Equivalent System of Equations I Interchange any two equations of the system I Multiply or divide each side of an equation by the same nonzero constant I Replace any equation in the system by the sum or difference of that equation and a nonzero multiple of any other equation in the system Method 3 Calculator I Step 1 Solve the equations for y I Step 2 Enter the equations in the 3 screen I Step 3 Graph the equations and adjust the window so that you can see the intersection if any I Step 4 Press CALC choose 5 7 The calculator will ask which of the two lines you want to nd the intersection of Since we only have two lines press ENTER twice Method 3 Calculator Step 5 Use the lef n39ght arrows to move the cursor to the suspected point of intersection press ENTER Section 47 Growth and Deday Based on 45 Exponential Growth and Decay Expuncuunl guAth l ACE l e u EVpuuLuilul decayquot 4 i l Note In the above A is amount 1 at I I Expunonllrd gruwlh nnd llolznv m slmn lnphlcnlly in tlw lollmnng nlmgmnle Exercise 1 3947 1aPTThe size P of a certain insect population at time t in days obeys the function Pt sow 2 Afmer how many days will line population reach 100039 Eli O in O a N E O E Me 0 Exercise 2 3947 1bPTThe population N in millions of a country may be approx imated by the formula Nt Noelquot If the population is 25 million millally and 5 million after 1 year whal will be the population after 5 ye 7 r 7 o 25n quoti O 35PMquot n o 359539quot 7 O 2565iquot Exercise 3 3947i1cPTIodine I 31 is at radioactive material that decays according to At A e 75 where A is the initial amount present and Ag is the amount present at time t in days What is the halflife of indine 317 Exercise 4 3947i1dPTThe voltage of a certain conductor decreases over time ac cording to the law of uninhibited decay V0 Vile k lt 0 If the initial voltage is 100 volts and 5 seconds later it is 20 volts then what is the voltage after 8 seconds Li 0 100g in 0 10000275 u I 0 100a M n 0 100m Exercise5 47ZaPTFind the exponential function Nt 39aekt that satis es the cunditiuns Mo 10 M5 2 15 in 3 O Nit miniE iv 1 0 Nu 154 O Nt mythQ5 O NU 10AM Section 113 Geometric Sequences Geometric Sequence De nition A geometric sequence x a sequence 11 WWEH saus es the ququg recurswe demnmun m 5e the sequence x essermaHy determmed by We parameters the we term a and the common a o r nth Term Formula Gwen the rstterm a and me eemmen ratm r the furmu afurthe new term 5 a MWquot e a m0quot a a2 maquot m d m Ex ercise 1 1131aPT1The 7 term of a ganmetric nquence with rst term a and common ratio T f 15 o 8 2 o c39l o Z 04 Exercise 2 a n term 0 a geomet sequence m m erm a 11 3 1bPT Th 1 f 39th H t 2 i and commun ratio r 7 0 None of these 2 Q 3 0 239 7 O 6 quot O Ziir 2 l O 3 Exercise 3 391132aPTH a geometric sequence has n 12 and a 13 what is the common ra i0 0 None of these 7 A O ia O E is 3 0 3 Sum of finite terms and infinite terms of a geometric sequence Finite Sum qaza3 a aararzarz m 1 o Swa1a2a3 ama m3 al IflrldOtherwise thesum 739 ofaiitermsisundefined Exercise 4 1133bPTFind the sum of the in nite geometric series 2L5 5 l 3M O O O 5 sln gr 0 Exercise5 11330PTFind the sum of the alternating in nite geometric series i i n r i 39 0 mg 5 O O 1 51 Writing Repeating Decimal as a Fraction Example Repeating Decimal Turn HI US39N39J as l nrtimL V 50 sh an Snluuult o 55939 WWTW serum wt 4 so u i xercise 6 391133dPTIf the repeating decimal 0387387387 is written as E i reduced form where m and n are integers then m o 11 O 387 o 29 Section 38 Polynomial and Rational Inequalities Inequalities Up to this point we have only solved linear inequalities isuchas 2x5 6 Our goal today will be to solve more complicated polynomial or rational inequalities The Basic Idea I We will reduce the inequality to the format fx30fx2Q x gt0 or x lt0 2 We will make a list called a sign chart that states where the mctionfx is positive or negative 3 We will use our chart to solve reduced the problem in l RationalPolynomialY l quot and the XaXis The only way for the graph of a rationalpolynomial function to change sides of the Xaxis and hence switch the sign offx is to 7 1 to cross the xaxis These are the zeros ofthe function 7 2 to encounter a vertical asymptote These are the undefined points ofthe function Partition Values The partition values of a function are the places where the function is either zero or unde ned Solving by PaperandPencil Step 1 Simplify ie obtain a common denominator the inequality so that one side is zero Step 2 Find the partition values 7 Find the x values that make the denominator zero unde ned points 7 Find the x values that make the mction zero zeros of the function Solving by PaperandPencil Step 3 Take a testpoint in each interval to see if the function is above the Xaxis positive or below the Xaxis negative Section 92 Parabola Manifesto In this chapter we are not studying functions We study the equations that represent twodimensional curves So we need to adjust our usual habitual understanding In section 31 we studied parabolic function whose graph is a parabola Now we are going to study more general parabolas Some notions are similar Some are NOT Application Definition Given a line D directrix and a point F focus not on the line D the set of all points P such that dPFdPD is called a parabola Note dAB denotes the distance between A and B If A xAyA and B xByB are two points dAB 1lxA x3 2 yA yB2 Examples of parabola y14ax Axis of symmetry D x ay I Pv 3 dF P The line through the focus F and orthogonal perpendicular to the directrix is called the axis of symmetry of the parabola The point where the parabola intersects with its axis of symmetry is called the vertex Four Fundamental Forms of Parabola F401 Dya T My 1 eqtutior Vertex Focus Direcsrix Equation Description 0 0 a ll x 393 y1 lax Parabola axis of symmetry is the xaxis opens to right 0 ll 3 0 x a y39 743x Parabola axis of symmetry is the xaxis opens to left 0 l 0 a y a x1 lay Parabola axis of symmetry is the yaxis upens up 0 0 0 a y a x1 4ay Parabola axis of symmetry is the y axis opens down Decipher the relation between the table and the graphs Let us pay attention to the column of Equation rst 1 If the square is on y then the axis of symmetry is the xaxis in which case the directrix is vertical If the square is on x then the axis of symmetry is the y axis in which case the directrix is horizontal Summing up the square tells us the equation is one of the rst two graphs or one of the last two graphs 2 The sign on the side that does not have the square tells us whether the parabola opens to the positive or negative direction of the axis 3 Now a which is always positive gives us the distance between the vertex and the focus It also tells us the distance between the vertex and the directrix Because the vertex is on the parabola its distance to the directrix is the same as the distance to the focus ATTN The 4a is associated with the linear part Extra Credit If the directrix is x1 and the focus is 10 use the de nition of parabola and the de nition of distance between two points to show that the equation of the parabola is yz4x ATTN We must draw the graphs to solve the problems in this chapter Example 1 92121131quot Select the graph of 12 475 b lt 0 O Exercise 2 921bPTSelect the equation of the following graph Oy2 czrcgt Oy2czccgti 2 OJ ycgti Or2 cycgt Exercise 2 92 10PTSelect the equation of Lhe following parahula Exe rclsea 9223PTSelecL the equation of the parabola wnh focus at 01 and vertex at 00 o 12 4 0 13912 41 o 12 1 o 12 1 Exercised 922hPTS l we mlumiun ur we parabula with Rum M 0 Y3 and direcmx the 11112 g T Exerc e5 922el l jb iud he dirccu39ix of the parabola given by 2 2y Translated Forms and Graphs 0 WWE repraee x by xrh hgtEI than the graph ufme eguanuh rs sm ed hght by h r We repraee x by gtlth WEI menthe graph enhe eguauuh rs sm ed heft by h o rrhuah WW2 repraeey by wk my menthe graph enhe eguauuh rs sm ed guwn by K ATTN The vemear trans atmn aehreveg by yeytk rs eguwareht m the vemear trahsrahuh We are fammar wwm yxtk am the yxik ahg yeytk have mrrereht mEam gS THE D RECT ON OF THE SH FT rs D FFERENT y D x n a Axis of V n 0 symmetry y k V h a k x a y mumn Axis of symmetry X FV h k a Dyka c Jr inf 46y k y DXha Axis of symmetry V II If y k Fh ak b y k2 4ax h Axis of symmetry X J lt gt Dyka V h k 39 F n k a a x h Maw k Vertex Focus Dlrectrix Equation Descripkinn n k n a k 0 k 411 7 n Panhnll axis nl xymmmy pl ll m x axis 0pm la rigm M n e a k x h a y A k 4zx n mm axis a yummy pmllll m mix npens m left It k M a y k 2 x n My 7 k Pinball axis nl xynlulury pirlllll m ruimpens up M nka yka xh tay k Panhula axis of symmth pmllel to rub npm dwn When we look at such translated problems it is MOST IMPORTANT to identify its fundamental form and the vertex Example 6 923aPTSelect the equation of the parabola with focus at 7 2 2 and vertex at 17 2 31 1 l2at 1 o y 22 o y 22 o 1 1V 3y 2 o y 22 12I 1 o w 132 12y 2 019 12 12f 2 Exampe7 923bPTSelect the equation of the parabola with directrix 3 E and vertex at 12 09 22 1 o12 y 2 0mI12 11092 o39y1293 2 09 221 Example 8 9236PTSelect the equation of the parabola with focus at 1 and directrix a g o y 32 21 1 0 1 3 2 8 y 1 0 9 1 83 5 3 O 3 2 2y 1 019 1 401quot 3 0y l 1 22L 3 Example 9 923dPT1Fmd the focus of the parabola with equation x 7 2 41 7 3 0 12 0 1 Example 10 9236PTFiud the directrix of the parabola with equation y 7 32 21 7 Section 42 Exponential Functions ALERT Check all files in this folder MASTERY of TRANFORMATION and EXPONENTS IS ESSENTIAL Exponential Functions 1 Definition An exponential function assume the form fx a a gt 061 1 Its domain is the set of ALL real numbers Its range is the set of all positive real numbers Remark What is the difference between yax and yx Which is exponential function Which is power function 2 Properties of exponentials agt0bgt0 I a01 a5 615 O abf asbs The Graph of fxaagt1 Figure i Graph uf fx2x Properties The graph of fgtc M 1 gt1 looks similar to fx 2 o Strictiyincreasingne if x gtx2then 11 gt1 0 y gteo as x gteo o y gt0 as Xarw ie y0isahorizontai asymptote The yeintercept isi The graph contains the points0 1and1 a The Graph of xa0 1 lt1 Ham 2 Gvaph u fx Properties The graph of fx an 1 lt1 looks slmllarto fx o Stncnydecreasmg e w x gtX2HEH a lta o yaw as Kaine o ygt0 as xamme y sahunzuntaasymptme o The wntercept s 1 The graph cumams me pmms an and 1 a Exercise1 421aPTSelect Lhe equation of Lhe following graph Oyb bgtl oy bquotbgt1 Oy7l739bgt1 Oyb mbgtl Examplez 4210PTSelecc Lhe equation of the following graph Oy71aw0ltalt1 O y 1 y1 1 O gn39 0lt lt1 Oy71am1lta Exercises 42 1dPTSelecL the graph of y 1 7 a l lt a J Section 11 5 The Binomial Theorem Recall Factorials For nonnegative integers a factorial is defined as follows 0 0 1 o 1 1 onnxn1XX3X2X1if n22 Combination quotC39quotL r rln7rl This is usually read as o n choose r 0 Or the combination of n things chosen r at a time Combination Practice 3 7 is 39 6 100 98 8 10001 K 0 Pascal s Triangle Pascal s Triangle 1331 14641 15101051 1615201561 Binomial Theorem x y fojxnija 10 J quotLn quotWHal quotW172112 Via K0 K1 1 K2 Kquot Binomial Practice 19 xzyz5 22 543 u 24 Waby4 Binomial Practice 25 o The coef cient of x6 in the expansion of x3m 27 o The coef cient of x7 in the expansion of2x112 33 o The third term in the expansion of 3x29 Section 102 Systems of Linear Equations in Three Variables Geometric Meaning of Three Linear Equations in Three Variables Reminder of linear equation in three variables A linear equation containing three variables x y and z is an equation of the form AxByCzD where A B C and D are constants The graph of such an equation is a PLANE in xyzspace 2 A system of three linear equations in three variables Thus the system of three linear equations containing the variables x y and z is a TRINARY of PLANEs in xyzspace Each x y z trinary that satis es the system of three equations must satisfy all three equations ie the trinary x y 2 must be on all three planes gtsuluuon a Conslsmmsyslem 50Wquot a Cnnslslanlsyslem c lncnnslxmnl sysmm una solullun Inllnltc numbur m snlullnns no salullnn 3 Possible solutions oflinear systems 0 Exactly ONE solution UNIQUE solution The solution is exactly the point where l l 39 vhi h l l 39 39 0 INFINITELY MANY solutions This is the second case where the three line overlaps or their intersection forms a line 0 NO solution This is the third case where the three planes have no point in common There is no point that could be on all three plains ATTN IN NO CASE can a linear system has exactly two or three solutions Solving System of Equations by Elimination Tile steps of the lVIethod of Elimination Selve Axl3yCzD i ExFyGzH 2 Ix V K7 L 3 o 1 Eliminate the variable r in a pair uf equations 1 anrl 2 Then eliminate the variable 139 in a pair at equations 1 and 3 Ynu mn picu my Turn pairs of Q lt TOlls as long IN the varCubic 139 will in uunwntrd o 2 Eliminate the variable 3 of the lust two equations in the equivalent system Then we win solve for z 3 BACK SUBSTITUTE by substituting the erunputed Values of z inte one of the last two equations in the equivalent system Then we mm 501m fur r 4 BACKSUBSTITUTE by substituting the 1 unputed Values of r and z lwk inte either equation 1 2 or Then we mu mle for 139 Note You do not necessarily need to eliminate X first The order of eliminating X y or z and solving for X y and 2 can be changed to your preference xample 1 1021PTSelect the type of solution for the following system 1 y z 4 21 3y 42 2 15 43 5y6 23 O In nitely many solutions 0 No solution 0 None of these 0 Unique solution 0 Exactly three solutions Equivalent Systems Revisited Recall Two systems of linear equations are equivalent if the two systems have identical solutions The idea hehjlnd the method at elimination is to keep replacing tho original equations the am with EQUWALENT eqlmnone until e svstenl ot equations with an obvious n39 matched Let us have anutherluuk attne rnetnen etellrnlnatlen we used ln example 1 R2lt2R1 1a 0 Flrststep R 74R1 R2 0 Secund step KleeRu K o eneralize and sum up the following Elementary Row Operations produce equivalent systems Elementary Row Operations lllllt39H llttltgt lwtt mlll Multiply both sides of nil mlutitn by a nonzero constant Add an equation to another equation IMPORTANT Note Operatluns We 1 eaRerM wnere 110 ls a eempeslte et tne 2quot and 3quotwa et Elementary ruvvuperatlurls Easlcally we nuld regard replaclng a raw wltn tne sum et a multlple et anetner raw and a multlple et ltselt as a type at Elementary ruvv uperatlun as well ExerclseZ 1021PTSeleut the type of solution for the following system a 8 2 39 l 15 0 In nitely many solutions 0 None of these 0 Exactly three solutions 0 Unique mlutinn o No solution