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# PRECALCULUS ALGEBRA MAC 1140

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This 25 page Class Notes was uploaded by Nathen Fadel on Thursday September 17, 2015. The Class Notes belongs to MAC 1140 at Florida State University taught by Staff in Fall. Since its upload, it has received 77 views. For similar materials see /class/205607/mac-1140-florida-state-university in Calculus and Pre Calculus at Florida State University.

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Date Created: 09/17/15

MAC1140 TEST1 REVIEWA 09152007 3135 Mr Fei Hua fhuamathfsuedul Due 09192007 Full Name 1 F H Ma Sec Instructions for preparing the test I You must have done all previous homework and quizzes You must have primed out all lecture notes homework solutions and quiz solutions Study or review them carefull r You should understand why and how we solve each problem before starting to do the review pro lems The following review problems should be regarded as a mock test Alter nishing the following review problems you need to check every solution especially where it is likely to err before submission The instructor will use review sessions to answer the problems the most of you may have dif culty with After the review session you all should reach such a state that you could use half of your brain to solve 90 problems correctly The other half will be used to handle pressure during the test You should be familiar with each problem including the steps to solve them the common pitfalls and rough time you need to solve them This standard is not high because we need lo make enough room for careless mistakes in the real test Then you are ready for the battle my dear warriors 1 311bPTlThe graph of the quadratic function 1392 6139 10 0 r as I 4 i a Z A 1739 oNoneoletese Nquot NJU n b HAC O 1 139 iniercept Z S 1 6 4 U 0 I intercepts i a 4 O O 392 rintercepts 2 311cP I Find Lhc quadratic equation whose graph has vertex M 12 quot1 39quot quot Iquot 4 1 it m m llt l i a l l gt7 y 2 6r 1 2 Oy 6J39122 Oy 2w 12 2 Oz I 221 3 312bPTThe quadratic function fx 12 21 8 has 0 a maximum value of 1 lmwx if if 0 a minimum value of 1 A L o a maximum value of W5 W l39 l g a minimum value of 9 9 4 AL7 312cPTz 1139 22 1 a gt 0 h H l I a minimum value of 1 WW 1 0 AT 0 V17 0 a maximum value of 1 l O a minimum value of 2 1 y o a minimum value of 1 o a maximum value of 2 5 313bPTSelect the equation 0139 the following graph 1 97449 MOMLy 2 TA Am of KgMML zTry V Nyall w 0 lt0 4quot W lamp A Wlwvduzgm 714 oy12 2m 2 oy m22x2 y v2 I2 OyI2x 2 3143cPTSelect the equation of the following graph Note 139 is a constant l Vfim39 1 b 1 L 7 Q glande l 1 i vi X 139 LM Jiffwwi 93M Oy 2m2 5L c Oy2x25mc oy2x2 5vc y 2x25ac 7i 332aPTSelecL the choice containing all the real zeros of fa x IWHW 4 m 1 my r w Wm l o LQA zmk J I 1 62 2 l O 2 12 0 There are no real zeros 391 2 y i O f fvvv39m b1 ln maw l T 339 L l I VanM 332bPTISelecl the correct statement for 32 2 22a 3x 4 7 f12 has a zero at w 2 with multiplicity 2 3m xq th Z panxvi 39 fz has a zero at 1 2 with multiplicity 1 mg 73 Man O fu has a zero at 5 2 with multiplicity 1 2 Id L l Q 2 V a has a zero at 1 2 ith multi licit 2 o f W P y kw 5 I L 1 V 9 S 3 333aPTThe graph of MI 22 n 9003 x Us gt Y m F WW 3 gt i L 0 none of these 39z eW ab 39 0 39 1 a 1 amp touches the maxis at r 30 and Crosses t e m axis at a 30 O r C crosses the m axis at u 30 and touches the maxis at a 30 o touches the rteaxis at at 30 and touches the xaxjs at u 30 crosses the maxjs at L 30 and crosses the JJaxis at a 30 393 13N1SPTSGIGCL ALL If the following that have zeros at 9 7 5 a y x 9x2 7762 5 Cl None of these my 1399w75 my 1 9w7m5 D 3 8I 9 7r 5 L39J LU 2 9r 73L 5 11 335aPTSeIect the statement that is elsefor 35 24152 o The graph off behaves like 1 L 3 for39f trge M a 3 1 o f has one ar intercept 70 Hi7 1 O The yintercept of graph is 100 a fpj h39eslda loeal a 5 O f has degree 4 7 7 r 7 I 12 335bPTSelect the equation of the following graph r if gquot pf k 3 quot IL x 39 w x x n 4 L39La LmLfH39V L 7 41quot r L oyz z2ltx 1ltm32 I I 0 yr2c 1r32 y 4 JV c gt o y m mam3 Oy121 l2l393 39 quotEMquot quot394 quot quot quot V quot 39J 13 L39 I 3411aPTQhoose the function that has a factor of 32 w M MIL ugh LL OfJ2x34v2 l8a 38 quot39 A r i 4 Ofx3m3 V3 2il mgtr723 V I a 39 ofa3m39 912 30x 72 th r quot Ofm213quot9i 8 14 wquot H W 3 4 1bPTTwu ofthe factors offz3x414za 9c2 38z24 are f W4 and z2 Another factor is fox311 39 quot 39 612 4354 I 1 I aw 7 lt m hum v r a r sxw l W al39 a 4 3 L 1 4 w A C I 39 V11 r F 15 I 34 1dP39l Find k such that f1L 2a3 7 km 7 39Im A 21 has a l39aIcLur of 1 H r A l i r39 r L L 02 quotL quot39 15 at 3442aPTSelecL the choice containing ALL according to the nnl Zeios Theoi39einl potential rational zeros of the polynomial fungiiou f151 48z3 311215 m O i1i3i5i8i15 iii 15 O ili3i5i15ii um um an i1i3i5i15igi O l1i3i5i15 17 34113PTFind the quotient if 41 22 4 is divided by 252 1 H V 0213 2 i quot LI i 397 212 1 v o32c H o221 l 7 18 y quot39 r t 34V4bPTFind thawmaindg if 4139 237 Ll is divided by 232 1 g 3 2x 0212 1 quotHi Hy 0 3 21 o 52l 19 0 I does not have a zero in 12 345aP39l39Application of the IntermegiagLYalue Theorem in fx 2x5 715 1 912 5x 3 on the interval 12 implies Q 110 conclusion about a zero of f in 1 2 O f has a zero in 12 7 20 k w k 139 345bMSPTSelect Ag he intervals for which the Illt Tbgt ieI Il implies the f 1 8m 1132 341 15gthas ro 121I 391 1M 394 mu m L4 lt i 4 39 rl l D 1v0 quotI l A gt I3x2 A39 39 L V 39 x Z L vquotvr quotquot 39 w39Lquot5 V 351aPT Write 2i22 1 in 12 bi form 2i 71 1 19117 H MAC114O TEST1 REVIEWA 09152007 3642 Mr Fei Hua fhuamathfsuedul Due 09212007 Full Name 39 39 39 quot L Secfl 1 36 1P39I Sclch the polynomial with real goef ciean ofdcgree 5 having zerosof5l i 2i w l v O w 5x1iz 1 ir2i 2i Oc 5mlir1 iJ 2 i 2i 5w5m lim l ic239ic2i wl iz2 i12i O I 5IIi 0 None of these 2 371nlquotl Choose the domain of the rational function m mail 7 i none of these a 0000 v l o 0030U0100 l 394 o 00 1 U 10 u 01 u 100 0 00 1U 11U1oo 3 372aPTSclecl the choice having ALL tlicxerticql gyinuqtgtgs of the function HI izfgn V g2m 2 oy2wwu2w 0 20 20 Oxr2zv 2a0 4c3 Oy212 O No vertical asymptotes V 4 r 37i2bPTSeIect the choice having ALL ihe horizontal asy39mptbms of 3quot 221413 7 M iv 77 U o 1 3 1 1 o x oy0 i i i 39 o 010 oymy ampy1 o No horizontal asymptotes 5 17 71 5 373aPTFind Lhe gbliqug symptom ol39 1 1 5 o No oblique asymptote a gymn1 L oyw1 quot oy 6 6 374aPTSelect the graph of y O 7 38laPTSolve 221 1l 1a 7 3 Z O o oo 1U 3 o 11U03 O lv U 3100 o 00 l U 3100 0 none of these 8 381BPTSolve mquot x 5 20 o 0041Ul5m 0 None of these w O quot0039 5 U 4 0 0 4v5 1 if o 5v4 9 382aPTSolve 43 2 o n 0 00 1 U 200 V o 131 o 200 0 12 0 l2 quotLuna 1 f M U I 1o Iv 39 J 4113MSPTSelect the formulas of ALL metaone functions 1 fwx2w1 BM x 1 a Nquot 7 7 7 1 quotTA 3f11 a 1x2 1 11 412aPTChoose the farmula or the inverse of the function fx 2 1 r PM quot33 H K of wT J 7 of riIfzf I of1mfi 12 gt A 4 u Al2bPTIf 339 ar 22 1 on a39nd iz B U then the inverse function IS y 0 PM 1 ma 2 011x 2on Loo or x 2 nlt v 2 of1x 23 20quot H072 7 7 L Of uxIHE 2onl0ltgt 13 41331 T Selecl the graph of the inverse of the function shown below z 4k v 1439 gtV 7 u il quot 9w 42 laPTSelecL the equation of the folllow mg graph A I 3 r v a 33 v Oy 0Vltalt1 a 39 I Y O y grz o lt a lt 1 WE 7 x Q 7 quot3 1E10lt alt11wvr n 1 5 gy 0ltaltl quot A T 15 13172 421cPTSelecv the equation of the follpiiviug graph L r y am 1 lt a O ya quot 0lta lt1 Oy la 1lta Oy 1a 0ltalt1 15 30L Hal 421dPTSelect the graph of y 11quot1 lt 1 g fk 1 39w J Mr 0am V A Ilta Review of Exponents and Radicals see page 967970 in the text Rules and de nitions X real numbers39 m n inte ers examples 1 x0 1 de nition for x 0 70 1 e0 1 1 2 1 3 1 2 x n de nition of negative exponents x 0 5 5 2 e 3 x e 3 xmxquot xmiquot product rule 2325 28 e4e3 e7 4 x quotquot xmquot power ofa power rule 325 310 e34 e12 xm 27 5 66 4 5 x quot quot uotient rule 2 e 6 xyquot xquotyquot power ofa product rule 3235 37 e3e4 e7 ri n 5 5 2 2 7 5 x power ofa quotient a 2 5 3 8 2 y yquot 3 3 3 3 l l l 8 xquot4 defn22andxgt00rnisodd 325 2 e3E 2 5 9 x7quotxm defn22andxgtOormevenornodd 32 2224 1 1 Notice that ifx 1 and n 2 then the expression xquot 12 is not de ned in the set of real numbers since the square root ofa negative number does not exist in the set of reas However if b is a postive real number then the expression bquot can be de ned for any real number n See the discussion on page 294 of your text Try evaluating 2 5 or 2 on your calculator When de ning the exponential function fx b the base b is required to be positive This guarantees that the function is de ned for any real x that is the domain offis oooo Section 31 Quadratic Functions and Models College Algebra Review Two common ways to write a quadratic function i xaxzbxc 7 x army k Vertex is located at h k You can get from the rst equation to the second by completing the square We will learn how to completethesquare in a moment College Algebra Review The importance of a 7 If a gt 0 the parabola opens up the vertex is a minimum point 7 If a lt 0 the parabola opens down the vertex is a maximum point h and k Form Example fx 2x32 6 7 Vertex 36 0 Remember The 3 inside the parentheses moves the graph to the left so the center is at 3 7 This parabola opens up because j V Completing the Square 0 A very clever factoring trick credited to the Babylonians Example fx 2x2 12xl6 Step 1 Add and subtract b24a to the leftside 122 122 fx 2x2 12x l6 42 42 2x2 12x18 16 18 Completing the Square 0 Step 2 Factor the rst three terms it should be a perfect Squajrfa 2x2 6x 9 2 2x32 2 Formula for the Vertex If you complete the square on fxaxzbxc you get 2 X d39 at m m e MW x E A V Ycoordinate This gives us a simple formula for the x coordinate of the vertex and a not so simple formula for the ycoordinate We will nd the ycoordinate a different way College Algebra Review When you are given fxaxzbxc you can nd the vertex without completing the square by using the following formula 774611 The maxmin ofthe pambolais located at the x wlue The maxmin ofthe parabola is given by the y wlue College Algebra Review Finding Intercepts Xintercepts 7 To cross the xaxis the yvalues must be zero Therefore x0 and solve forx Yintercepts 7 To cro s the y axis the xvalues must be zero Therefore nd the value 0 0 College Algebra Review Number of X Intercepts 0 A shortcut using a piece of the quadratic equation for nding how many xintercepts a quadratic function has 7 b2 4ac gt 0 Two xintercepts 7 b2 4ac 0 One xintercept 7 b2 4ac lt 0 Zero xintercepts Examples bobX3x2 6x 7 o How many xintercepts does bob have 0 Does bob have a maximum or a minimum 0 What is the maximum or minimum value of bob 0 Where is the max or min of bob College Algebra Review 0 Example 7 Graph fx x 42 5 U More Examples Find the equation for a parabola with a vertex at 15 and a y intercept of4 College Algebra Review Some Transformations 7 fx xZ 7 fxxZ k kgt 0 The gmph moves up k units add k to the yvalues 7fxxZ kkgt0 The mph moves down k units subtract k from the yvalues 7 fx x k2 kgt 0 k from the xvalues College Algebra Review 7 x x kz kgt0 The graph moves right k umts addk to the x values 7 x axz Multiply the yvalues by a the gmph is vertically stretched or compressed Section 103 Determinants Determinants Remark Determinants are scalars related to square matrices They kind of serve the role as the abosolute value of matrices although determinants could be a negative number Determinants are important mathematical tools that can frequently be used to aiialyz llll l39i quotteiii s There are various methods for nding determinants Some of the sinip only apply to 212 and 313 matrices Determinants of 232 matrices b 4 It D EL d is a 2 by 2 matrix then the determinant 0t D 15 7 7 a b 7 7 o detD 7 E d 7 quot1 br 4 1 fl Example Find the determinant ot D 2 73 Solution 1 71 2 73 lt1737lteigtlt2ei Determinants of 333 matrices 1 ii 12 013 If D 121 L22 123 is n 313 iiiati39ix then the detei39ii iiiiant of D is 131 032 quot33 111 12 113 721 22 723 731 32 33 ilerD liiuzzdas 1i2 123031013021032 i 13122aBl112021033 011n23 32 The following is a good device for i39eiiienilering how to evaluate 313 determinants rubyi lui pug r39lnghg craby39l hu y rl rmng Exercise 1 Example Find the letorlninant of fl I D Z 1 1 fl 72 71 Solution 1 71 l 2 1 l 11Jlt11171quot2lt2 WXUEIH1lt2J1gt112 711ui02720 Important note The 111etlioltlgt shown above for 212 and 31f determinants loos NOT apply to 414 or liighoriordei39 dotel niinunts Exercise2 1 2 3 1041aPT 3 4 a a y 22 O r2y 2z O2x 4y 42 Oar Qy Zz 0 2x 43 12 Cramer s Rule Remark Cramer s rule is a formally elegant way to solve a linear system But in real life people do not use it to solve linear system because it is computationally costly Runner s Rule ur39s rule is it lilt lllUIl leL L eLn lJL usml lul hUl ing linear hy LeJJl 39VlJL39ll LlJL number 01 LLnkuLm n the 1llllllgt0139 of equations illlll the determian of th Lents coc 39icient matrix is not in 1 10 Mill r qiml tn mm Cramer s Rule for the case of two unknowns Consider the following sylttein 1139 by s CT My 1 c b 1 ll L 1 7E U 111 the above systunii then the above system has a unique solution given 11 Example Solve the following system using Trainer39s rule 3139 7 2y 1 I 2y 2 Solution By Cranier39s Rule 1 72 793 8 4 319711111 5 3072 1 Cramer s Rule for the case of three unknowns Consider the following HystonL 1111 any 1132 b1 U211 sz 0232 52 3931 13w 1332 b3 If in the above systcnn 11 112 013 D 121 122 123 U 3931 32 G33 than the AM 0 system 11215 21 uniun solution given by 1 1 quot12 113 11 bl quot13 1121 b2 L23 L31 3 133 D M an an 3 I33 132 1r Exercise 3 Example Use Gunner s rule to solve the following system 139 7 1 z 0 2139 y 7 z 1 l 7 y 22 72 SolutimL Let 1 D21713 1 0 71 1 1 1 71 2 71 2 1 I D 1 U 1 2 1 71 1 72 2 5 1 D E 71 J 2 1 1 7 72 z 1 7 76773 Exercise 4 10142aPTSelecL the solution given by Cramer s rule for Lhe following system1 where 1472 0472 1072 D3723A47233343C 21 3 711 3 71 3 140 3 24 21 3m72y324 I4y72120 2xy73z 1

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