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### Ontaphamsobac3(2)

Course: MATH 1002, Spring 2011
School: University of Western...
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Word Count: 2930

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TAP ON VE HAM SO BAC 3 (Trung tam Luyen thi ai hoc Vnh Vien) Gia s : y = ax3 + bx2 + cx + d vi a 0 co o th la (C). y = 3ax2 + 2bx + c, y = 6ax + 2b b 1) y = 0 x = (a 0 ) 3a b x= la hoanh o iem uon. o th ham bac 3 nhan iem uon 3a lam tam oi xng. 2) e ve o th 1 ham so bac 3, ta can biet cac trng hp sau : i) a &gt; 0 va y = 0 vo nghiem ham so tang tren R (luon luon tang) ii) a &lt; 0 va y = 0 vo nghiem...

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TAP ON VE HAM SO BAC 3 (Trung tam Luyen thi ai hoc Vnh Vien) Gia s : y = ax3 + bx2 + cx + d vi a 0 co o th la (C). y = 3ax2 + 2bx + c, y = 6ax + 2b b 1) y = 0 x = (a 0 ) 3a b x= la hoanh o iem uon. o th ham bac 3 nhan iem uon 3a lam tam oi xng. 2) e ve o th 1 ham so bac 3, ta can biet cac trng hp sau : i) a > 0 va y = 0 vo nghiem ham so tang tren R (luon luon tang) ii) a < 0 va y = 0 vo nghiem ham so giam (nghch bien) tren R (luon luon giam) iii) a > 0 va y = 0 co 2 nghiem phan biet x 1, x2 vi x1 < x2 ham so at cc ai tai x1 va at cc tieu tai x2. Ngoai ra ta con co : + x1 + x2 = 2x0 vi x0 la hoanh o iem uon. + ham so tang tren (, x1) + ham so tang tren (x2, + ) + ham so giam tren (x1, x2) iv) a < 0 va y = 0 co 2 nghiem phan biet x 1, x2 vi x1 < x2 ham at cc tieu tai x 1 va at cc ai tai x 2 thoa ieu kien x 1 + x2 = 2x0 (x0 la hoanh o iem uon). Ta cung co : + ham so giam tren (, x1) + ham so giam tren (x2, + ) + ham so tang tren (x1, x2) 3) Gia s y = 0 co 2 nghiem phan biet va y = k(Ax + B)y + r x + q vi k la hang so khac 0; th phng trnh ng thang qua 2 iem cc tr la y = r x + q 4) (C) cat Ox tai 3 iem phan biet 2 ph b x1 y' = 0 conghieman iet , x2 y(x1).y(x2) < 0 5) Gia s a > 0 ta co : i) (C) cat Ox tai 3 iem phan biet > 2 ph b thoa y' = 0 conghieman iet < x1 < x2 y( ) < 0 y(x1).y(x2) < 0 ii) (C) cat Ox tai 3 iem phan biet < 2 ph b thoa y' = 0 conghieman iet x1 < x2 < y( ) > 0 y(x1).y(x2) < 0 Tng t khi a < 0 . 6) Tiep tuyen : Goi I la iem uon. Cho M (C). Neu M I th ta co ung 1 tiep tuyen qua M. Neu M khac I th ta co ung 2 tiep tuyen qua M. Bien luan so tiep tuyen qua 1 iem N khong nam tren (C) ta co nhieu trng hp hn. 7) (C) cat Ox tai 3 iem phan biet cach eu nhau y = 0 co 2 nghiem phan biet va y(x0) = 0 (x0 la hoanh o iem uon) 8) Bien luan so nghiem cua phng trnh : ax 3 + bx2 + cx + d = 0 (1) (a 0) khi x = la 1 nghiem cua (1). Neu x = la 1 nghiem cua (1), ta co ax3 + bx2 + cx + d = (x - )(ax2 + b1x + c1) nghiem cua (1) la x = vi nghiem cua phng trnh ax 2 + b1x + c1 = 0 (2). Ta co cac trng hp sau: i) neu (2) vo nghiem th (1) co duy nhat nghiem x = ii) neu (2) co nghiem kep x = th (1) co duy nhat nghiem x = iii) neu (2) co 2 nghiem phan biet th (1) co 3 nghiem phan biet iv) neu (2) co 1 nghiem x = va 1 nghiem khac th (1) co 2 nghiem. v) neu (2) co nghiem kep th (1) co 2 nghiem BAI TAP ON VE HAM BAC 3 Cho ho ng cong bac ba (C m) va ho ng thang (Dk) lan lt co phng trnh la y = x3 + mx2 m va y = kx + k + 1. (I) PHAN I. Trong phan nay cho m = 3. Khao sat va ve o th (C) cua ham so. 1) Goi A va B la 2 iem cc ai va cc tieu cua (C) va M la iem bat ky tren cung AB vi M khac A , B . Chng minh rang tren (C) ta tm c hai iem tai o co tiep tuyen vuong goc vi tiep tuyen tai M vi (C). 2) Goi la ng thang co phng trnh y = 1. Bien luan so tiep tuyen vi (C) ve t E vi (C). 3) Tm E e qua E co ba tiep tuyen vi (C) va co hai tiep tuyen vuong goc vi nhau. 4) nh p e tren (C) co 2 tiep tuyen co he so goc bang p, trong trng hp nay chng to trung iem cua hai tiep iem la iem co nh. 5) Tm M (C) e qua M ch co mot tiep tuyen vi (C). (II) PHAN I I.Trong phan nay cho tham so m thay oi. 6) Tm iem co nh cua (C m). nh m e hai tiep tuyen tai hai iem co nh nay vuong goc nhau. 7) nh m e (Cm) co 2 iem cc tr. Viet phng trnh ng thang qua 2 iem cc tr. 8) nh m e (Cm) cat Ox tai 3 iem phan biet. 9) nh m e : a) ham so ong bien trong (1, 2). b) ham so nghch bien trong (0, + ). 10) Tm m e (Cm) cat Ox tai 3 iem co hoanh o tao thanh cap so cong. 11) Tm ieu kien gia k va m e (D k) cat (Cm) tai 3 iem phan biet. Tm k e (Dk) cat (Cm) thanh hai oan bang nhau. 12) Viet phng trnh tiep tuyen vi (Cm) va i qua iem (-1, 1). 13) Chng minh rang trong cac tiep tuyen vi (C m) th tiep tuyen tai iem uon co he so goc ln nhat. BAI GIAI PHAN I : m = 3 Khao sat va ve o th (oc gia t lam) 1) Goi n la hoanh o cua M. V ham so at cc tieu tai x = 0 va at cc ai tai x = 2 nen 0 < n < 2; y' = 3x2 + 6x he so goc cua tiep tuyen tai M la k 1 = 3n2 + 6n (0, 3] (v n (0, 2)). ng thang vuong goc 1 vi tiep tuyen tai M co he so goc la k 2 = (vi k1 0 < k1 3). Hoanh o cua tiep tuyen vuong goc 1 vi tiep tuyen M la nghiem cua 3x 2 + 6x = (= k1 1 k2) 3x2 6x = 0. Phng trnh nay co a.c < 0, k1 k1 (0, 3] nen co 2 nghiem phan biet, k1 (0, 3]. Vay tren (C) luon co 2 iem phan biet ma tiep tuyen o vuong goc vi tiep tuyen tai M. 2) E (e, 1) . Phng trnh tiep tuyen qua E co dang y = h(x e) + 1 (D). (D) tiep xuc (C) he x3 + 3n2 3 = h(x e) + 1 co nghiem. 3x2 + 6x = h Phng trnh hoanh o tiep iem cua (D) va (C) la : x3 + 3x2 3 = ( 3x2 + 6x)(x e)+ 1 (1) 3 2 x + 3x 4 = x( 3x + 6)(x e) (x 2)(x2 x 2) = 3x(x 2)(x e) x = 2 hay x2 x 2 = 3x2 3ex x = 2 hay 2x2 (3e 1)x + 2 = 0 (2) 2 (2) co = (3e 1) 16 = (3e 5)(3e + 3) (2) co nghiem x = 2 8 2(3e 1) + 2 = 0 e = 2 5 Ta co > 0 e < 1 hay e > . 3 Bien luan : 5 i) Neu e < 1 hay < e < 2 hay e > 2 3 (1) co 3 nghiem phan biet co 3 tiep tuyen. 5 ii) Neu e = 1 hay e = hay e = 2 3 (1) co 2 nghiem co 2 tiep tuyen. 5 (1) co 1 nghiem co 1 tiep tuyen. 3 Nhan xet : T o th, ta co y = 1 la tiep tuyen tai (2, 1) nen phng trnh (1) chac chan co nghiem x = 2, e. 3) V y = 1 la tiep tuyen qua E (e, 1), e va ng x = la khong tiep tuyen nen yeu cau bai toan. (2) co 2 nghiem phan biet x 1, x2 thoa : y'(x1).y'(x2) = 1 5 e < 1 e > 3 nghiem (2) cua x1,x2 la (3x2 + 6x )(3x2 + 6x ) = 1 1 1 2 2 5 e < 1 hay e > 3 x + x = 3e 1 1 2 x .x = 1 2 12 9x1.x2(x1 2)(x2 2) = 1 5 e < 1 hay > e 3 9[1 (3e 1 + 4] = 1 ) 55 55 e= . Vay E ,1 27 27 4) Tiep iem cua tiep tuyen (vi (C)) co he so goc bang p la nghiem cua : y' = p 3x2 6x + p = 0 (3) Ta co ' = 9 3p > 0 p < 3 Vay khi p < 3 th co 2 tiep tuyen song song va co he so goc bang p. Goi x3, x4 la nghiem cua (3). Goi M3 (x3, y3); M4 (x4, y4) la 2 tiep iem. Ta co : x3 + x4 b = =1 2 2a 2 y3 + y4 (x3 + x3 ) + 3(x3 + x2) 6 3 4 4 = = 1 2 2 Vay iem co nh (1, 1) (iem uon) la trung iem cua M3M4. 5) Cach 1 : oi vi ham bac 3 (a 0) ta de dang chng minh c rang : iii) Neu 1 < e < M (C), ta co : i) Neu M khac iem uon, ta co ung 2 tiep tuyen qua M. ii) Neu M la iem uon, ta co ung 1 tiep tuyen qua M. Cach 2 : Goi M(x0, y0) (C). Phng trnh tiep tuyen qua M co dang : y = k(x x0) x3 + 3x2 3 (D) 0 0 Phng trnh hoanh o tiep iem cua (D) va (C) la : 3 2 x 3 + 3 x 2 3 = (3 x 2 + 6 x)( x x0 ) x0 + 3 x0 3 (5) 3 3 2 2 2 x x0 3(x x0) + (x x0)(3x + 6x) = 0 x x0 = 0 x2 + xx0 + x2 3x 3x0 3x2 + 6x = 0 0 2 x = x0 hay 2x (3+ x0)x x2 + 3x0 = 0 0 x = x0 hay (x x0)(2x + x0 3) = 0 3 x0 x = x0 hayx = 2 Do o, co ung 1 tiep tuyen qua M (x0, y0) (C) 3 x0 x0 = x0 = 1 2 Suy ra, y0 = 1. Vay M(1, 1) (iem uon). Nhan xet : v x0 la 1 hoanh o tiep iem nen pt (5) chac chan co nghiem kep la x0 Phan II : Tham so m thay oi. y' = 3x2 + 2mx 6) (Cm) qua (x, y), m y + x3 = m (x2 1) , m x2 1= 0 x = 1 x = 1 hay 3 y = 1 y + x = 0 y = 1 Vay (Cm) qua 2 iem co nh la H(1, 1) va K(1, 1). V y' = 3x2 + 2mx nen tiep tuyen vi (C m) tai H va K co he so goc lan lt la : a1 = y'(1) = 3 + 2m va a2 = y'(1) = 3 2m. 2 tiep tuyen tai H va K vuong goc nhau. 10 a1.a2 = 1 9 4m2 = 1 m = . 2 7) Ham co cc tr y' = 0 co 2 nghiem phan biet. 3x2 = 2mx co 2 nghiem phan biet. 2m x = 0 va x = la 2 nghiem phan biet. 3 m 0. Khi o, ta co : 1 2 1 y = m2x m + x my' 9 9 3 va phng trnh ng thang qua 2 cc tr la : 2 y = m2x m (vi m 0) 9 8) Khi m 0, goi x1, x2 la nghiem cua y' = 0, ta co : 2m x1.x2 = 0 va x1 + x2 = 3 2 2 2 2 y(x1).y(x2) = m x1 m m x2 m 9 9 2 4 = m2(x1 + x2) + m2 = m4 + m2 9 27 Vi m 0, ta co y(x1).y(x2) < 0 42 m +1 < 0 27 27 33 m> 4 2 Vay (Cm) cat Ox tai 3 iem phan biet. 2 phan x y'= 0 co nghiem biet1,x2 y(x1).y(x2) < 0 m2 > m> Nhan xet : 33 2 33 th phng trnh y = 0 co 2 nghiem am va 2 1 nghiem dng. 33 Khi m > th phng trnh y = 0 co 2 nghiem dng va 2 1 nghiem am. a) Ham ong bien tren (1,2) 3x2 + 2mx 0, x (1,2). Neu m 0 ta co hoanh o 2 iem cc tr la 0 2m va . 3 2m ,0 . Vay Neu m < 0 th ham ch ong bien tren 3 loai trng hp m < 0 Neu m = 0 ham luon nghch bien (loai). i) Khi m < ii) 9) i) ii) 2m iii) Neu m > 0 th ham ch ong bien tren 0, 3 2m , Do o, ycbt m > 0 va [12] 0, 3 2m 2 m 3 3 b) T cau a, ta loai trng hp m > 0. 2m Khi m 0 ta co ham so nghch bien tren , va 3 ham so cung nghch bien tren [0, + ). Vay e ham nghch bien tren [0, + ) th m 0. Ghi chu : nen lap bang bien thien e thay ro rang hn. y" = 6x + 2m , y" = 0 x = 10) m 3 (Cm) cat Ox tai 3 iem cach eu nhau. y = 0 co 3 nghiem phan biet va iem uon nam tren truc hoanh. 33 33 m> m> 2 2 m m3 m2 y = 0 27 + m. 9 m = 0 3 33 m> 3 6 2 m= 2 2 2m 1= 0 27 11) Phng trnh hoanh o giao iem cua (C m) va (Dk) la x3 + mx2 m = kx + k + 1 m(x2 1) = k(x + 1) + 1 + x3 x + 1 = 0 m(x 1) = k + 1 x + x2 x = 1 hay x2 (m + 1)x + k + m + 1 = 0 (11) a) Do o, (Dk) cat (Cm) tai 3 iem phan biet (11) co 2 nghiem phan biet khac 1 1+ m + 1+ k + m + 1 0 )2 ) (m + 1 4(k + m + 1 > 0 k 2m 3 2 (*) k < m 2m 3 4 b) V (Dk) qua iem K(1,1) (Cm) nen ta co : (Dk) cat (Cm) thanh 2 oan bang nhau. m 2m3 m cua (Cm) (Dk) qua iem uon ; 3 27 2m3 m m = k + 1 + 1 27 3 2m3 27 27 m k= (**) 9(m + 3) Vay ycbt k thoa (*) va (**). 12) Phng trnh tiep tuyen vi (Cm) i qua (1,1) co dang : y = k(x + 1) + 1 (Dk) Vay, phng trnh hoanh o tiep iem cua (D k) va (Cm) la : x3 + mx2 m = ( 3x2 + 2mx)(x + 1) + 1 (12) 2 2 3 m(x 1) = ( 3x + 2mx)(x + 1) + 1 + x x + 1 = 0 m(x 1) = 3x2 + 2mx + 1 x + x2 x = 1 hay 2x2 + (1 m)x m 1 = 0 (13) m+ 1 x = 1 x= 2 y' (1) = 2m 3 2 1 m + 1 m+ 1 m + 1 y' = 3 + 2m = (m2 2m 3) 4 2 2 2 Vay phng trnh cua 2 tiep tuyen qua (1, 1) la : y = (2m + 3)(x + 1) + 1 1 y = (m2 2m 3)(x + 1) + 1 4 Nhan xet : Co 1 tiep tuyen tai tiep iem (1, 1) nen phng trnh (12) chac chan co nghiem kep la x = 1 va phng trnh (13) chac chan co nghiem la x = 1. 13) Cac tiep tuyen vi (Cm) tai tiep iem cua hoanh o x co he so goc la : h = 3x2 + 2mx bm Ta co h at cc ai va la max khi x = = (hoanh 2a 3 o iem uon) Vay tiep tuyen tai iem uon co he so goc ln nhat. 2 m m2 m2 Nhan xet : 3x + 2mx= 3 x2 + 3 3 3 Ghi chu : oi vi ham bac 3 y = ax3 + bx2 + cx + d, ta co : i) Neu a > 0 th tiep tuyen tai iem uon co he so goc nho nhat. ii) Neu a < 0 th tiep tuyen tai iem uon co he so goc ln nhat. 2 PHAM HONG DANH (Trung tam Luyen thi ai hoc Vnh Vien)
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Axia College MaterialAppendix CPsychotherapy MatrixCognitiveBehavioralPsychodynamicSummary of ApproachThis perspective suggests thatabnormal behavior is a symptom ofdamaged thoughts and beliefs.Rational-emotive behavior therapycould be used her
University of Phoenix - PSY - 202
Long- TermMemoryPresentationBy: Angel BinghamHow information is stored inour long term memoryTransferring information from short-term to long-term memory isdone through rehearsal Rehearsal:repetition of information located inshort-term memoryEl
University of Phoenix - PSY - 202
Running head: NORMAL VERSUS ABNORMAL BEHAVIOR1Normal Versus Abnormal BehaviorAngel BinghamPSY 202October 12, 2011Rebecca MarekNORMAL VERSUS ABNORMAL BEHAVIOR2Normal versus Abnormal BehaviorThe definition of normal varies from person to person. A
New Mexico - BIO - 201
Jessica Richards-ElliottBiology 201-011HW 91. (A) Why does FADH-2 enter the electron transport chain at complex II, instead of complex I, which iswhere NADH enters. Your answer should reference the different amount of energy in FADH-2 &amp; NADH,and why
New Mexico - BIO - 201
Chapter 8 Questions for Review1. What is binary fission and what types of cells use this process for their reproduction?3. What is the difference between chromatin and chromosomes? Why does the DNAcondense (super-coil) prior to cell division?4. What i
UCLA - ENGLISH - 161C
Crusoe and Travel Narrative 1st person detailed descrip. of environment, particular to physical appearance ofindividuals Epistolary/journal form Reputation of travel writers as liars. Some authors try to satirize this. Traveler responds with wonder t
UCLA - ENGLISH - 161C
Crusoe 80-160 Time suddenly speeds up; seasons are passing, Journal more about his crop cultivations.He is adapting to the island. Preoccupied with basketweaving, agriculture, rearinganimals, building more shelters Despite no companionship, he is taki
UCLA - ENGLISH - 161C
What is a novel?o Historically: readings that first appeared in Europe in early 18th cenuryo Formally: &quot;prose fiction of a reasonable length&quot; - Terry EagletonNovels are characterized by conscious experimentation. Therefore the most typical novelmay a
UCLA - ENGLISH - 161C
Henry Fielding (1701-1754) Combined 3 professionsLawJournalismProse fiction First foray into fiction is Shamela (1741), parody of Pamela. One of the 3 authors that pioneered the Novel. Among others are Defoe and Richardson. Mock tragic/epic in natu
UCLA - ENGLISH - 161C
Fielding elevates charity above pietySee Joseph's speech, p204Echoes of Richardson's critque off upper-class corruptionSee p.238Joseph as Hero Chapter 11 (p229-232) Does Joseph's display of emotion augment or diminish his heroism? Open ended questi
UCLA - ENGLISH - 161C
Horace Walpole (1717-1797) Son of Robert Walpole Re-models his home in the gothic style Gentleman-writerOronoto's framing devices What do you notice about the title pages to the 1st and 2nd edition?o 1st edition: originally translated from an Italia
UCLA - ENGLISH - 161C
Receptionp117: reviews and criticismsWhy was Otranto considered so offensive?oThe educated at the time believed the supernatural to be ridiculousoIt is a reason why Walpole is anxious to reveal himself as the author of acontroversial novel.Neoclas
UCLA - ENGLISH - 161C
PamelaSamuel Richardson (1689-1761)Middle class, owns printing businesso 1st work is advice manual for apprentices (who are of lower classes). Sawapprentices as more open to reform than their social superiors.o Pamela grows out of volume of exemplary
UCLA - ENGLISH - 161C
Perhaps Pamela supports the idea of meritocracy, where those with merit and virtue riseto the top of societyPamela's merit enables her to reform the morally degenerate aristocracyp299, 300. How does Pamela imagine fulfilling the role of wife?Note Mr.
UCLA - ENGLISH - 161C
Charles Brockden Brown (1771-1810) Philadelphia Quaker classical education at a Quaker school Politically progressive values linked to Quakerism: gender equality, abolitionism Influenced by William Godwin, Mary Wollstonecraft period of counter-revol
UCLA - THEATER - 120a
3 key elements to a film (Ex: Dustin Hoffman in The Graduate)o Objective: What each character wants or needs; a goal. Every character needs anobjective and must be motivated to obtain it. Actor needs to realize that objectiveo Obstacle: Something that
UCLA - THEATER - 120a
hayes code: 1934-1968. guideline that determined what was acceptable on screen. As aindirect result, the screwball comedy subgenre of the romantic comedy was created. Tobypass the haze code by disguising sexuality in comedy. In 1968, replaced by the rat
UCLA - THEATER - 120a
Monster Film Romantic Adventure Musical: heavily dependent on sound Studios have a fear of bringing up the depression. Arc of a scene: beginning to end Gold diggers of 1933 (a musical): after the famous dance scene &quot;We're in the money&quot;, thefilm intr
UCLA - THEATER - 120a
Frame-in-a-frame Ongoing physical activity Neo-Realism Subjective RelationshipFilms of 1940s Grapes of Wrath (John Ford, 1940)o Starring Henry Fonda as Tom. Very real/natural style of performance. Tom is verycalm, ex-convicto Scene 1: Fonda is tr
UCLA - THEATER - 120a
Neorealism (Italian): characterized by stories of poor/working class that parallel thedifficult economic times of post WWII Italy.ooolow budget way of asserting creativityA counter against Hollywood films.used real lighting, typically outdoors. If
DeAnza College - PHIL - 01
STUDY GUIDEMidterm, Phil001T. RamirezThe exam will be given in multiple choice format. Please bring a Scantron 882-Eform (the green one) and #2 pencil to class on the day of the exam.Make sure that you understanding what each of the following terms m
DeAnza College - BUSINESS - 10
10/4/2011Chapter 1:Taking Risks andMaking Profitswithin theDYNAMICBUSINESSEnvironmenthttp:/www.youtube.com/watch?v=XF8nRC6YAUTrue or False?The only way a firm canincrease its profits toincrease its profits is toincrease its sales revenue.11
DeAnza College - BUSINESS - 10
10/4/2011Chapter 2:UnderstandingHow Economicsaffects BusinessBusinesshttp:/www.youtube.com/watch?v=0C3XQ3BTd4o1. The share of all productive assetsthat are owned and managed by thegovernmentgovernment2. The degree to which governmentregulates
DeAnza College - BUSINESS - 10
9/13/2011Chapter 3:Doing Business inBusinessGlobalGlobalMarketsTrue or False?The continent of North Americais home to the largestpercentage of the worldspopulation.19/13/2011True or False?When the U.S. provides foreignaid to Pakistan andA
DeAnza College - BUSINESS - 10
10/7/2011Chapter 4:Demanding EthicalEthicalSociallySociallyResponsibleResponsibleandBehaviorTrue or False?According to the Association ofCertified Fraud Examiners,employee fraud causes 30% ofall business failures.Enron shareholders lost \$74
DeAnza College - BUSINESS - 10
9/16/2011Chapter 5:How toFormForm aBusinessBusinessa. C Corporationsb. Partnershipsc. Sole proprietorshipsd. S Corporationse. Limited Liability Companies19/16/2011True or False?If a sole proprietorship fails,the owner may lose whateverwas
DeAnza College - BUSINESS - 10
9/17/2011Chapter 6:EntrepreneurshipandStarting aSmallSmallBusinessTrue or False?Successful entrepreneurs areexclusively motivated byexclusively motivated bythe desire to become rich.True or False?State governments create newstartup business