Precalculus Spring 2017


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1 Preclculus Spring 2017 Exm 3 Summry (Section 4.1 through 5.2, nd 9.4) Section P.5 Find domins of lgebric expressions Simplify rtionl expressions Add, subtrct, multiply, & divide rtionl expressions Simplify complex frctions Rewrite difference quotients Section 4.1 Find the domin of rtionl function. Find the verticl symptotes of rtionl function. Find the horizontl symptotes of rtionl function. Section 4.2 Sketch the grph of rtionl function. Find slnt symptote of rtionl function. Sketch the grph of rtionl function with slnt symptote. Section 9.4 Divide polynomils using polynomil long division. Fctor polynomil. Find the prtil frction decomposition of rtionl expression tht contins distinct or repeted liner fctors nd/or distinct or repeted qudrtic fctors. Section 4.3 Write the eqution for prbol, ellipse, or hyperbol with center or vertex (0,0). Grph circle, prbol, ellipse, or hyperbol with center or vertex (0,0) Find ll the pertinent informtion (vertices, foci, endpoints, vertex, p, symptotes, directrix) bout prbol, ellipse, or hyperbol from the grph or the eqution with center/vertex (0,0). Section 4.4 Write the eqution for prbol, ellipse, or hyperbol in stndrd form by completing the squre. Determine the type of conic section once the conic is in stndrd form. Write the eqution for prbol, ellipse, or hyperbol with center or vertex (h,k). Grph prbol, ellipse, or hyperbol with center or vertex (h, k). Find ll the pertinent informtion (vertices, foci, endpoints, vertex, p, symptotes, directrix) bout prbol, ellipse, or hyperbol from the grph or the eqution with center/vertex (h,k) Section P.2 Use properties of exponents Use scientific nottion to represent rel numbers Use properties of rdicls Simplify nd combine rdicls Rtionlize numertors nd denomintors Use properties of rtionl exponents Section 5.1 Recognize n exponentil function Grph n exponentil function with bse or bse e Evlute n exponentil function with bse or bse e Solve rel world problem using n exponentil function Section 5.2 Recognize logrithmic function Grph logrithmic function with bse or bse e Use the properties of logrithms nd nturl logrithms to simplify Evlute logrithmic function with bse or bse e Solve rel world problem using logrithmic function
2 P.5 Rtionl Expressions I Domin Domin: Rtionl expressions : Finding domin. polynomils: b. Rdicls: keep it rel! i. sqrt(x2) x>=2 [2, inf) ii. cubert(x2) ll rels since cube rootscn be positive nd negtive c. rtionl expressions: cn t divide by zero (undefined) so look t denomintor nd set it equl to zero. i. x+ 1 ( x 2)( x 4) ii. 2 x x x+ 3 II Simplifying rtionl expressions *simplifying rtionl expression cn hve n impct on domin helps determine grphs in this course nd clculus. #36 2 x x 2 x x III Opertoins with Rtionl Expressions remember frction opertions from P.1 notes A) Multiply B) Dividing
3 C) Combining rtionl expressions + + IV Complex Frctions Complex frction: seprte frctions in numertor nd/or denomintor EX x x 2 methods: stright division, multiply numertor nd denomintor by LCD Fctoring out negtive exponents / / sign 2 wys: fctor out smllest exponent, multiply top/bottom by the smllest exponent with the opposite V Difference Quotients: gol is to eliminte the originl denomintor Ex 11 x+ h x h rtionlize numertor
4 I Intro 4.1 Rtionl Functions nd Asymptotes Rtionl function: = given tht nd re polynomils. Ex1 Domin of rtionl function = II Verticl nd Horizontl Asymptotes Definitions p.333 blue box 1) The line = is verticl Asymptote of the grph of if or. 2) The line = is horizontl symptote of the grph of if s or. Find the domin (from p. 333) = f(x)= f(x)= = P. 334 Verticl nd Horizontl Asymptotes of Rtionl Function f(x)= = with no common fctors 1) The grph of hs verticl symptotes t the zeros of 2) The grph of hs one or no Horizontl Asymptote determined by compring the degrees of nd. If, the grph of hs the line (xxis) s horizontl symptote. b. If, the grph of hs the line (rtio of leding coefficients) s horizontl symptote c. If, the grph of hs horizontl symptote. Find the symptotes Ex 2: = 2b) = Ex 3 f(x)= =
5 Ex 4 & 5 re Appliction problems only going to do ex 5 For person with sensitive skin, the mount of time hours the person cn be exposed to the sun with miniml burning cn be modeled by =.. 0< 120 where is the Sun sore Scle Reding (bsed on intensity of Ultrviolet rys). d. Find the mounts of time person with sensitive skin cn be exposed to the sun with miniml burning when =10, =25,& =100. e. If the model were vlid for ll >0, wht would be the horizontl symptote of this function, nd wht would it represent? Clss Discussion: cn horizontl symptote be crossed? How do we find tht? 1) = 2) = 3) h= I Anlyzing grphs of rtionl functions 4.2 Grphs of Rtionl functions Guidelines Let = /, where nd re polynomils Steps to grphing rtionl functions: Grph the following functions Ex1 = #26 =
6 Ex2 = #20 = Ex3 = #37 = = #40 =
7 II Slnt Asymptotes Only occurs when the degree of is bigger thn the degree of. Use to find the slnt symptote Ex 5 = Similr: #58 = III Appliction finding minimum re Ex6 A rectngulr pge is designed to contin 48 squre inches of print. The mrgins t the top nd bottom of the pge re 1 inch deep. The mrgins on ech side re 1 inches wide. Wht should the dimensions of the pge be so tht the lest mount of pper is used?
8 9.4 Prtil Frctions Decomposition into prtil Frctions (p. 690): 1) If degree of numertor is gret thn denomintor, the divide first, then use reminder to pply steps 2 nd 3 2) Fctor denomintor (over the integers) to get fctors of or + + 3) Liner fctors: for ech fctor of the form, the prticl decomposition must include the following sum of fctors: ) Qudrtic fctors: for ech fctor of the form, the prticl decomposition must include the following sum of fctors: Ex 1: Ex 3: Ex 4:
9 4.3A Prbols (p ) Imge credit: From previous sections: = + + cn become = h + If >0 then opens If <0 then opens Conic section formt: h =4 verticl =4 h horizontl Focus: equidistnt from this point, point Directrix: equidistnt from this line, line P: the distnce from the vertex to the focus, number Vertex: is hlf wy between the focus nd directrix, point D of O: direction of opening (up/down/left/right if >0 then it opens right/up A of S: xis of symmetry, line FD: focl dimeter, found by finding the bsolute vlue of 4 Bsic Verticl Prbol =4 : D of O: Vertex: A of S: Focus: Directrix: FD: Focl dimeter is 8; so t the focus we know tht ech side there is point 8/2=4 wy. Then points re (4,2) nd (4,2) to help us get the shpe of the prbol. Bsic Horizontl Prbol =4 : D of O: Vertex: A of S: Focus: Directrix: FD: Exmple = 2 D of O: Vertex: A of S: Focus: Directrix: FD: Exmple: Find the eqution of prbol tht hs vertex t 0,0 nd its focus t 5,0 D of O: Vertex: A of S: Focus: Directrix: FD: Exmple: A prbol s vertex is (0,0) nd its FD=10 nd it opens verticlly. Wht is its eqution? D of O: Vertex: A of S: Focus: Directrix: FD: Exmple 6 =0 D of O: Vertex: A of S: Focus: Directrix: FD: Exmple: Find the eqution of prbol tht hs its vertex t (0,0) nd the directrix is =6 D of O: Vertex: A of S: Focus: Directrix: FD:
10 4.3B Ellipses nd circles Circles (p. 349) h + = mens the circle hs its center t (h,k) nd its rdius is r. Ex: =1 Ellipses (p ) Assume > Horizontl + =1 Center: (0,0) Foci,0 Vertices,0 Mjor xis 2 Minor xis 2 To get : = Imges credit: Verticl + =1 Center: (0,0) Foci 0, Vertices 0, Mjor xis 2 Minor xis 2 To get : = Eccentricity of Ellipse is how stretched it is. We mesure it by =. If is 0, then the reltion is circle. If is 1, then it is very elongted/stretched. Exmple: =100 Stndrd form: Center = b= c= Vertices Foci Mjor Axis Minor Axis Eccentricity Sketch Exmple: 9 +4 =1 Stndrd form: Center = b= c= Vertices Foci Mjor Axis Minor Axis Eccentricity Sketch Exmple: 9 +9 =81 Stndrd form: Center = b= c= Vertices Foci Mjor Axis Minor Axis Eccentricity Sketch Exmple: The mjor xis length is 6; the minor xis length is 4. The foci re on the xxis. Wht is the eqution of the ellipse? Exmple: The foci re 8,0 nd the eccentricity is 0.8; wht is the eqution of the ellipse?
11 4.3C Hyperbols Imge credit: The first term indictes the type of opening. Find c: = + Horizontl =1 Verticl =1 Center (0,0) Foci,0 Vertices,0 Brnches (this is the hyperbol itself) Trnsverse xis (vertex to vertex distnce) = 2 Conjugte xis =2b Asymptotes = Center (0,0) Foci 0, Vertices 0, Brnches (this is the hyperbol itself) Trnsverse xis (vertex to vertex distnce) = 2 Conjugte xis =2b Asymptotes = Exmples: 9 16 =144 Stndrd form: Center Find c: Foci Vertices Trnsverse xis Conjugte xis Asymptotes Sketch Exmple: 9 +9=0 Stndrd form: Center Find c: Foci Vertices Trnsverse xis Conjugte xis Asymptotes Sketch Exmple: Wht is the eqution of the hyperbol with 0, 6 nd symptotes =?
12 Circle h + = with center (h,k) Ellipse + =1 With center (h,k) Vertices t h, Foci h, h + =1 With center (h,k) Vertices h, Foci h, 4.4 Moving Conic Sections Hyperbol h =1 Center (h,k) Foci h, Vertices h, Asymptotes = h =1 Center (h,k) Foci h, Vertices h, Asymptotes = h Prbol h =4 verticl Center (h,k) Focus (h, k+p) =4 h horizontl Center (h,k) Focus (h+p, k Conic generl eqution =0 (Usully B is zero) 1. If or is zero, then it s usully 2. If A nd C hve the sme sign then. A=C mens it s b. mens it s n 3. If A nd C hve different signs then it s Nme tht conic section! Conic: Center: =16 Notble chrcteristics/sketch: Stndrd form: =0 Center: Notble chrcteristics/sketch:
13 =0 Conic: Stndrd form: Center: Notble chrcteristics/sketch: Conic: Stndrd form: =0 Center: Notble chrcteristics/sketch: Degenerte conics: give reson why ech conic is degenerte = = =1
14 Exponent Rules ) Properties (p. 15) m n m+ n = m n n n 0 1 = n 1 = n = 1 m ( b) = b ( ) = m n m m n = = b b m m m* n m m P.2 Exponent properties EX 3 & c 4 3 ( 3 b )(4 b ) 12x y x y b) Scientific nottion n extension of the exponent rules (cn tke ny number nd crete it s power of 10) = 4 x 10^4 b. # lwys between 1 & 10 Divide by Exponentil Functions nd their grphs I Exponentil Functions The exponentil function with bse is denoted by where >0, 1, nd is ny rel number. =2 for x=2 =2 for x=2 =0.6 for x=2/3 II Grphs of exponentil functions EX2 = =2 EX3b = =4 Mke tble nd plot points (3 to 3)
15 Generl conclusion on pre 382 see if we cn get these from students! = = Domin rnge yint Incresing/decresing Horizontl Asymptote Continuous These re lso onetoone Onetoone functions help us relize tht they hve nd tells us tht we cn! EX4 9=3 EX4b =8 Trnsformtions of exponentils : =2 +5 = 5 +1 III The nturl bse know the first 5 digits minimum! EX 6: use clcultor to evlute = t = 2,& =.25 EX 7 Grphing =2. =. p. 384 for exct picture
16 IV Appliction: Continuously compounded interest compounded once yer Yerly = 1+ p. 386 N compoundings per yer = 1+ P = principl Continuous = nnul interest rte r (Proof on pge 385) Ex 8,c A totl of $12000 is invested t n nnul interest rte of 9%. Find the blnce fter 5 yers if it is compounded () qurterly (c) continuously Rdioctive decy : Hlflife uses bse of where h is the number of yers in hlflife for prticulr element. The generl eqution is =. Exmple: If n element hs n initil vlue of 25 grms nd its hlflife is 1599 yers, how much of the smple if left fter 2000 yers? Extr one to one property to solve equtions = =81 2 =8 =16
17 5.2 Logrithmic Functions nd their Grphs Logrithmic Functions The logrithmic function of bse is the inverse of the exponentil function with bse. For x> 0, > 0, nd 1, y= log xifndonlyif x= The function given by f( x) = log x is the logrithmic function with bse. y *The logrithmic function with bse 10 is clled the common logrithmic function. It is denoted most often by log. The Nturl Logrithmic Function f( x) = log x= ln xx, > 0 e The nturl logrithm is the inverse function of the nturl exponentil function Evluting Logrithms f( x) = log 1. 4x, when x = 1 64 f( x) = log x f( x) = log, when x = x, when x =49 Properties of Logrithms/ Properties of Nturl Logrithms 1. log 1 0 = becuse 0 = 1. / ln1 0 0 = becuse e = log = 1 becuse 1 1 =. / lne= 1 becuse e = e. x 3. log = x nd log x = x. (Inverse Properties) / lne x = x nd lnx e = x. (Inverse Properties) 4. If log x= log y, then x = y. (OneToOne Property) / Iflnx= lny, then x = y. (OnetoOne Property) Simplify or solve using properties of logrithms. log713 log log lne 8. 7ln1 9. ln 10. e 2 ln8 e
18 11. log 4(5x 9) = log 4(15x+ 29) 12. log(2 x ) = log Grphs of Logrithmic Functions *Logrithmic functions re the inverse of exponentil functions. It is possible to obtin the grph of logrithmic function by finding its inverse nd then reflecting it over the line y = x. Trnsformtion of logrithmic function: f( x) = k± log ( x h) Prent function: f( x) = log x * The verticl symptote trvels with the horizontl shift. 13. Sketch the following grphs. (Use n x/y chrt nd shifting from the prent function.) ) y = log x b) f( x) = log 2( x+ 3) c) g( x) = 4+ log3x d) h( x) = log( x 3) + 5
19 Trnsformtion of the Nturl Logrithmic Function: y = ln (x h) + k Prent Function: y = ln x Grph nd find the domin of ech function. 14. y = ln x 15. y = ln (x+6)  1 Appliction of Logrithmic Functions 16. Students prticipting in psychology experiment ttended severl lectures on subject nd were given n exm. Every month for yer fter the exm, the students were retested to see how much of the mteril they remembered. The verge scores for the grph re given by the humn memory model f(t) = 75 6ln(t+1), 0 t 12, where t is time in months. ) Wht ws the verge score on the originl exm? b) Wht ws the verge score t the end of 4 months? c) Wht ws the verge score t the end of 8 months?
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