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Week 3 - Chapter 4 Linear Functions(4.1 Definition ‘...

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Unformatted text preview: Chapter 4: Linear Functions (4.1) Definition: ‘ notion (always degree 1) is defined by an equation of the form — where A and B are constants is called the Standard form. +EY: C; For a Line passing thrgfigh [x1,y1) and (>12, yzy Slope=m :32 —‘a l a Z *1 l 1) Slope-intercept form of Line: . fizm'x-i-lo ) rm: glare} bt‘a'm 2) Point-slope Form: M’fim ‘Uu UM CD 9 3) Standard Form: W . “r: M 4— 859714? 6): Me MW 613%)» ‘ (XML) EX: If f[3)= 2 and f[- 3) — -,4 find the linear function defining f[x] 1n the slope- intercept form. ”E r“ 09’9”: (3 513 m: SW?“ {[31 ”jg—"é: (X9492); (3) Til) I. - Mir“ . - CAL \ ll Ff.- (:0 9629 + I “C . ‘f ‘“ VOLULL A factory owner buys a new machine for $20 000. After eight years the machine ' has a salvage value of $1000. Find a formula for the value of the machine after t years when 0 < it- < 8. - Application of linear function concept. is given by Cost functio :C[x] = 220)} + 4000 - (y = mX +b form) .where m = 220 and b = 4000 :SlOPE’. 0&— m limewr Malina MMran (Legr- “: $Lo?€ 2‘37: 110 a) Find the marginal cost = additional cost to produce b] Compute the cost of500 motorcycles C(X‘) ”I, zwx Jr L1 0 O O U500) 1: 22,0651) 0’) 4—H o 00 :\\0, 000 +510 00 : Ill-f, 000 c) Use the results from a] and b] to find the cost of 501 motorcycles $010135; Haas 069+)”; WML “Nora M‘l“ eathe fixed cost [the cost before starting production] ”tagl ugt Y~m+ m-b fixed Cos-+— j4oQo éVMnexf used? \fiCO/Q’FEdeC oi} how mama, WW 0301ng corresponding temperature on the Fahrenheit scale. " a) Find the linear function relating x and y; use the facts that 32°F corresponds to 00C and 212°F corresponds to 1000C. Write the function" 1n standard form. (XI) Y1) : CD) 52) m: lag—L5! :rllIZ ‘52 “' l X ~>< 00 ~— 0 "' " 949—) 1. 1 CXZ)Y1) ([00) I lgD :_ lag ‘6 %; :MCK Kl) lOo W132: WY 0) ‘47 fisazlsx b) What Celsius temperature corresponds to 98 60F? p 31 + 31 j (42% (p :- l 36 +31 W 1" mac: -_— coo 1%@ in awn-01ml «Cam l c] Findanumberzforwhich L0F=20C l l 'W ”a '"' __..r~32éii t::lsc+szm ‘3C F: PM +55% 2 Q, Homomm gm— \ferlw‘cd SUE“ .7C~'—0\Xl‘€ «QHechm_ W 3 Quadratic Function (42) I Miniti ' adratic function [always degree 2) is defined by an equation of the form ‘ ‘ -—~‘ , 2 bx+ 0 here a, b, care constants, with a i 0 he aph . quadratic functions is a parabola. ‘ Graphs representing a quadratic equation: Axis of symmetry ; X __--: h M W W W‘h-{Jhm tN-fm'ooh. . Vertex form of a parabola: '_w Ex: Rewrite the function f(x) = x2 + 6x — 3 in the vertex form. Completing the square method: Perfegf- g-LW. rm -: 6 xi (0x15) «4 5 «1 0+5) s mu W °~ . In general given a quadratic function of the form y = f(x) = we2 + bx + c.) express it in the vertex form by completing the square. , fl 2-. Q0t1+bvt+ C; VQijrex ‘fvrm golf 6\ %M£€mhb mffifiemlta 0x firm-271W x/ [email protected]( x—- 1») H44 (MK) : Vertex How does the constant a plays a role? “52) "'2 3) Ll: loll 4l {1014! ,9 (ti-'1 L .LM 215) 3". _ 3 Now let’s go back to the first examp] 0 find the X- intercept y- intercept and the L1 [“5 $ general shape of the parabola. Graph the parabola and note how the value and sign - tolw of the constant a determine the shape. gm %: M 1+97 3 Ldm‘l’ (91:0) . Gréfia .. the parabolafiéfli) —2x2 + 2016 +15 by determining its vertex, y—intercept, "a axis of s misery. Do’ef’ii‘rthe raph open up or down and why? i \ x..‘,?§.‘:;‘-2“ ”2—3 “‘1 Find the vertex by compieting the square method: \. 1:2“ > +5”? (L +y§ +50 T, L;— 1 9‘5 7‘ 2:— Wm" W 2 C ) > Lj; Jflvhff‘i’ ‘4 I (WK) :2. VMM Check your answer from above by using the vertex formula: Urfiiwkwfli’f/W 4};me ‘. Using the vertex formula: For Vertex 00.14) 1 In general given a guadratic function of the form y = f(x) 2 cm2 + bx+ c, express it in ' the vertex form by completing the square. _ Vera/C X {lawman 0&3 0\ @flumflmh‘cs YQfTa/wnhnfi a Nada : 1 . ' ' fix) :QCXE%\+‘% . T -19 'Verwx :1. MK . “the /2-0\ C) )O‘DVEXW k:é_}ic;£ '40» l) H? a>_0, ‘U/ua Fmtmu (990115 W19 U a Mime, W PWL“ ammfl 3) W“ \Q‘ 7‘ (—IVUKVQ ?WO[A (‘5 mmwer 4 W \01\. 4 \ ' “I if“? MW ‘5 ”Gide” {BM 44! H?“ 61 23,6123 mgr) Now let’s go back to the first exampl to find theix-intercept, y-intercept and the general shape of the parabola. Graph the parabola and note how the value and sign of the constant 3 determine the shape. t6: Xl+w~3 , ”MN“ (‘1: O) - O : x+tox~3 tbs {Marmara W +0 salt/*9 a a :1 \- x... 40 a: W b: T a :"3 (X: _ (0 —_\_— filtth—MDC-B’) I 3 *(Q 4— {3394-17— 10) 9. Ex: Graph the parabola f(x) = —2)c2 + 20x + 15 by n r A. Rnng it ertex y- intercept axis of symmetry. Does the graph open up or downlandnH why? Find the vertex by completing the square method: *0“ : ~2X1+20x 445‘ Check your answer from above by using the vertex formula: \fwlex {armla : \n :— “b/w a: ‘" H I “no—.9 - K '2: 403:9 51> a 46-93(15) "'" (”like DC x—inter rcept: W 3: O RDWV ij mgkuhm O @x +10K® MHIHFlmsflgy 23 ‘ _2(x— 53+(05‘ -._____, Mob *7)?ng 20 w :bfflé UOJr fat/W . “‘2, I 75 ngiff fifilfijg Y'intercept: $2M x: O ‘3: ._t‘-'—i_m__.l.\;fi-.._m._'..;.m'_1__‘_A.‘ ' ' Setting up Equations that define functions (4.4) Steps: 1] After reading the problem carefully 'draw a picture or make a table that conveys the given information. Figure out what is given? 2] Identify what the problem is you to find. Most of the time the prOblem will ask you to find a particular quantity or a formula for a quantity. Assign a variable to denote the key quantity. Figure out what are you looking for? 3) Label any other quantities that appear relevant. Jot down what you already know systematically. Figure out if there are expressions that relate these quantities. You want to express the resultant expression in terms of the key variable identified in step 2. Answering 4) Use substitution to express the final equation in terms of theake ' ‘ ariable. W cit/122% Overall, find a strategy that will connect what you know with what you are looking for to solve. maid ”€— Ex: A rectangle is @in a circle of diameter linch_s. Ex ress the perimeter and the area of—the rec angle as a functiontffits widthx. Follow the'logical thought process-as describedmwe. 9 \Vl (Fm? want? i Final ECU M N12 3 mmfllol-W ' VS 'Vt‘r/KCK' ‘VM 07w 0&3 (TN W JFWS Egix: 'The‘product 0f- W'QEXE‘JFESS the sum of thfe'"squares of the twong " - numbers as a functio'n‘bf a Single variable. '5'— ”5"." A (X +%)2— Blair, Gwyn/lame, SkqQS 1M?) i>wmelflil "* 251 “"be away): (w—xofl (Hi-t “'2 Pg; mm. 2 GWFWM Maximum and Minimum Problems [4.5) In this section we will learn how to set up equationsof degree 2 that will help us find the maximum and minimum. W Recall: A standard quadratic function of the form y = me2 + bx + c can be rewritten into the vertex form 32 = a(x — h)2 + k where (h, k)'= vertex of the parabola and Qua h =;—:, k = 432:2 = f(h) w—l—‘omia If a>0, then the parabola opens upward and the vertex in that case represents the w minimum point. verl‘fix ;: minim ‘3?" "Jr (“1 L) If a<o, then the parabola opens downward and the vertex in that case represents the maximum point, /.-\Clnll4) : MAX ‘7‘“ Theorem: For y z m2 + bx + c representing the p ' a the x—coordinate of the maximum or the minimum takes place Strategy for solving max/min problems whose model equation is of the form y = axz + bx + c (quadratic expression in one variable) ' Steps: 1) Express the quantity to be maximized or minimized in terms (if a of a single variable. The equation should be a quadratic. —b 2) Use the vertex formula 1: : 2— to locate the x-coordinate of the vertex. If a a " the parabola opens up, you got a minimum at x = _— and if the parabola 2a _ b 1 opens down, you get a max1mum at x = — 9‘?ng ‘9 b “Wm?“ ”M Wm M ' 7W éivtaie Vmiofle Lit ts\ ATLM'SMGHS’) ‘2\ :lWfl WAX/mih b3 X:__b/Zé\ Wt \W VMIM 3] After you determined the x—coordinate of the vertex, relate the information to the problem at hand; ofteniyou will need to findthe function value at the given x—value. . 37m;- “1 J .7/1 EX: Suppose you have 600 ft. of fencing with which to build three adjacent _ rectangular corrals as shown below. Find the dimensions so that the enclosed area is as large as possible. Ex: A baseball is thrown straight upward and its height as a function of time t is given by Mr) 2 —16t2 + 321. Find the maximum height of the bail and the time £1th “"1 (0 1g: 32 which the height is attained. ‘ "'” I W6- . .. ' Give/{Al ' h(‘E) :sf‘a‘filfi OUSWCQ bah/Jew (mi?) (AA/«Q (4,0) ...
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