College Algebra LinC
College Algebra LinC MAC 1105
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This 67 page Class Notes was uploaded by Deshawn Dickens on Thursday October 29, 2015. The Class Notes belongs to MAC 1105 at Valencia College taught by Fitzroy Farquharson in Fall. Since its upload, it has received 16 views. For similar materials see /class/231222/mac-1105-valencia-college in Calculus and Pre Calculus at Valencia College.
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Date Created: 10/29/15
Solving Linear SVstems of Two Variables Consider the following two equations Let C lOb 100 represent the cost per loaf R 90b represent the revenue per loaf A possible application problem represented by the former two equations and graph is as follows The Bread Alone company spends 100 each day for operational expenses ie lights payroll insurance etc The company also spends 10 each time it makes a loaf that same day The revenue generated each day is only from selling bread at 90 a loaf How many loaves would have to be sold to break even with the operational cost Graphing keeps up with both equations simultaneously the revenue and the cost At the point where both lines intersect we have a common number of loafs at the same dollar value Systems of Equation Technical When a second value number is obtained from an initial value introduced the relationship is called an equation ie C lOb 100 represents the cost per loaf When other equations are added that depend on the same number being introduced all the equations involved are called a system of equations C lOb 100 R 90b represents the cost per loaf represents the revenue per loaf III A B Solving the SVstem bV Steps 1 Optio Optio Optio Optio The intersection of the graphs of these equations represents where a common value results We are interested in finding the same results for the revenue and the cost C This meeting point is called the solution of the system Methods for finding the solution of a system Modeling Graphing calculator Algeb raically RIJI LetC ylRy2andbx Now yl le 90 y2 90x Window Setting WINDOW Xmin lO Xmax 10 Xscl l Ymin lO Ymax 10 Yscl l Xres 1 Press GRAPH wait for graph to complete On Screen Choose 11 Calculate 5 intersect n 1 Curve Enter n 2 GI Curve Enter 11 Guess Enter 5 Solutions appear as x 1125 y 1015 Therefore b 1125 is the number of loaves sought and C R 10150 represents where the operational cost is the same as the revenue Solving the SVstem A 39 LetC R Now 10b 90 90b Therefore C 10b 90 R 90b 90 80b C 101125 90 R 901125 1125 b C 10125 R 10125 Solving Nonlinear SVstems of Two Variables 1 Follow the case see page 17 Let h 16t2 1600t b 1000 represents the height of a safety line represents the height of a building A possible application problem represented by the former two equations and graph is as follows A safety line is fired by a canon towards a building to keep it from toppling The building is 1000 ft tall The safety line is fired upward at 1600 ft times each second but gravity pulls it down at 16 ft times each squared second When will the line reach the top of the 1000 ft building Graphing keeps up with both equations simultaneously the height of the line and the height of the building When both graphs intersect we have found a common height for the line and building at a common time Systems of Equation Technical When a value number is obtained from the value of another number being introduced the relationship is called an equation ie h 16t2 1600t represents the height of a safety line When other equations are added that depend on the same numbers being introduced all the equations involved are called a system of equations h 16r2 1600t represents the height of a safety line b 1000 represents the height of a building The intersection of the graphs of these equations represent a common height at a common time This meeting point is called the solution of the system 20 III Methods for solving for the Solution of a system A Modeling Graphing calculator B Alng raically Solving the SVstem bV J quot Steps 1 Lethylby2andtx Now y1 16t2 l600t y2 1000 3 Window Setting WINDOW Xmin 25 Xmax 25 Xscl l Ymin l8OO Ymax 1800 Yscl l Xres l 4 Press GRAPH wait for graph to complete On Screen Choose Option Calculate 5 intersect Option 1 Curve Enter Option 2 GI Curve Enter Option Guess Enter 21 5 Seluueus appear asx 7 3 9 y 7 1000 TherefDrE 1 7 3 91 are ume 11 takes me Inn to reach the top DthE 1000 n bulldmg Sol g Lh Systcm Algebr Cally Lac 7 11 Now 7163 1600 71000 7163 1600t71000 7 0 r 7 me quadraue formula Dmele both 5113 by 74 41 7400 250 7 0 meLhe quadraue equaan above we see that a 7 4 b 7 7400 and c 7 250 Now by usmg me quadraue formula we can solve me equaan for r b vbz 4ac 23 t NOV quot 499050 24 7 400 1 50000 7 4000 8 400 t e 56000 f Therefore r 7 3 9 Direct and Inverse Variations All linear functions where m is a real number can be described by an equation of the formy mx b In the special case where b O the equation becomesy mx Such an equation which meansy varies directly as x ory is directly proportional to x is called a linear variation Example Given a gallon of premium unleaded gas costs 122 at the pump find the cost of a purchase of 5 75 10 and 17 gallons of premium unleaded gas Find a rule for the function that describes the cost of premium gasoline in terms of the number of gallons purchased NOTE The term direct variation indicate that the output can be obtained by multiplying the input by a constant value here the value is 122 The inputoutput table is given below Input number of gallons Output cost of gasoline 610 75 915 10 122 17 2074 A rule for the function is y l22x We notice from this equation as expected from a direct variation that as the input increases the output increases as well constant is positive In general y varies directly with x or y is directly proportional to x If there is a nonzero number k such that ykx The number k is called the constant of proportionality The graph below shows the relationship between y and x if y varies directly with x and k gt O x Z 0 NOTE the constant of proportionality is the slope of the line Y ykxkgt0sz 23 Guidelines for solving variation problems 1 Write a general formula that involves the variables and a constant of variation or proportionality k 2 Find the value of k in guideline l by using the initial data given in the statement of the problem 3 Substitute the value of k found in guideline 2 into the formula of guideline l obtaining a specific formula that involves the variables 4 Use the new data to solve the problem Example For a certain gas enclosed in a cylindrical container the pressure P in newtons per square meter varies directly with temperature T in kelvins Note that P is dependent on T If the pressure is found to be 50 newtons per square meter Nmz at a temperature of 200 k find a formula that relates pressure P to temperature T Then find the pressure P when T 240 K Solution Guideline 1 Guideline 2 Guideline 3 Guideline 4 Write a general fromula P kT is the constant of proportionality Find the value of k We know that when T 200 when P 50 so 50 k200 divide both sides of the equation by 200 i the constant of proportionalityk gt k 2 Substitute k i into the formula in Guideline 1 P T this is the equation of proportionality Using P T and T 240 k we get 1 P a240 60 Nmz 24 Graph showing the relationship between the pressure P and temperature T 100 200 300 400 Using the TI graphing calculator We can use the equation P T and the TIcalculator to find the new pressure when T 240 k Steps 1 LetPyandTx We gety ix 2 E 4 3 Press and enter the values as displayed Display x min O x max 400 x scl 20 yrnin O ymax 100 y scl 5 25 4 Press GRAPH Display To find the new pressure when T 240 k 5 Press then enter the value 240 Next press Display x yl 26 Invexse Vaxlat39mn Exam le 50 65 7o 90 and Fmd NOTES constant Value hm th Valu S 258 by the mput The mpmDutput table ls gwen below Input 7 speed Output a ham so 5 16 houxs as 3 97 houxs 7o 3 69 houxs 90 2 s7 houxs 120 2 15 houxs Amle fox the functan 15 y 7 253m constant ls posltlve n genexal y vanes lnvezsely with x ox y ls lnvezsely pxupuxtiunal to x If them ls a nonzexo constant It such that 7 k y x The numbex It ls callecl the constant of pxoponlonallty The yaph below shows the xelatlonshlp between y anelx as y vanes mvexsely with x andk gt o x gt o Exam le v of a len thhe smng L If a sumg ts 42 mches long and Vlbxates 240 mm pa second what ts the length of a stung that Vlbxates 420 mm pa second7 Solutlon Gmdehne 1 Wnte a genexalfomula V e 5 L Gmdelme 2 Fmd the value oflt We know thatwhen L a 42 v a 240 so 240 e L 4 m It a 10030 constant of pxoponlonallty Gmdelme 3 Substltutek a 10030 mto the formula m guldelmel V g equatlon of pxoponlonallty Gmdehne 4 Usmg v a 1080 and v a 420 240 7 10080 L Le M 724mches 420 solve usmg the TI yaphmg calculath follow the same pxoceduxe outhned In le To the pxevmus exam nest Va ex Invexse nut Funct ns that axe N Exam le salmxv n M om nd v u v she settled during the month Fmd ha mcome fox a gwen month 1 she sold 30000 140000 350000 ox 1000000 womb of pxopenles Dvex a foul month pexlod ls thts n example of a duect Vaxlatlon ox an mvexse Vaxlatlon7 Input 7 sales Output 7 Income 80000 22400 140000 24200 350000 30500 1 000000 50 000 NOTES The rule for the funcurm 15 y 7 20000 0 03x Is dus a dnecl vanauon Check dowehopey 7 In my 7 1087 22400 7 1430000 k 7 22499 7 80099 50000 7 141000000 k7 759999 i L 005 1099999 100 20 k1 unconsan so y 15 n01 a slum wnaunn ng Graphing Rational Functions A function f is a rational function if m f X goo where gx and hx are polynomials The domain consist of all real values of x except for those values where gx 0 Recall the domain of a rational expression For instance let39s consider 2xl 3x 4 Domain All reals except where 3x 394 0 this mean x i 2x The same thing would apply for rational function For instance if fx 2 4 X Domain All reals except where x2 394 0 this means x 39 39 2 or 2 and x 2 and 2 are called the vertical asymptotes of the function However it is important to answer a couple of questions when graphing rational functions 0 What is the behavior of the function as x get closer and closer to the vertical asymptote 0 What is the behavior of the function as x get larger through positive values or smaller through negative values 0 When using the calculator to graph y l is calculator sketching in the x asymptotes 0 What is a way calculators can correct this problem 53 A Hrm nntal A of a Rational Function Let us graph fX i using the graphing calculator We use the standard X window push zoom 6 and obtain a graph similar to the one below One notices that the curve is getting close to the XaXis line y 0 When X is large whether it is positive or negative The graphing calculator actually is not able to show the curve distinctively for values greater than 6 or smaller than 6 when we use zoom 6 However an inspection of the inputoutput table 2Dd table shows that large values of X correspond to yvalues that are certainly not zero Let us set the table to start at X 100 X 1000 and X 1000 Push WINDOW for TBLSET and change the value for TblStart to 100 We can also change TBL which is the value of increments between any two inputs Let us obtain a screen such that TblStart 100 rb1500 and 54 We now push TABLE to inspect the inputoutput table 100 600 1 100 1600 2100 2600 3100 TbIStart 1000 Tb1 1000 01 00167 91 E4 63 E4 48 E4 38 E4 32 E4 001 5 E4 33 E4 25 E4 2 E4 17 E4 14 E4 Recall 91 E4 is scienti c notation 91 134 9110394 9100001 000091 We notice that as X grows the value of fx diminishes as it approaches 0 Now Do the same to check the behavior of y when X is large and negative What do you notice One would set the table to start at 1000 and go by increments of 500 for instance This is what one obtains when using Tblstart 2000 Tbl 1000 VI 55 Clearly the outputs are getting closer and closer to zero In this case though as the absolute value of X increases the output y is increasing to get closer and closer to zero YA Summa Q ola lar d 39 A 1 d 39 ger an sta s os xeis growmg s X gets arger an stays negative 3251 mfg t in positive fx is increasing toward zero direction x is growing Origin in the negative direction 3 QUESTION Is the curve crossing the XaXis anywhere N Is this possible The answer is amp ANSWER Well if there existed an Xintercept that is a point of intersection between the curve y l and the Xaxis this point would have a ycoordinate equal to 0 x That is 0 1 Is this possible x Could we divide l by a number and obtain zero 56 No we cannot Therefore the curve does not cross the xaxis 57 The XaXis is called a horizontal a ptote We can recognize the horizontal asymptote from the graph because the curve is getting closer and closer to the line as X increases or decreases These are a few other examples of a ptotes i j y 2 is a horizontal asymptote ii y 0 the XaXis is a horizontal asymptote Note that in this case the curve crosses the XaXis at X 0 However since the curve gets closer to 0 X 0 without touching the line when X grows larger y 0 is still an asymptote In this case the curve gets closer to the XaXis only when X grows large and positive 2 This is not the case when X is large or small and negative However since we approach the line y 0 at least on the righthand side it quali es as an asymptote 58 PRACTICE Use your graphing calculators to nd the horizontal asymptotes with the following cases You can use the graph of the function and the table of inputs and outputs fx X i The xasymptote isy 1 X 2X l gx 3 The horizontalasymptote is y 2 x 2 hx 2 i The horizontalasymptote isy 1 X2 2 rx The horizontalasymptote is y 0 x 2 x x 2Xl X3 4X 1 The horizontalasymptote is y 0 59 B A Quick WaV to Find l lori mtal 4A of a Rational Function if the Awptotes Exist The quick way out is to compare the degree of the polynomial in the numerator and the degree of the polynomial in the denominator These are the three possible situations 1 w No horizontal asymptotes degree on the bottom is smaller x3 2x2 3 3x32x2 4xl x2l7x27 2x23x5 Example Verify that these rational expressions do not have horizontal asymptotes by 9 graphing them using your calculators 2 w The horizontal asymptote degree at the bottom is larger is the XaXiS line y 0 x 4x35x2 2 1 x23x l73x5 3x34x2573x 2 Example Check that these expressions have y 0 as a horizontal asymptote 3 degree on top and degree at the bottom are identical Emle 3x2 2x2l p 39 y 4x 5 y 3x25x6 5x4 2 y 6x4 3x37x2 1 There is a horizontal asymptote y Where a is the coef cient of the term with the largest exponent on top and b is the coef cient of the term with the largest exponent at the bottom 60 3x 2 2 Example y 4X 5 has ahorizontal asymptote y 4 2 y A has ahorizontal asymptote 3x25x6 5x4 2 has for horizontal asym tote y 6x4 3x32x2 1 P C Vertical Aw ptote s y If we examine again the graph of y i by using x a calculator we notice that when X approaches zero the values of y are increasing without bound X and the curve is getting closer and closer to the 0 yaXis The yaxis acts as a vertical awptote Let us also use an inputoutput table to illustrate the presence of the vertical asymptote Example 1 TblStart 0 Tbl 01 We push TABLE to inspect the inputoutput table 61 Example 2 TblStart 0 Tbl 001 l In the case of y the vertical asymptote X 0 corresponds to setting the X denominator to zero Indeed X 0 is the one value that is not in the domain of the function In general vertical asymptotes are obtained by setting the denominator to zero 2 X 3X l f has a horizontal asymptote at X X 2X y 0 and three vertical asymptotes corresponding to the equation For instance the function fX X3 X2 2XZO XX2 X 2O XX 2Xl0 xO X l It is important to note that on the graphing calculators the three vertical lines X 0 X2X lareshown 62 D Summag In order to nd horizontal asymptotes of a rational expression one just inspects the degrees of the polynomials in the numerator and the denominator 1 There are no horizontal asymptotes when the degree of the numerator is larger than the degree of the denominator 2 There is a horizontal asymptote the line y 0 when the degree of the denominator is larger than the degree of the numerator 3 There is a horizontal asymptote y when the degrees of the denominator and numerator are identical a is the coef cient of the term with the highest exponent in the numerator and b is the coef cient of the term with the highest exponent in the denominator 4 The vertical asymptotes correspond to the values that mallte the denominator go to zero They are of the form x c where c is a zero of the denominator 63 1 In the preVious section we examine the behav10r of y and made some general conclusion X about the behavior of the equation as it approaches the vertical asymptotes and as X increase or decrease Let39s use this information to graph a few of these rational functions by hand Example Sketcharough graph of the equation fx 2 X 1 6 X X Solution 1 The Xintercepts Recall the values of X causing numerator only to be zero are the Xintercepts X 391 0 X 1 II The yintercept 0 1 1 LetX0 yzng III Horizontal Asymptote Degree of the polynomial in the numerator is smaller than the degree of the polynomial in the denominator therefore y 0 is the horizontal asymptote IV Vertical Asymptote X2 39X 396 0 X 3X2 0 X3 2 V Sub interval RHRZERSER ERS 210123 VI Test Values 64 R1 3 R2 1 Ry R4 2 R54 65 VII Table fx gt 0 21 3 7 fx lt 0 o 3 2 1 3 VIII Graphing a IA A See Section VII 66 2 Example Sketch a rough graph of the function fx 2X 2 x x Solution 1 The Xintercept X2 0 Recall or 111 VI X 0 multiply 2 The yintercept y0 Horizontal Asymptote Degree of the polynomial in the numerator is equal to the degree of the polynomial in the denominator therefore y l is the horizontal asymptote Vertical Asymptote x2 39x 2 0 x 2x l 0 X2 l Subintervals R1 lel R3 l R4 4321012 34 5 Test Values R139 2 R2quot R3l R44 67 VII Table fx gt 0 o 32 2 7 fx lt 0 1 0 0 2 VIII Graphing a Cquot In R1 To determine if fx crosses the horizontal asymptote test a value f 3 L 2 lt l 9 3 2 11 In R2 Use the same procedure to determine if fx crosses the horizontal asymptote It doesn39t 68 Example Sketch the graph of the function X2 9 2X 4 fx Solution 1 III The Xintercept x2 90 X 3x3 0 X3 3 The yintercept 0 9 0 4 Letx0 y 2 No horizontal asymptote Since the degree of the polynomial in the numerator is greater than the degree of the polynomial in the denominator therefore an oblique asymptote exist Let39s nd it xl 2X 4 X2 OX 9 X22X 2X 9 2x4 5 AsX gt iooy xl Vertical Asymptote 2x 394 0 X 2 69 V VI VII VIII 5 ub intervals 111311141115 54 21012 45 Test Values R19 1R2 2 R31 3142 R54 Table Graph 21 70 0 0 0 0 As X gt 0 fX gt oo As X approaches zero from the left y increases without bound As X gt 00 fX gt 0 As X gets larger approaches positive in nity y is decreasing towards zero As X gt O fX gt ea As X approaches zero from the left y decreases without bound As X gt 00 fX gt 0 As X gets small approaches negative in nity y is decreasing toward zero 71 Linear Approximations In many scientific investigations we may have a collection of data for which we would like to find a mathematical model linear model that relates them In many cases the methods used are linear interpolation estimation between known data points or extrapolation making predictions beyond the range of known data as shown by the graphs below Known Data Estimated Point Point Known Data xPoint Linear Extrapolation Linear Interpolation The collection of points are called a scatter plot Once the points are plotted we will attempt to find the line that most closely represents the plotted point Technology Most graphing calculators have built in programs that will calculate the equation of the bestfitting line for a collection of points 2 or more For instance to find the bestfitting line on a TI83 follow the following procedures delineated below Steps 1 Press STAT Display EDIT CALC 1 Edit 2 SortAC 3 SortDC 4 ClrList 2 Press Display L1 L2 L3 3 Enter x coordinatess of the data points in L1 Column and ycoordinates in L2 Column 4 Press MODE Display I Press CLEAR To clear home screen Once the home screen has been clear go to Step 5 5 Find the regression equation for our data Press STAT then us g to get CALC menu 6 Press forTI82 or Press forTI83 Display Lin Reg ax b 7 Press Display LinReg ax b y ax b a some number b some number 8 Press Display y1 I Y2 y3 Clear all equations Y4 YS 9 Press Display Xy E EQ Box PTS ln 2zg 3Sx 40x 5 6Sy 7l0y 10 Use to select EQ ll Press for RegEq on TI83 or press 01 Press for RegEQ on TI82 or press 12 Equation should appear at y1 Reg Equation in the form y ax b 13 Setup window dimension 14 Press Display 15 Press Example STAT PLOTS 1 Plot 1 On N LILZo Plot 2 Off N LSL Plot 3 Off N L5L6 4i Plots Off 3 w GRAPH Turn on the appropriate STAT PLOT l The temperature in Orlando dropped from 86 at 300 pm to 81 at 600 pm Use linear interpolation to estimate the temperature at 400 pm and extrapolation to estimate the temperature at 700 pm Solution Let y represent the temperature at time x pm We will look for a linear equation using the graphing calculator that relates x and y Steps 1 Press STAT 2 Choose 1 or Press 3 Clear data in L1 and L2 4 Enter x coordinates of the data points in L1 Column and ycoordinates in L2 Column as shown below L1 L2 L3 3 86 6 81 L26 5 Press MODE 4 make sure the home screen is clear 6 To find the Regression Equation for our data Press STAT use to get CALC menu 7 Press for LinReg ax b TI82 or Press for LinReg ax b TI 83 Press ENTER Display LinReg y ax b a l666666667 b 91 r l 8 Press Clear all equation 9 Press then use to get EQ menu 10 Press to select RegEq on TI83 01 Press to select RegEq on TI82 11 Press WINDOW Display TABLE SETUP TblMin 38 Use to select Ask for the Indpnt A Tbl 1 independent variable Indpnt Auto Ask Depend Auto Ask 12 Press then enter the times as shown below Display x yl 4 8433 7 7933 x Enter the number 4 then press ENTER Follow the same procedure for the next number Example The points shown below shows best times for various women running 100 meters and 100 meters 111 524 115 547 119 574 122 579 127 613 130 630 We can use the graphing calculator to find the least square regression line Solution Steps 1 Press then choose 2 Clear data in Column L1 and L2 3 Enter xcoordinates of the data points in L1 column andycoordinates in L2 Column as shown below Display L1 L2 L3 11 1 524 11 5 547 11 9 574 12 2 579 12 7 613 13 O 4 Press 5 Press then use g to get CALC menu 6 Press for TI82 then press 01 Press for TI83 then press Display LinReg y ax b a 5489637306 b 8458290l55 r 9956567355 7 Press then for STATISTICS menu 8 Select EQ then press or for RegEq Display y1 5489637X 84 5829 Y2 Y3 y4 YS 9 Press WINDOW the enter the following window settings WINDOW FORMAT xmin 1000 xmax 1400 xscl 1 ymin 5100 ymax 6400 yscl 1 10 Press then to the appropriate STAT PLOT on 1 1 Press GRAPH Display Quadratic Functions A function of the form ax2 bx c a 7k 0 is a quadratic function Equations of Quadratic Functions The equation of a quadratic function may be written in the form yax2bxc or fxax2bxcwhereat0 Examples of some quadratic functions are y2x23a y2x20x3 y 3x2 7x 2 The graph ofy ax2 bx cwhere a gt O The graph of a quadratic function is a parabola that opens down or up depending upon the value of a If a gt O the parabola opens up Otherwise if a lt O the parabola opens downward Let39s examine the graphs of the following functions below with a gt O 2 Ngtlt Klt lt Klt gtltNgtlt n By observing each graph we notice Since a l a gt 0 all graphs open upward The graph of y x2 l is the graph ofy x2 shifted up 1 unit The graph of y x2 l is the graph ofy x2 shifted down 1 unit 30 Now let39s examine the graphs of the following functions with a gt O yx x2 12 E12 yxl2 E yXI Notice All graphs open upward The graph ofy x l2 is the graph ofy x2 shifted to the right 1 unit The graph ofy x l2 is the graph ofy x2 shifted to the left 1 unit The graph ofv ax2 bx cwhen a lt 0 If a lt O the graph of a quadratic function opens downward For example let39s examine the following graphs of these parabola where a lt 2 y 2x2 y 2x2 l y 2x2 1 1 Notice All graphs open downward The graph of y 2x2 l is the graph ofy 2x2 shifted up 1 unit The graph of y 2x2 1 is the graph ofy 2x2 shifted down one unit 31 Intercepts The yintercept The yintercept of a function is the point where the graph crosses the yaxis that is the point where x O A quadratic function of the form y ax2 bx c will have only one yintercept O c To find the yintercept set x O and solve for y the constant value in ax2 bx c the c value is the yintercept For example to nd the yintercept of y x2 x 2 a Setx O y O2 O 2 b Solvefory yOO 2 y 2 c The yintercept is the ordered pair 0 2 Note this value was given in x2 x 2 The xintercepts The xintercept of a function is the point where the graph crosses the xaxis that is the point where y O A quadratic function may have 2 l or no xintercepts Example to find the xintercepts ofy x2 x 2 a SetyO Ox2x 2 b Solveforx Ox2x l x2O or x lO x 2orxl c The xintercepts are the ordered pairs 2 O and l O m The completed square form of the quadratic function y ax2 bx c is given by y ax h2 k standard form Standard form where the point h k is the vertex 32 The venex h k Dthe quadrauc func Dn can he found by compleung me squaxe Dr usmg me formulas 2 2 h gandk c7 c7 dsmevmae aghx ao For example graph the uheueh y 7 2 8x 5 Fmd us vmex and x and ydhtexeepu 1 Letusgxaphy 2x13x Sby rstfmdmgthevmex ldmufyabandc a7 2h78e7 5 a TD ndh let h7 i 23 3 3 72 74 2 b b TD ndk letk 7 c7 7 4g 52 s4 7 7 7 7 5 7 7 5 s 7 3 5 42 5 c The Venex 1 me ordered pamz 3 2 xmtexcept Lety 7 o o 7 2 8x 5 muluplyby 1 o 7 2 8x 5 he facmxable smce h 7 43 mm square number h 74 7 73V 7 4mm 7 64 74o 7 Usmg Quadratic formula 7 b 1132 7 43c 22 7W 7 W32 7 425 m2 5164740 4 Eam aaxmf 4 4 The xrintexcepts axex 7 2 and 2 e E 0 and The decimal appxoxima ons axe 3 xrintexcept Letx e 0 y 2m 30 5 y s The yrintexceptis a 5 4 Crxaph Dpensdown hoxizontal Venicalshi shm Summaxy uf Gxaphical Txansfuxxnat39mn The table below summaxizes me pxindples developed A221 inat39mn 0f Quadxat39lc Funct39mns The helght of the ball fox tune t ls glven by Mt Tatzbt c Whemalt0 The path ofthe ball ls a paxabola an xeslstance anel Dthex fumes axe negllglble as shown below Tm unntd to reach tht ball 11kt Lh I mluan 0i mamum hught gmund x mmttpt tht thud X 7 eb 23 Fox example a ball ls tloown mtD the an by a chlld The helght h In feet ls glyen by the f nnula Mt e 166 alt 4 H4 downw xd e a upwaxd mltlal accelexatlon Kate he Whexe t 15 the tlme In seconds aftex the ball has been tluown a Fmd the tlme xequlxed fox the ball to attam lts manmum helght b State the manmum helght of the ball c Fmd an appxopxlate Vlewlng xectangle fox the gxaphmg calculath 1 At what tlmes wlll the ball be 5 5 feet above the gxound7 e How long wlll the ball be m mgha S 01mm To nd the ume requ ued for the ball to attam It mammum hexght let re 4 2 16 2sec b To fmd the maxn num hexght 13me ball let 1 68fL c Wmdow Settmg xrmm o xmax 4 25 xscl 1 yrmm o ymax 69 yscl 2 Now use the graphmg calculator to answer the questmns m pan 1 and e Calculator Procedures Steps 1 Press WINDOW than ante wmdow settmg shown In pane 2 Press 3 Press Dlsplay 4 Press then use to enter 55 for Y2 y1 16x2 64x 4 Display y2 55 Y3 y4 YS Y6 Y7 Y2 5 Press then 6 Press to find the intersection for the two lines or Press Display First curve xsome ysome 7 Press Display Second curve xsome ysome 37 8 Press Display Guess xsome ysome 9 Use g to move x close to the point intersection on the right hand side then 10 Display Intersection x 39764235 y 55 Follow the same procedure shown in Steps 19 to find the other point of intersection Therefore after at t 3976 seconds and 0236 seconds the ball is 55 feet above the ground 38 Linear Models In this section we will investigate the relationship between two variables by using equations graphs and tables of values This process is called mathematical modeling In this section we will examine applications involving linear equations of the form mcbyc where a and b are not zero These equations can often model quantative relationships This relationship can be represented by a table of values an algebraic equation a graph For instance Purcell is selling tickets to his son39s basketball game to anyone at his job who has the time to attend the game and 275 to spare To avoid tedious calculations he has made a table TABLE OF VALUES tickets t cost c l 275 2 550 3 825 4 1100 5 1375 Example 1 Yumiko39s long distance telephone company charges a 300 access fee for a call to Tokyo and 200 for each minute of the call t Z 0 a Write an equation that expresses the cost of the call to Tokyo in terms of the length of the call b Graph the equation of part a Solution a Labels Let t length of a call in minutes c cost of a call in dollars TABLE OF VALUES Ordered pairs 0 3 l 5 and 2 7 Equation for cost c 2t 3 Lets use the TIgraphing calculator to sketch the graph Obviously we can graph the equation by plotting the point shown above b TIgraphing calculator procedures Steps 1 Letcyandtx Wegety2x3 2 Press keystroke sequence 3 Enter Window settings Display x min I 5 Press Example 2 In 1970 a threebedroom house in Middville cost 30000 The price of a home has increased by an average of 4000 per year since then a Write an equation that expresses the price of a threebedroom house in Middville in terms of number of years since 1970 t Z 0 b Graph the equation in a c Use the graph to calculate the increase in the price of the home from 1980 to 1982 Solution a Labels let t the number of years p the price of the home TABLE OF VALUES Time t Price p 0 30000 1 30000 40001 34000 2 30000 40002 38000 3 30000 40003 42000 t 30000 4000t Equation p 30000 4000t Ordered pairs 0 30000 1 34000 2 38000 and 3 42000 b Tl graphing calculator procedures Steps 1 Press 30000 4000 keystroke sequence 2 Enter window settings Display x min I x max x scl y min y max y scl 3 Enter the values as displayed Display x min 0 xmax 10 xscl l y min 0 yrnax 100000 yscl 1000 4 Press C Using the calculator to calculate the increase in the price from 1980 to 1982 Steps 1 Press Display TABLE SETUP Tbl start 0 A Tbl l Indpnt Auto Ask Depend Auto Ask Q g 2 Use to select Ask for Indpnt independent variable Do not change the Depend dependent variable 3 Press Display x l yl 4 Enter the first date 1980 then press enter Enter the next date 1982 5 Find the difference between the two values Example 3 In a lake the concentration of toxic chemicals in water is currently 285 parts per million ppm Local environmental officials would like to reduce concentration by 15 ppm each year a Write an equation that expresses the concentration of toxic chemicals 1 years from now b Graph Solution a Labels Let t number of years Let c concentration of toxic chemicals TABLE OF VALUES time t toxic chemical c o 285 1 285 15 2 285 152 3 285 153 t 285 15a Eguaunn 15 235 or y 15x 235 Now yDu graph me equation above usmg your graphmg calculator Intucagts Coordmate of x7 and yrmtemepts a D and D b Examgle 4 a Fmd the x7 and ydntexcep s DthE graph of lSDDx 1300 9000 In Use Lhemtexcepts m gxaph me equaum Soluuon Method Algebra Pmceduxe a To nd me x7 and yrmtacept 1 Letx 0 yrmtexcept 15000 1800 9000 1800 9000 The yintercept is O 5 n Lety 0 xrmtexceyt 1500x 15000 9000 1500x 9000 1500 9000 1500 1500 x 6 The xdntaxcept 1 6 0 lntamepts are 60 and 0 5 b mm me two mtexcepu y Memeel 2 Usmg me Terraphmg Calculatm a To nd the x and yrmtemepts Slaps 1 Solve the equamn for y ISDDx 1500 9000 ISDDx ISDDx 1500y ISDDx 9000 1500y 1500x 9000 4500 4500 9000 y x 4500 4500 4500 Exe 5 s 2 Press x6 tn Enter 3 Using x and y intercepts to setup window Press and Enter the following Display x min 8 x max 8 xscll ymine7 ymax7 yscl2 4 To calculate x intercepts i Press ii Press Display y15x16 5 left bound 2 iii Use to move 1 on calculator to a position to the left of the point of intersection then press Next move 1 to a position to the right of the point of intersection then press Enter Press Enter Display 5 To find the yintercept make sure the calculator is setup to prompt you to enter the xValue in the table m Display 7 5 Factoring Polynomials We begin our study of factoring polynomials with factoring out common factors and factoring by grouping using the distributive law Next we factor general trinomials and consider special cases of perfect squares difference of two squares and sum and difference of two cubes It is important to recognize the various forms of a polynomial and identify the procedures necessary to factor these polynomials Let39s begin with the first step in factoring a polynomial finding the greatest common factor for the terms of a polynomial The greatest common factor for a polynomial is the largest monomial that is afootor of each term of the polynomial Example 1 Factor the greatest common factor 8x4 2x3 Solution 8x11 2x3 4x 2x3 2x3 Note 2x3 is the greatest common factor of both terms 2x34x l Factoring Trinomials Expressing a trinomial as the product of two binomials is one of the most common types of factoring used in algebra For instance when we multiply binomials x 2x 4 the result is the trinomial below x 2x 4 x2 2x 8 We call this trinomial a quadratic polynomial In general a quadratic polynomial in a variable x can be written in the form of bx o where a O The term of is called the quadratic term bx is called the linear term and the o is called the constant term The coefficient a of the quadratic term is called the leading coefficient To factor these trinomials we will consider the following two cases a l and a 1 When factoring a polynomial look for a common factor first Case 1 Factoring Trinomials of the Form x2 bx c a 1 We want to consider trinomials where leading coefficient term is 1 that is a trinomial of the form 3 bx 0 Let us consider the product x mx n and use the distributive property to show how each term of the resulting trinomial is formed x mx n xx xm xn mn x2 m nx mn 39 In general to factor x2 bx c we look for the pair of integers whose product mn is c and whose sum m n is b Therefore when we factor the quadratic trinomial x2 bx c the factors should be the form x m andx n Lets discuss some rules that can be used to determine the sign value for m and n Rules for Determining Sign Values of m and n Consider the quadratic polynomial of the form a bx c where b c gt O and a Z 1 We can use the following rules to determine the sign positive or negative of the two numbers m and n l ax2 bx c 2 two positive numbers 2 ax2 bx c 2 one negative and one positive 3 ax2 bx c 2 two negative numbers 4 ax2 bx c 2 one negative and one positive Example 1 Factor x2 5x 6 Solution Based on 1 above we need to find two positive numbers whose sum is 5 and whose product is 6 x25x6xmxn Since the product of 3 and 2 is 6 which is c and the sum of 3 and 2 is 5 which equals b we write x25x6x3x2 Example 2 Factor x2 2x 15 4O Solution Based on Rules of Signs 2 we need two numbers one negative and one positive number whose sum is 2 and whose product is lS x22x 15 xmxn mn2 m n lS 532 5 3 15 Since the product of 5 and 3 is lS which is c and the sum of 5 and 3 is 2 which equals b we write x2 2x 15 x 5x 3 Example 3 Factor x2 8x 16 Solution Based rule 3 above we need to find two positive numbers whose sum is 8 and whose product is 16 The product of 4 and 4 is 16 which is c and the sum of 4 and 4 is 8 which equals b therefore x2 8x 16 x 4x 4 x 42 Case 2 Factoring Trinomials of the Form ax2 bx c a 7k 1 We want to consider trinomials where leading coefficient term is a 7k 1 that is a trinomial of the form shown above Let us begin by rewriting our trinomial in the form ax2bxcax2mxnxc 41 To factor a bx c we find the product ac then look for the pair of integers whose product mn is ac and whose sum m n is b The rules for determining the signs of m and n in case 1 can be applied in case 2 Next rewrite the quadratic trinomial in the form do 6 then factor by grouping Let39s apply this technique Example 1 Factor 3x2 7x 2 Solution We notice that there are no common factors to the terms 3x2 7x and 2 Since a 3 b 7 and c 2 the product do 6 Two integers m and n must be located whose product is 6 and whose sum is 7 3x27x23x2mxnx2 3x27x23x2x6x2 Factor by grouping 3x2 x 6x 2 x3x l 23x 1 3x2 7x2 3x lx2 Example 2 Factor 6x2 x 15 Solution Identifying the terms of the trinomial as 6x2 x and lS we notice that there are no common factors to these terms Since a 6 b l and c lS the product do 90 Therefore we need to find two integers m and n whose product is 90 and whose sum is l Keep in mind the signs of the two integers m l39n 1 m n 9O l9 lO l 9190 42 6x21 15 6x2 10x 9x 15 Now factor by grouping 6x2 10x 9x 15 2x3x 5 33x 5 Therefore 6x2 x 15 3x 52x 3 Example 3 Factor 12x3 14x2 20x Solution The terms are 12 14x2 and 72x The common factor to these terms is 2x Factor this out 12314x2 20x 6 x2 7 g x g lO g 2x6x2 7x 10 Then factor 6x2 7x 10 we notice that a 6 b 7 c 10 and the product ac 60 Therefore we need to find two integers m and n whose product is 60 and whose sum is 7 Do not forget to write the common factor in front 6x2 7x 10 6x2 mx nx 10 mn m n 6O 1257 12 5 60 6x2 12x 5x 10 Factor by grouping 6x2 12x 5x 10 6xx2 5x2 x 26x 5 Do not forget to write the common factor in front Therefore 12x3 14x2 20x 2xx 26x 5 43 Example 4 Factor 8x2 22xy lSyZ Solution The terms of the polynomial are 8x2 22g and 153 We notice that there are no common factors to the three terms Here a 8 b 22 c 15 therefore product do 120 Therefore we need to find two numbers m and n whose product is 120 and whose sum is 22 8x2 22xy 15y2 8x2 mg n 15y2 mn22 I m n120 12 10 22 12 1 120 8x2 12 10 15y2 Factor by grouping 8x2 ley lOXy lSyz 4x2x 3y 5y2x 3y We get 8x2 22xy 15y2 2x 3y4x 5y 44
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