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# MATH 240 Exam 1 Notes Math 240

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This 62 page Bundle was uploaded by AnnMarie on Thursday October 1, 2015. The Bundle belongs to Math 240 at Louisiana Tech University taught by Jonathan B Walters in Fall 2015. Since its upload, it has received 64 views. For similar materials see Precalculus in Mathematics (M) at Louisiana Tech University.

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Date Created: 10/01/15

Lo ic A logic statement is a sentence that is either true or false Example Leopards have spots Yes this is a logical statement Belts are made from leather Yes this is a logical statement I am hungry No this is not a logical statement due to I being unde ned This sentence is false No this is not a logical statement Mangos are delicious No this is not a logical statement Logical statements are represented with letters P Q R which are propositions Connecting Logical Statements 1 Logical OR V or U Notation PORQ PVQ PUQ This compound logical statement is true if one or both of the propositions is true If both are false the compound logical statement is false This can be represented on a truth table The truth table format using 0 s for false and 1 s for true is shown below for P v Q P Q PVQ 1 1 1 1 O 1 O 1 1 2 Exclusive OR XOR or lt9 Notation P XOR Q P 0 This statement is true only when one of propositions is true This can be seen in the truth table below P Q P Q 1 1 O 1 O 1 O 1 1 O O O 3 Logical AND AND or A Notation P Q This statement is true only when P and Q are both true This can be seen in the truth table below P o PAQ 1 1 1 1 o o o 1 o o o o Conditional Statements Implications have the form if P then Q Where P is the hypothesis or assumption and Q is the conclusion Notation P gt Q or P gt Q This statement is false only when P is true and Q is false This can be seen in the table below P Q P gt Q 1 1 1 1 O O O 1 1 O 1 1 Example If you get 500 points then you get an A If you wash white and red clothing together then your clothing will turn pink Converse of an implication P gt Q is the conditional statement of Q gt P Example If you earn an A then you scored 500 points If your clothing turned pink then you washed white and red clothing together Contraposition of an implication P gt Q is the conditional statement of If not Q then not P Notation Q gt p Example lfA not earned then you did not score 500 points If your clothing is not pink then you did not wash red and white clothing together An implication and its contrapositive have the same truth tables This is shown in the table below P Q Q nP P gt Q Q gt P 1 1 O O 1 1 1 O 1 O O O O 1 O 1 O O 00 O 1 1 1 1 Biconditional Statements P iff Q Notation P ltgt Q The statement P if and only if Q is true only when P and Q have the same truth values This can be seen below in the following truth table P o PltgtQ 1 1 1 1 o o o 1 o o o 1 You can check to see if P gt Q is true and Q gt P is true also Example Students pass MATH24O if and only if student scores are at least 350 points Negations The negation of a logical statement P is the statement P does not occur Example It is 1230 Negationgt It is not 1230 A statement and its negation have opposite truth values Example 3 lt 5 Negationgt 3 2 5 quotx 5 Negationgt x lt 5 orx gt 5 Universal Statement A universal statement is of the form For all property P holds Example For all real numbers x X2 1 H For all engineering majors solidworks is required Existing Statement An existing statement is of the form There exists property P holds Example There exists a car which is receiving a ticket today There exists an integer x such that x S O There exists a fish such that cannot swim Neoation of Universal Statement The negation of a universal statement is of the form There exists a such that property P does not hold Example There exists a real number x such that x2 1 There exists an engineering major such that solidworks is not required Neoation of Existing Statement The negation of an existing statement is of the form For all property P does not hold Example For all cars on campus they are not receiving a ticket For all integers x x gt 0 For all fish fish can swim Neoation of AND OR and IfThen De Morgan s Laws 1P AQ P V Q 1P VQ P Q Examples Jacob has cats and dogs Negationgt Jacob does not have cats or dogs 4 is the product of 2 gtlt 2 and 1 gtlt 4 Negationgt 2 gtlt 2 4 or 1 gtlt 4 4 If P then Q Negationgt P A SQ If the professor shows up to class then we have class Negationgt If the professor shows up to class and we do not have class T T u 39 lfx IS the solution to ax2bxc O then x V22 4 or b V2 4 NegatIongt lfx IS not 2 2 the solution to ax2bx c O and is 1 0r 1 Notation Universal Quantifier gt V means for all Existential Quantifier gt El means there exists orthere is X e D gt x is in the set D or X is an element of set D Sets E gt Set of real numbers Q gt Set of rational numbers 12 910 Z gt Set of integers 3 O 4 N gt Set of natural numbers 1 2 3 4 5 C gt Set of complex numbers Example Vx e D x2 1 Negationgt Elxe R x2 1 Lo ic A logic statement is a sentence that is either true or false Example Leopards have spots Yes this is a logical statement Belts are made from leather Yes this is a logical statement I am hungry No this is not a logical statement due to I being unde ned This sentence is false No this is not a logical statement Mangos are delicious No this is not a logical statement Logical statements are represented with letters P Q R which are propositions Connecting Logical Statements 1 Logical OR V or U Notation PORQ PVQ PUQ This compound logical statement is true if one or both of the propositions is true If both are false the compound logical statement is false This can be represented on a truth table The truth table format using 0 s for false and 1 s for true is shown below for P v Q P Q PVQ 1 1 1 1 O 1 O 1 1 2 Exclusive OR XOR or lt9 Notation P XOR Q P 0 This statement is true only when one of propositions is true This can be seen in the truth table below P Q P Q 1 1 O 1 O 1 O 1 1 O O O 3 Logical AND AND or A Notation P Q This statement is true only when P and Q are both true This can be seen in the truth table below P o PAQ 1 1 1 1 o o o 1 o o o o Conditional Statements Implications have the form if P then Q Where P is the hypothesis or assumption and Q is the conclusion Notation P gt Q or P gt Q This statement is false only when P is true and Q is false This can be seen in the table below P Q P gt Q 1 1 1 1 O O O 1 1 O 1 1 Example If you get 500 points then you get an A If you wash white and red clothing together then your clothing will turn pink Converse of an implication P gt Q is the conditional statement of Q gt P Example If you earn an A then you scored 500 points If your clothing turned pink then you washed white and red clothing together Contraposition of an implication P gt Q is the conditional statement of If not Q then not P Notation Q gt p Example lfA not earned then you did not score 500 points If your clothing is not pink then you did not wash red and white clothing together An implication and its contrapositive have the same truth tables This is shown in the table below P Q Q nP P gt Q Q gt P 1 1 O O 1 1 1 O 1 O O O O 1 O 1 O O 00 O 1 1 1 1 Biconditional Statements P iff Q Notation P ltgt Q The statement P if and only if Q is true only when P and Q have the same truth values This can be seen below in the following truth table P o PltgtQ 1 1 1 1 o o o 1 o o o 1 You can check to see if P gt Q is true and Q gt P is true also Example Students pass MATH24O if and only if student scores are at least 350 points Negations The negation of a logical statement P is the statement P does not occur Example It is 1230 Negationgt It is not 1230 A statement and its negation have opposite truth values Example 3 lt 5 Negationgt 3 2 5 quotx 5 Negationgt x lt 5 orx gt 5 Universal Statement A universal statement is of the form For all property P holds Example For all real numbers x X2 1 H For all engineering majors solidworks is required Existing Statement An existing statement is of the form There exists property P holds Example There exists a car which is receiving a ticket today There exists an integer x such that x S O There exists a fish such that cannot swim Neoation of Universal Statement The negation of a universal statement is of the form There exists a such that property P does not hold Example There exists a real number x such that x2 1 There exists an engineering major such that solidworks is not required Neoation of Existing Statement The negation of an existing statement is of the form For all property P does not hold Example For all cars on campus they are not receiving a ticket For all integers x x gt 0 For all fish fish can swim Neoation of AND OR and IfThen De Morgan s Laws 1P AQ P V Q 1P VQ P Q Examples Jacob has cats and dogs Negationgt Jacob does not have cats or dogs 4 is the product of 2 gtlt 2 and 1 gtlt 4 Negationgt 2 gtlt 2 4 or 1 gtlt 4 4 If P then Q Negationgt P A SQ If the professor shows up to class then we have class Negationgt If the professor shows up to class and we do not have class T T u 39 lfx IS the solution to ax2bxc O then x V22 4 or b V2 4 NegatIongt lfx IS not 2 2 the solution to ax2bx c O and is 1 0r 1 Notation Universal Quantifier gt V means for all Existential Quantifier gt El means there exists orthere is X e D gt x is in the set D or X is an element of set D Sets E gt Set of real numbers Q gt Set of rational numbers 12 910 Z gt Set of integers 3 O 4 N gt Set of natural numbers 1 2 3 4 5 C gt Set of complex numbers Example Vx e D x2 1 Negationgt Elxe R x2 1 21 Functions definitions evaluating piecewise difference duotentt domain Funcdon A tunction troim a set D to a set P is a rule t that assigns each element x trom D exactly one elernent x from R ID gt domain R range Basically it means that a function r is a rule that assigns to each x element in a set A only one element called x in a set B Notation x 2 expression depending on x x independent variable y gt dependent variable Example Evatuate tx 3x2 4 at a x f 2 b x f o oh i o 11 Solution a t2 3 22 4 1H 34 4 t2 12 4 1i2 2 8 b 1ia sung 4 ta 36 4 c ta h 2 3o h 2 4 1ia h 3a ho h 4 Hat0i d2ohohh 4 ta h 2 302 20h r12 4 ta h 1 3o2 can if 4 i1 m2 Solution We are unable to evaluate x at x f 2 because A 2is not within the domain 0t fo Example Evaluate fx at x 2 When finding domains we exclude the tollowing 1 Points that make the demoninator 0 2 Points that produce a negative number under an even radical Exampie Evainate fx V i Solintion x xix it cannot be evaluated at x a ti x f 1 x Exampie Find the domain of x Vx 1 Solitition Need x it 2 i x E 1 This means that the domain is Loo i Where means inoiluded and means not inculidiediquot Exampie Find the domain ot the foilowingi a x ib gx Soliution a From the numerator we find that 2 ta 0 2 22K From the demonator we find that H t e O gt t 1 This means that the domain is g 1 iU 5 i 2 Note The U is used to say ailso include t hr From the demonator we find that 3x 2 1 ID so anything under the radioai muist striotiy greater than 0 9 3 2 gt 0 9 3x gt 0 gt x gt 2 This means that the domin is 00 Piecewise Functions x X2 1 X2 1 ix 2 x lt 1 is a piecewise function Evaluate x at a 2 bx 11 cx 1 as Since 2 lt 1 we use x 26 N Z 2 2 4 ib Since 1 re e 1 we use 22 1 H1 122 2 c Since 1 392 1 we use x2 1 fii 121 2 22 Graphinq Piecewise Functions A piecewise function is defined by different formulas on different part of it s domain Example Sketch the graph of the function fx x0 ifxlt2 x 1 ifx Z 2 Solution L i i is i f litquot T 3 at l As you notice ifx lt 2 the part of the graph to the left x O concides with the graph of y 0 If x 2 2 the part of the graph concides with the graph y 1 You will also see the solid dot at 21 which means that the point is included in the graph At 10 there is an open dot which means that the point is not included in the graph How to determine if an equation defines y as a function of x 1 If the equation is not in the form ofy solve for y 2 If you are given a graph use the vertical line test 3 Count the values of y for a given value of x If there are more than one value of y it is not a function of x Example Does the equation define y as a function of x a3x 5y7 bxy2 Cxy3 Solution y wWENO Q xy3 y cYes Example Use the vertical line test to determine whether the curve is a graph of a function of x a b l I L l Flirtingquot l r I lifmw r l 391 i l l l u i 1 Solution a Yes b No 26 Trwformations of Functions Vertical Shiftinq When adding a constant to a function it will shift the graph vertically upward if the constant is positive and if the constant is negative it will shift the graph vertically downward Suppose c gt 0 To graph y fx c shift the graph of y fx upward c units To graph y fx c shift the graph of y fx downward c units Example Use the graph of fx x 2 to sketch the graphs of each function a gxx21 hXx22 Solution 55 flhltj Ht ll HM if 3 Horizontal Shiftinq When adding a constant into a function it will shift the graph to the left and if the constant is positive and if the constant is negative it will shift the graph to the right Suppose c gt 0 To graph y fx 6 shift the graph of y fx to the left c units To graph y fx 6 shift the graph of y fx to the right c units Example Use the graph of fx x2 to sketch the graphs of each function a gx x4 2 bhx Jr 2 2 Solution E ill 3 fXX2 if glAllxrllquot 5 all 2 t1 3 1 a ITA A gig You will sometimes come across combined vertical and horizontal shifting Example Sketch the graph of fx x 3 4 Solution 20 Reflecting Graphs When you take the negative of a function the graph will reflect on the xaxis And when a negative is placed inside a function the graph will reflect on the yaxis Example Sketch the reflection of gx x 2 Solution Suzi1 Ell Example Sketch the reflection of gx V Solution 543 arr Vertical Shrinkinq and Stretchinq of Graphs When multiplying function by a constant the graph will either stretch or shrink depending on the constant To graph y fx If C gt 1 stretch the graph vertically by a factor of c If 0 lt c lt 1 shrink the graph vertically by a factor of c Example Use the graph of fx x 2 to graph the following functions a gx 3x2 b hx x 2 Solution 1 ll url j 2 539 In the following example you will see combined shifting stretching and reflecting Example Sketch the graph of the function fx 1 2x 3 2 Solution Horizontal Stretching and Shrinkinq of Graphs When the multiplying constant is inserted into the function the graph will either stretch or shrink depending on the constant To graph y fx If C gt 1 shrink the graph horizontally by a factor of lf 0 lt c lt 1 stretch the graph horizontally by a factor of Example Use the graph of y fx to graph the following functions 1 a yf2x b y f x 11 10 439 39 l m3 ll M 435 M 133 112 DJ Iquot I Iquot39I Iquot39I Iquot I Elsi it 2 111 115 118 1 El 12 14 Solution 3 11 10 119 08 DE 116 115 ail 113 112 111 H IIII39JEHII39 112 5 1151 GTE 113 12 5 151 1 1 5 21 115 11 15 21 2 5 31 Mil Even Function If a function fsatisfies f x fx for every number x in its domain then fis an even function Odd Functions If a function fsatisfies f x fx for every number x in its domain then fis an odd function Example Determine whether the functions are evem odd or neither a fx x 4 b fx2x Q fx3 x Solution a f x x 4 x 4 since f x fx the function is even b f19 x2 x x 2x since f x fx the function is odd 0 fva x3 x x3x x3 x since f x fx the function is odd The graph of an even function is symmetric with respect to the yaxis Whereas the graph of an odd function is symmetric with respect to the xaxis 27 Combininq Functions Sum of the functions f 900 fx gx Domain A B Difference of the functions f gx fx gx Domain A 113 Product of the functions ngx fxgx Domain A n B Quotent of the functions 2 52 Domain x E A DBgx 0 Example Find f gf gfg andJ g and their domains fx 2969 gx 2x Solution a fx gx x 2x 3x Domain 0000 fx gx x 2x x Domain 0000 fxgx XXZX 2x2 Domain 0000 L gx 2x Domain 000 U 000 Composition of Functions Given two functions fandg the composite function f0gaSO called the composition of fand g is denoted by f 0 8000 fgx Example Find the functions offog gof fof andgog and their domains fx 2x 3 gx 4x 1 Solution ngXx f4X1 24x 1 3 8x 2 3 8x1 Domain 0000 goXx g2x 3 42x 3 1 8x 12 1 8x11 Domain 0000 fofXx f2x 3 22x 3 3 4x 6 3 4x 9 Domain oo oo 24 Ave rat e Rate of Change of a Function The average rate of change of the function y x between x o and x b is l 39 39 b average rate of change Example if your height at birth was 18 inches and your height at thirteen years of age was 59 inches About how many inches per year did you grow Solution Change in height 595 18 41 inches Change in years 13 0 13 years Average change inches per year or about 315 inches per year Nlote Professor Waiters only wants you to give exact numbers unless its an easy fraction such as 2 Othenvise simplify the fraction if possible fibi HG b7 o 39 in general the average change in x over ab is Difference Quotient A difference quotient is the rate of change between x a and x a h la a h Nlote In the text assigned for MATH 240 Section 8 this defination appears on the side of page 152 next to exam ple Example if an object is dropped from a high cliff then the distance it has fallen is given by do 162 in feet Find the difference quotient between t 2 1 sec and t 2 1 1 h sec Solution 7 drirlrledll 16111137 16 df if H i i t h 16i1ih14h7 16 h renown if h 1613314 16113 16 h amuletquot h 32 16h feet per second Nlote lDiffernet values of h give different rates of change 28 OnetoOne Function and Inverses Afunctionfis oneto one iffx1x2 implies fx1 fx2 Examples of onetoone functions fx x fx 2x 3 fx x 3 fx x 5 fx lie and x V37 Graphically E i ll Hi i T i 1 E 1 1 1 7 lje i l h l 4 Example x is not onetoone fx1andx 1thenf1landf 11 Horizontal Line Test Draw horizontal lines for each y value if the function is onetoone then each horizontal line will hit the graph at most once Example Determine if the following functions are onetoone a b It 2 if 2 C l 1 1 Solution a No b Yes c No If we were to restrict the domain of fx x 4 to 0oo then it is a onetoone function i 1 Example s fx 3x3 a onetoone function Solution First check the contrapositive of the definition for a onetoone function The contrapositive states fx 1 fx 2 implies x 1 x2 fx1fx2gt3x133xzt3 3x1 3x2 x1x2 Yes fx 3x 3 is a onetoone function Inverses Let fbe onetoone with the domain A and the range B Then its inverse f1 has a domain B and the range A 2 x are yfor any yeB fltf 1ltx x ff1v y 2 Example Verify fx x and f1x c on 000 are inverses Solution fltf 1ltx V702 mm x or fltf 1ltx x 2 i x This means that these two functions are inverses How to find inverse when given fx2 1 Write yfx and solve for X 2 Swap y and X in newly solved equation 3 Write f1x y Example Find the inverse of fx 3x 3 Solution 1st y 3x 3 y 33x T x 2nd y 3rd 1 x f x T3 Check 1 f fx x 1 x f fx T3 3x3 3 3 x Example Find the inverse of fx 3 Solution 4 ltgty6x 110x3 y x6 6xy y10x3 3 y10x6xy 3 yx10 6y 3 y x10 6y 3 x 3 10 6x f 1ltxgt 1332 Example Show fx 16 is onetoone using the contrapositive of the de nMon Solution fX1fX2 10x 13 10x 23 6x1 1 6x2 1 6x2 1 10x2 3 6x 2 110x1 3 10x 2 36x1 1 28x228x1 xzle Graphing Inverse Functions To obtain the graph Off 1x given fx reflect fx across the line yx Example E Example 4 Example 31 Quadratic Functions and Models Def fx ax2 bx c where a 79 0 is called the quadratic function ax2 is the quadratic form bx is the linearform c is the constant ab andc are all real numbers The standard form of the quadratic function is fx altx hf k k fh The point hk is called the vertex The graph of a quadratic function is called a parabola Opens upward if a gt O The vertex is the minimum point Opens downward is a lt O The Vertex is the maximum point The line x h is called the axis of symmetry Example Convert the following to standard form a fx 2x2 16x 3 b fx 4x2 5x 4 Solution 9 a First we find h by h 2 a 16 h 2lt 2gt h4 Then we find k by fh f4 242 164 3 f432 643 f4 29 Then we can find fx 2x h2 k fx 2x 42 29 b 337 h k lt gt 4 gt25 gt 4 k3 16 fx 4x 92 Example Find b and c so that y 2x2 bx c has the vertex of 109 Solution y 2x 102 9 y 2x240x 200 9 y 2x2 40x 209 b 40 and c 209 Example Determine the standard and general forms of the quadratic function that has a vertex of 34 and goes through 15 Solution First insert h and k into the standard form of the quadratic function to get fx ax 32 4 and solve for a Since f15 then 5 a 1 32 4 5aaf4 54a4 5 44a 94a a The standard form of the quadratic function is fx x 32 4 Second use a h and k to find the general form of the quadratic function 3f4 fx x26x94 fx x2 22 7x 4 9227 65 x 2x 7x I Thus the general form of the quadratic function is fx x2 22 7x Completing Squares To make x2 bxa perfect square add 32 the square of half the coefficient of x This gives the perfect square x2bxgt2 x 2 Example Complete the square of 2x2 12x 13 Solution 1 Make coefficient x2 1 by factoring out 2 from the quadratic and linear terms y 2x2 6x 13 2 Add 32 inside the parentheses and subtract the factored out value times 2 from the constant term y2o max ag13 m y2x2 6x9 5 3 Factor the terms inside the parentheses to complete the square y2x2 3x 3x9 5 y2d 3f 5 Note that y 2x 32 5 looks like the standard form of the quadratic formula Graphing Parabolas When plotting parabolas you will want to plot the following key points 1 Vertex 2 yintercepts 3 xintercepts might use quadratic formula to obtain 4 Indicate which direction the parabola is opening Example Graph fx x2 2x 3 Solution First find the vertex by the following h 2 1 and kfh h 1 k 2 4 3 k 4 Second find all the yintercepts by solving f0 y022O 3 y 3 Third find all the xintercepts by factoring the quadratic formula 2 x2 2x 3 3x x 3O QZ3 Kx30 Mx3 l3y0 Q IXWH9O x10rx 3 Finally graph the function using the vertex yintercepts and the xintercepts 5 E5 Example A farmer wants to build two adjacent rectangular pins of the same size There is 300 feet of fence available Find the dimensions of the pins with the largest possible total area Solution 4x 3y 300ft A 2xy 4x3y300 3y300 4x y100 x A 2x100 x Ax 200x gx The area function contains the maximum at the vertex Maximizing Width b h JN 0Q 20lOOgtlt72 50r375ft UJloo Maximizing Length y 100 72 5 y 100 50 y 50ft 32 Polynomial Functions and Their Graphs Def A polynomial is a function of the form px a nx a n1x 1 a 1x a 0 72 Z a M k0 Note 2 means sum Add all terms after subsituting k values a n is called the leading coefficient a n 79 0 n is called the degree of the polynomial a 0 is c the constant term Example What is the leading coefficient degree and the constant term of 3x56x2 x2 Solution The leading coefficient is 3 The degree is 5 The constant term is 2 You will notice that linear and quadratic function are polynomails Basic Graphs H I I 5 5 5 I 5 5 as y 2 x 5 l 5 5 Shifting stretching and reflection rates are the same Domains of any polynomial is 0000 Range depends on the degree and leading coefficents End Behaviors of Polynomials angt0 anlt0 When n is even When n is odd Example Determine the end behavior of px x5 6x4 7x7 x2 x Solution The degree is 7 and a n 7 As x gtOO y gtOO and when x gt oo y gt OO 34 Real Zeros of Polynomials Theorem The following are equivalent 1x a is a zero of px 2 x a is a solution of pO 3 x a is a factor of px 4 a0 is an x intercept of the graph pf px Example Find all the real zeros of px 2x4 x3 3x2 Solution 190 2lt0gt4 03 30gt2 190 0 Using factoring we get the other zeros px x22x2 x 3 px x22x 3x 1 x20 x andx1 Notice that x acts like a double zero of the polynomial We say that x O is a zero of multiplicity two Def Multiplicity of a zero of a polynomial px is the largest positive integer m such that x am is a factor of px Plotting Polynomials To plot polynomials we need to know how functions behave around their zeros The multiplicity helps us with this Let pc 0 where multiplicity of c is m misoddmgt1 m is even mlt1 If multiplicity is 1 the graph will look like a straight line Steps to sketching Find zeros and multiplicities Mark zeros and indicate behaviors around zeros Indicate end behaviors Mark yintercept Plot a few intermediate points to capture oscillates Draw smooth curve through plotted points SnrhSDN Example Sketch px x 32x 2 Solution Zero Multiplicity x 2 1 x 3 2 yintercept gt px 18 end behaviors degree 3 a 1 as x gt ooy gtoo and x gtooy gt oo x m You will notice that around x 2 px looks like y x and around x 3 px looks like y x2 A and C are called local maximum values This means for x values for a Xfx lt fa B and C are called relective local minmum values This means for all x value for x b we have fx Zfb 33 Dividing Polynomials Review Divide 10903 by 29 375 r28 mm 87 We could say 375 or 10903 37529 This concept introduces the division algorithm which is explained below Division Algorithm If px and dx are polynomials with dx 79 O then there exists unique polynomials qx and rx where rx is either 0 or of degree less than the degree of dx such that 1 100 or W doc x qltxgt M The polynomials px and dx are called the dividend and the divisor respectively qx is the quotient and rx is the remainder ExampleFind the quotient and remainder when dividing px 6x4 5x3 2x 4 by dx 2x2 1 Solution 2x20x 16x4 5x30x2 2x 4 You will notice that I included a space holder for x and x2 in the dividend and the divisor I did this to allow number to line up correctly 13x2 x 2x20x 1V6x4 5x30x2 2x 4 6x4 0 3x2 5x33x2 2x 4 5x3 0 x 3x2 x 4 3x2 0 l 2x 4 qx3x2x and rx x 6x4 5x3 2x 43x2 x2x2 1 x Remainder Theorem If a polynomial pX is divided by Xk then the remainder in the division is pk Synthetic Division A quick method of dividing polynomials it can be used when the divisor is of the form x c The coefficients of the polynomial are inserted into a straight line and then the constant in the divisor is put on the outside c I use the to symbolized that there were coefficients listed Example Use synthetic division to divide 3x3 7x2 5x 10 by x 3 Solution 3 l3 7 5 1O The quotient is in green 3x22x1and the remainder is in orange 7 The coefficients of the polynomial of the degree n1 if n is degree of the original polynomial degree Remainder Theorem XC is a zero of pX if and only if Xc is a factor of pX Example Determine if x1 is a zero of px x3 7x6 using synthetic division 11076 x1 is a zero and furthermore x3 7x6 x2x 6x 1 can be factored further to get x3 7x 6x 1x 2x 3 This shows that X 1 2 and 3 Example Find the polynomial of degree 4 having zeros X2 x0 x1 x3 and the coefficient of x3 is 4 Solution Guess px x 2x Ox 1x 3 x2 2xx 1x 3 x3 x2 2x3 2xx 3 x4 2x3 5x26x Multiply 2 through to get px 2x4 4x3 10x2 12x 34 Real Zeros of Polynomials Rational Zeros Theorem If p and q have no common factors and 1 is a zero of a polynomial fx anx alx do Where all coefficients are integers then 1 p is a factor of a0 2 q is a factor of an Example Find all zeros of fx2x3 5x3 Solution Possibilities for p i1 i3 Possibilities for q i1 i2 Possibilities for g 1 i2 ig Starting with whole numbers perform synthetic division 2x3 5x3 x 12x22x 3 Using the quadratic formula to get the zeros of 2x2l2x 3 2i 4 42 3 xzwsxzwsxz 23mm 13w Thus x1 x2 andx3 1 Example Find all real zeros of 2x4 19x29 Solution Possibilities for p i1 i3 i9 Possibilities for q i 1 i2 Possibilities for g i1 i3 i9 1 1 1 2 O 19 O 9 2 2 17 17 2 2 17 17 8 3201909 6183 9 26130 2x4 19x2 9 x 32x3 6x2 x 3 Possibilities for p i 1 i3 Possibilities for q i 1 i2 Possibilities for g i 1 i3 1 3 2 6 1 3 2x4 19x2 9 x 3x 32x2 1 Using factoring we get the last two zero values 2x2 1 O 35 Complex Zeros and Fundamental Theorem of Algebra Theorem of Algebra Every non constant polynomial With complex coefficients has at least one complex zero Complete Factorization Theorem If px is a polynomial of degree n with complex coefficients then it can be factored in the form px ax clx 62x cm with clcn complex numbers and a is the leading coefficient of px Basically this means With every nth order polynomial has exactly n zeros Note 1392 1 and Vj 139 Example Find all the zeros of px x3 3x2 x 3 Solution Using the rational zeros theorem 531 j3 3 1 3 1 3 pxx3x21 3 o 3 1 O 1 O Zeros of the polynomial are x 3 x i x i Conjugate Pairs Theorem If p is a polynomial with real coefficients and at a bi With b 75 O is a zero of p then a bi is also a zero ofp Example Find all the zeros px x4 4x3 7x2 50x 50 given 3 z39 is a zero Solution From CPT we know x 3iis also a zero Then px x 3 ix 3 iqx px x2 x3 i x3 x 3 i3 iqx 1936 x2 x3 i x3 i 9 3i 3i i2qx px x2 x3 i x3 i 9 1qx px x2 3x xi 3x xi 10qx px x2 6x 10qx x44x37x250x50 qx x2 6x 10 Using long division we find what qx is x2 2x 5 x2 6x 10x4 4x3 7x2 50x 50 x4 6x3 10x2 2x3 17x2 50x 2x3 12x2 20x 5x2 30x 50 5x2 30x 50 0 qxx22x 5 Now we can say that px x2 6x 10x2 2x 5 To find the other zeros use qf x 24 24l 5 x M xz 2 2 x liV6 Thus me x lt3 w lt3 igtgtltx lt 1 1m lt 1 16 Zeros of the polynomial are x 3 i 3 i 1 V5 and 1 V5 Example Find a polynomial with the given properties Degree 4 real coefficient zeros at 12i 1i and p11 Solution 1906 Z 6136 12ix 1 2ix 1 ix 1i px ax2 x1 2i x1 2i12i1 2ix2 x1i x 1 i1 i1i px ax2 x 2xi x2xi 1 2i22i 4i2x2 x xi xxi 1 i i i2 px ax2 2x 1 41 2x2 2x 1 i2 px ax2 2x2x2 2x5 p1 ax2 2x2x2 2x 5 1a1 221 25 1 a4 px x2 2x 2x2 2x 5 36 Rational Functions Recall y Transformation properties still apply L x2 Example Graph gx Solution If fx 916 then gx fx 2 Example Sketch hx Solution Try to put h into the form h bfx a c Where fx lc Using long division 8 5x 2Vm hx 5x 2 8x156 hxX5xI 2 hx 21 5 hx shifts t0 the right units re ects over the y axis shrinks vertically 2 15 and 3 shifts down The x intercept is g and y intercept is 5 its

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