1 Chapter 1: Graphs, Functions, and Models

Transcription

1 1 Chapter 1: Graphs, Functions, and Models 1.1 Introduction to Graphing Know how to graph an equation Eample 1. Create a table of values and graph the equation y = 1. f() f() The Distance Formula The Midpoint Formula Equation of a circle Eample. Find the center, radius and equation of the circle that has a diameter with end points (1, 3) and (3, 1) d = (3 1) ( 1 ( 3)) = 4 4 = So the radius of the circle is. The center of the ( circle is the midpoint ) of the line segment ( 1), = (, ) Using these two facts gives the equation of the circle as ( ) (y ) = 1. Functions and Graphs 1..1 Functions and Function Notation: Eample 3. The Domain is grades {A,B,C,D,F} and the Range is Students in a class {Bob, Sue, Fernando, Olga }. The rule assigns a grade to each student. D Bob, A Sue, A Fernando, and

2 A Olga. Is this relation a function? No the input A is sent to 3 different outputs 1.. The Vertical line test Eample 4. Is this a function? f() f() No. A vertical line can be drawn that intersects the graph more than one time Domain and Range of a function Eample 5. Find the domain of f() = 3 and g() = 1 1

3 Set the radicand greater than or equal to zero [3, ) Set the denominator equal to zero to find the ecluded values 1 = 0 = 1 { 1} Eample 6. Find the Domain and Range of the following function. f() The domain is the setup of inputs that have an out put. We can read the domain from the graph [ 4, 4]. The Range is the set of outputs [0, 3] 1.3 Linear Function, Slope, Applications and models Horizontal and vertical lines 1.3. Slope Eample 7. Find the slope between ( 4, 5) and (5, 4). m = ( 4) = 9 9 = Equations of a line Eample 8. Find the equation of the line that contains the points (1, ) and (4, 1).

4 m = = 3 = 1 = 1(1) b 3 = b y = 3 3 Eample 9. Find the equation of the line that contains the point (1, ) and has slope 3. y = 3 b = 3(1) b 5 = b y = Linear Equations and Modeling Linear Regression Eample 10. The data in the following table shows the decrease in the number of drive-in movie sites since Year, Drive in Movie Sites, y 1990, , , , , , Use your graphing calculator to find a linear regression of the data and use it to estimate the number of drive-in movie sites in 015. Find the residual (correlation coefficient) and decide if the line is a good fit. Please round your answer to the nearest whole number. The equation of the line is y = The year 015 corresponds to = 5. This gives y = 18.7(5) = So in 015 There will be about 434 Drive in Movie Sites. r.95 so the line is a good fit. Eample 11. Is the following data set linear?

5 f() No the points do not look like they are on a line Parallel and Perpendicular lines Eample 1. Find the equation of the line that is perpendicular to f() = 1 and contains the 5 point ( 3, ). Since perpendicular lines have slopes that are negative reciprocals the slope of the new line is m = 5. Using the point slope formula gives y = 5 b = 5( 3) b 17 = b This gives the equation of the line as y = Zeros of a function and Applications Zero s of linear functions Eample 13. Find all zero s of f() = 3. To find a zero of a function set y = 0. This gives 0 = 3 = 3 = Application problems Eample 14. Erica invested a total of $5,000 in two accounts. The 1st account pays 3% simple interest per year and the nd account pays 4% simple interest per year. After one year she earned $176 in interest, how much did she invest in each account?

6 Construct a table to organize the information. I = P r t 3% %.04(5000 ) total N.A. 1 The interest column gives the equation.03.04(5000 ) = 176 = 400. So the 3% account has $400 and the 4% account has $ Solving Linear Inequalities Solve linear inequalities Eample Eample 16. Solve 1 > 3 or 1 < 3 1 > 3 > 4 > 1 < 3 < < Eample < < 7 4 < 8 <

7 Chapter : More on Functions Eample 18. Determine where the function is increasing decreasing and constant. Write your solution in interval notation. f() The function is increasing on ( 3, 1), decreasing on (, 3) ( 1, 1) (4, ) and constant on (1, 4)..0. Relative Maimum and Minimum Eample 19. Find the local ma and min on the graph below. f() The graph has a local ma at ( 1, 5) and a local min at ( 3, 1)

8 Eample 0. Use a graphing calculator to find the maimum(s) of f() = 4 1 graphing on a TI gives The ma occurs at (±1, 0) and the local min is at (0, 1).1 The Algebra of Functions.1.1 Sums, Differences, Products and Quotients Eample 1. If f() = 1 and g() = ( ) f, find () and state its domain. g The domain of f() is [ 1, ) and to domain of g() is (, 0) (0, ). Since g() is never equal to zero we only need to find the intersection of the domains. This gives [ 1, 0) (0, ).1. Difference Quotient f( h) f() h Eample. Find the difference quotient of the function f() = 1 ( h) 1 ( 1) h = h h =

9 . The Composition of Functions...1 Function Composition Eample 3. If f() = 1 and g() = 1 find (f g)() and (f f)() and state the domain. We must restrict the domain of g() so its range is in the domain of f(). We must eclude the zero(s) of g(). This gives ±1. (f g)() = 1 with domain { ±1} 1 Since f() is never equal to zero we do not need to eclude any etra values from the domain, but the domain of f() is { 0}. So (f f)() = 1 1 = with domain { 0}.. Function Decomposition Eample 4. Write h() = 6 as the composition of two functions. f() = g() = 6 (f g)() = h().3 Symmetry and Transformations.3.1 -ais, y-ais, and the origin Eample 5. Determine the symmetries of the equation: y 9 4 = 1 Replacing with gives y 9 ( ) = 1 y = 1 Since we get the same equation back the graph is symmetric with respect to the y-ais Replacing y with y gives ( y) 9 4 = 1 y 9 4 = 1 Since we get the same equation back the graph is symmetric with respect to the -ais. Since the graph has both and y ais symmetry it is also symmetric with respect to the origin..3. Even, Odd, and Neither Eample 6. Determine if the function f() = 3 is even, odd or neither.

10 f( ) = ( ) 3 ( ) = 3 = ( 3 ) = f() f( ) = f(). So the function is odd..4 Translations of Basic Graphs Eample 7. Graph the function f() = ( 5) 3 and g() = 1 ( 1) Both f and g are translations of the basic parabola y = (dashed graph). f is shifted 5 units to the right and down three units. g is shifted 1 unit to the right,up two units, and shrinks by a factor of one half. f() Eample ( ) 8. If (4, ) and (1, 5) are on the graph of y = f() Find a point on the graph y = 1 f 4 1 ( ) 1 y = f 4 (4) 1 = f(1) 1 = (5) 1 = 11 Eample 9. The graph of f() is given below. Use it to graph g() = f( 4) 3

11 10 y The graph has one reflections and two shifts. The negative in front of f is a reflection across the ais. The 4 composed with f() shifts the graph four units to the left and the 3 shifts the graph up 3 units. y Variation.5.1 Direct Variation Eample 30. y varies directly as, when = 1 y = Find the constant of variation and y when = 10

12 y = k = k(1) y = y = (10) = 0.5. Inverse Variation Eample 31. y is inversely proportional to, when = 3 y = 1 Find the constant of variation and y when = 10 y = k 1 = k 3 k = 3 y = 3 y = Joint Variation Eample 3. z varies directly as and y, when = 3, y = 1 and z = 6. Find the constant of variation and z when = and y = 3. z = ky 6 = k(3)(1) k = z = y z = ()(3) = Applications of Variation Eample 33. The gravitational force between two objects is directly proportional to their masses (m 1 and m ) and is inversely proportional to the square of the distance (r) between them. What is the equation of variation? F = k m 1m r

13 3 Chapter 3: Quadratic Functions and Equations; Inequalities 3.1 The Comple Numbers The Comple Number System Eample 34. Simplify = i 7 16 = i 16 = 4i 48 = i 16 3 = (4 3)i 3.1. Addition and Subtraction Eample 35. Add (3 4i) ( 4i) (3 4i) ( 4i) = (3 ( )) (( 4) 4)i = 1 0i = 1 Eample 36. Subtract (4 5) (4 64) (4 5) (4 64) = (4 5i) (4 8i) = (4 4) ( 5 ( 8))i = 0 3i = 3i Multiplication Eample = (3i)(8i) = 4i = 4 Eample 38. (5 6i)(1 i) (5 6i)(1 i) = 5 10i 6i 1i = 17 4i Powers of i Eample 39. Simplify i 99

14 4 divides into 99 4 times with a remainder of 3. This gives i 99 = i 3 = i Conjugates and Division Eample 40. Find the conjugate of i i Eample 41. Find the conjugate of 6 6 Eample 4. Find the conjugate of 4 4i 4 4i Eample 43. Find the conjugate of 16i 16i Eample 44. i i i i ( i i ) = i i i = 1 i = i 3. Quadratic Equations, Functions, Zeros, and Models 3..1 Quadratic Equations and Functions Eample 45. Solve 5 = 6 First the equation must be put into standard form by subtracting 6 from both sides 5 6 = 0.

15 The quadratic can be factored as ( 6)( 1) = 0. 6 = 0 = = 6 or 1 = 0 = = 1 Using the zero product principle gives Eample 46. Solve 64 = 0 First isolate the square term by subtracting 64 from both sides = 64. Taking the square root of both sides gives = 64 = = ±8i 3.. Completing the Square Eample 47. Use completing the square and the square root principle to solve 4 11 = 0 ( ) b Isolate the and term by adding 11 to both sides and then add to both sides of the equation 4 = = = 15 Factor the perfect square trinomial and solve using the square root principle. ( ) = 15 = ± 15 = ± 15 Eample 48. Use completing the square and the square root principle to solve 3 9 = 0 Isolate the and term by adding 9 to both sides and then divide by 3, and finally add to both sides of the equation 3 = 9 3 = = = 8 9 Factor the perfect square trinomial and solve using the square root principle. ( 1 ) = = = ± 9 = 1 3 ± 7 3 = 1 ± 7 3 ( ) b 3..3 Using the Quadratic Formula Eample 49. Solve: 3 7 = 3 7 = 0 = a = 3, b 7, and c = This gives = ( 7) ± ( 7) 4(3)() (3) = 7 ± 5 6 = = = or = = 1 3

16 3..4 The Discriminant Eample 50. Determine the number and type of solutions of = 0. b 4ac = ( 70) 4(49)(5) = = 0. Since the discriminant is zero the equation has one repeated real solution Equations Reducible to Quadratic Eample 51. Solve ( 3) 5( 3) 6 = 0 The equation is quadratic in form. Let u = 3, then u 5u 6 = 0 Factoring gives (u 6)(u1) = 0 Using the zero product principle gives and substituting out u gives 3 6 = 0 and 3 1 = 0 Solving = 3 and = Application Problems Eample 5. A marble is dropped from the top of the Willis Tower, 44 meters above the ground. The height of the marble is given by s(t) = 4.9t 44. How long does it take the marble to reach the ground? (Round you answer to the nearest 10th) 0 = 4.9t 44 t = = t = Analyzing Graphs of Quadratic Functions Graphing Verte form f() = a( h) k Eample 53. Find the verte of the parabola f() = 5( 17) 6. The function is already in verte form. f() = a( h) k f() = 5( 17) 6 Identifying a = 5,h = 17, and k = 6 gives the location of the verte at (h, k) = ( 17, 6) Eample 54. Convert to verte form. f() =

17 f() = 3 [ 6 ( ) 6 ( ) ] 6 4 = 3[( 3) 9] 4 f() = 3( 3) 3 Eample 55. Find the verte, the ais of symmetry, the ma or min and the range of f() = Convert to verte form (see above) f() = 3( 3) 3 a = 3, h = 3, and k = 3. The verte is at (h, k) (3, 3) a = 3 < 0 so the quadratic has a maimum. The ais of symmetry is = h = = 3. The range is (, 3] 3.3. Graphing Standard form f() = a b c Verte Formula Eample 56. Find the verte of f() = h = 6 ( ) 3 = 6 3 = k = f(h) = 3 () 6() 5 = 1 (h, k) = (, 1) Application Problems Eample 57. A model rocket is launched with an initial velocity of 11 feet per second from a launch pad at a height of 0 feet. The equation of motion is s(t) = 16t 11t 0. Determine when the rocket reaches its maimum height and find the maimum height. The maimum is at the verte. h = b a = 11 ( 16) = 7. The rocket reaches its maimum ( ) ( ) 7 7 height after 3.5 seconds. The maimum height is s(t) = = 16 Eample 58. A rubber ball is dropped from the top of a hole. Eactly.5 seconds later, the sound of the rubber ball hitting the bottom is heard. How deep is the hole. Hint: The distance that a dropped object falls in t seconds in represented by the formula s = 16t. The speed of sound is 1100 ft/sec.

18 distance in time one s = 16t 1 time for the ball to reach the bottom distance in time two s = 1100t time for the sound to reach the top From the diagram we see that t 1 t =.5. The distance the object falls in t 1 is d = 16t 1. The distance the sound covers in t is d = 1100t. The distance is the same in both directions and gives the equation. 16t 1 = 1100t Using the equation t 1 t =.5 t =.5 t 1 to eliminate t from the equation gives 16t 1 = 1100(.5 t 1 ) 16t 1 = t 1 Putting the equation in standard form gives 16t t = 0 formula with a = 16,b = 1100, and c = 750 gives Using the quadratic t 1 = (1100) ± (1100) 4(16)( 750) (16) = 1100 ± Rejecting the negative solution gives t 1 = The distance to the bottom of the hole is s = 16(.41515) = ft 1100 ± Solving Rational Equations and Radical Equations Rational Equations Eample 59. Solve = 3 3 Factoring gives ( 3)( 3) 5 3 = 3. The LCD is ( 3)(3). Clearing the fractions 3 by multiplying by the LCD gives 5( 3) = 3( 3) 5 15 = 3 9 = 6 = = Radical Equations Eample 60. Solve 1 4 = 1 Isolate the radical by adding 4 to both sides. 1 = 3. Square both sides 1 = 9

19 and solve for. This gives = 4. The solution needs to be checked in the original equation. 1 ( 4) = 1 8 = 9 = 3. So = 4 is the solution to the equation. 3.5 Solving Equations and Inequalities with Absolute Value Equations with Absolute Value Eample 61. Solve 7 1 = 6 Isolate the absolute value. 1 = 1 = 1 = 1. This gives the two equations 1 = 1 1 = 1. Solving each linear equation gives = 0 and = 1 respectively Inequalities with Absolute Value Eample 6. Solve 8 < 9 This gives an and compound inequality 9 < 8 < 9 subtracting 8 from all three parts gives 17 < < 1 = ( 17, 1) Eample 63. Solve 8 > 9 This gives an or compound inequality 8 < 9 or 8 > 9 subtracting 8 from both sides of both equations gives < 17 or > 1 (, 17) (1, )

20 4 Chapter 4: Polynomial Functions and Rational Functions 4.1 Polynomial Functions and Modeling Eample 64. Determine the leading term, degree and classify the polynomial by type. (a) f() = π 5 f() has leading term π 5 and it is degree 0 because it is a constant. (b) g() = g() has leading term 3 and it it degree, so it is a quadratic The Leading-Term Test Eample 65. Sketch the end behavior of the polynomial p() = The leading coefficient is a 11 = 1 and the degree is n = 11. This gives y 4.1. Find Zeros of Factored Polynomial Functions Eample 66. Determine whether = and = 5 are zeros of p() = If c is a zero of p(), then p(c) = 0. This gives p() = 3 17() 15 = 7, So = is not a zero. p( 5) = ( 5) 3 ( 5) 17( 5) 15 = 0, So = 5 is a zero.

21 Eample 67. Find the zeros of f() = 3( 3) 3 ( ) Using the zero product principle gives 3 = 0 3 = 0 3 = 0 = 0 = 0 Solving each equation gives the two zeros = 3 of multiplicity 3 and = of multiplicity Finding Real Zeros on a Calculator Eample 68. Use a calculator to find the zero(s) of f() = Rational Functions The Domain of a Rational Function Eample 69. Find the domain of f() = Setting the denominator equal to zero gives 5 6 = 0 ( 3)( ) = 0 = = 3 or = This gives the domain { 3 and } or (, ) (, 3) (3, ) 4.5. Vertical Asymptotes Eample 70. Find the Vertical Asymptote(s) of f() = 5 4 4

22 Setting the denominator equal to zero and solving will give the vertical asymptotes 4 4 = 0 ( 4) = 0 = ( )( ) = 0 The function has 3 vertical asymptotes = 0, =, and = Horizontal Asymptotes Eample 71. Find the Horizontal Asymptote of f() = Since the degree of the numerator is 5 and the degree of the denominator is, the function has no horizontal asymptote. Eample 7. Find the Horizontal Asymptote of f() = Since the degree of the numerator is and the degree of the denominator is, the horizontal asymptote is the ratio of the lead coefficients. y = 3 5 Eample 73. Find the Horizontal Asymptote of f() = Since the degree of the numerator is 3 and the degree of the denominator is 7, the horizontal asymptote is the line y = Oblique (Slant) Asymptotes Eample 74. Find the oblique Asymptote of f() = Using polynomial long division gives: ) The oblique asymptote is the whole part without the remainder. This gives the slant asymptote

23 y = Polynomial Inequalities and Rational Inequalities Polynomial Inequalities Eample 75. Solve 5 Setting the equation equal to zero and solving for the zeros gives ( 3)( 1) 0 Making a sign chart gives f() f() is positive on (, 1] [3, ) 4.6. Rational Inequalities Eample 76. Solve 0 Setting the equation equal to zero and solving for the zeros of the numerator and denominator gives = and =. Making a sign chart gives f() Since = is a zero of the denominator we must eclude it. f() is negative on [, ) Eample 77. Solve > 4 4 Setting the equation equal to zero and then factoring gives > 0 3 ( 4 4) > 0 3 ( ) > 0 The zeros of the inequality are = 0 or = listing all of the factors and making a sign charts

24 gives. f() Zeros The solution set is (0, ) (, ). 0

25 5 Chapter 5: Eponential Functions and Logarithmic Functions 5.1 Inverse Functions Inverses Eample 78. Find the inverse of the relation = y 3y To find the inverse interchange the and the y to get y = Inverses and One-to-One Functions Eample 79. Show y = is not 1-1. f() This horizontal line intersects the curve twice Since a horizontal line intersects the curve more than once, it is not one to one Finding Formulas for Inverses Eample 80. Find the inverse of f() = 3 7 Replace f() with y y = 3 7 Interchange and y = 3y 7 Solve for y y =

26 replace y with f 1 () f 1 () = Inverse Functions and Composition Eample 81. Use function composition to show that f() = 5 4 and f 1 () = 4 5 are inverses and So the functions are inverses. (f f 1 )() = (4 5) 5 4 = 4 4 = ( ) 5 (f 1 f ) () = 4 5 = 5 5 = 4 5. Eponential Functions and Graphs 5..1 Graphing Eponential Functions Eample 8. Graph the function f() = 5 f() is shifted two units to the right and down 5 units. This gives f() Applications Eample 83. $100 is invested at 3% interest compounded quarterly. Find a function for the amount in the account after t years. Use the model to find the amount of money in the account after 0 years.

27 A = P ( 1 r ) nt P = 100 r = 3% =.03 n = 4 = A = 100(1.0075) 4t n Using the value t = 0 gives A = 100(1.0075) 80 = Graphing Eponential Functions, Base e Eample 84. Graph the function f() = 1 e 1 3 f() is shifted one unit to the left, up three units, and reflected across the -ais. This gives f() Logarithmic Functions and Graphs Finding Certain Logarithms Eample 85. Find each of the following: a) log 100 b) log π π 3 c) log a) log 100 = because 10 = 100 b) log π π 3 = 3 because π 3 = π 3 c) log = 4 because 3 4 = 1 81

28 5.3. Converting Between Eponential Equations and Logarithms and Logarithmic Equations Eample 86. Convert the equation to logarithmic form 65 = 5 log 5 65 = Eample 87. Convert the equation to logarithmic form e t = 1000 ln 1000 = t Eample 88. Convert the equation to eponential form log 8 = = Eample 89. Convert the equation to eponential form ln 5 = t e t = Changing Logarithmic Bases Eample 90. Find log 7 9 Using change of base on a calculator gives log 7 9 = log 9 log 7 = 3 This can also be done without a calculator if the logarithm is changed to base 3 log 7 9 = log 3 9 log 3 7 = Graphs of Logarithmic Functions Eample 91. Graph the function f() = ln( 3)

29 f() is shifted three units to the left, and reflected across the -ais. This gives f() Properties of Logarithmic Functions Applying Logarithmic Properties Eample 9. Epress in terms of sums and differences of logarithms log 3 y 7 log 3 y 7 = log ( 3 y 7) 1 ( ) = log 3 7 y = log 3 log y 7 = 3 log 7 log y Eample 93. Epress as a single logarithm ln( 5) ln( 5) ln( 5) Working from left to right gives ( ln( ) 5 5) ln( 5) ln( 5) = ln ln( 5) 5 ( ( ) ( 5)( 5) ( 5) ) ( 5) ln = ln = ln( 5) Simplifying Epressions of the Type log a (a ) and a log a () Eample 94. Simplify 7 log 7 (5t) and e 3 ln

30 7 log 7 (5t) = 5t e 3 ln = e ln 3 = Solving Eponential and Logarithmic Equations Solving Eponential Equations Eample 95. Solve 5 = 5 Writing each side of the equation as an eponential base 5 gives 5 = 5 = = = 0 ( )( 1) = 0 This gives the two solutions = and = 1 Eample 96. Solve 1 = 3 ln 1 = ln 3 (1 ) ln = ln 3 ln ln = ln 3 Isolating gives ln = ln 3 ln ln = (ln 3 ln ) ln = ln 6 Solving for gives = ln ln 6 Eample 97. Solve 3e 7 = 8 3e 7 = 8 = 3e = 15 = e = 5 = = ln Solving Logarithmic Equations Eample 98. Solve log ( 3) log ( 7) = 5 log ( 3) log ( 7) = 5 log [( 3)( 7)] = 5 ( 3)( 7) = 5 = 10 1 = = 3 = = 0 = ( 11)( 1) = 0

31 This gives = 11 and = 1. The solution = 1 must be rejected because the logarithm of negative number is undefined (not a real number) 5.6 Applications and Models: Growth and Decay; Compound Interest Population Growth Eample 99. In a research eperiment, a population of fruit flies is increasing according to the law of eponential growth. The eperiment starts with 100 flies, and after days there are 300 flies. How many flies will there be after 10 days? Identifying P 0 = 100 and P () = 300 gives 300 = 100e k = 3 = e k = ln 3 = k = k = ln 3 This gives P (t) = 100e ( ln 3 )t. So P (10) = 100e ( ln 3 ) 10 = 4, 300 After 10 days there will be 4,300 flies 5.6. Interest Compounded Continuously Eample 100. Aeris invests $ 100 compounded continuously. After years, her account balance is $300. How long will it take for her account balance to be $ 4,300? Identifying P 0 = 100 and P () = 300 gives This gives P (t) = 100e (ln 3 )t. 300 = 100e k = 3 = e k = ln 3 = k = k = ln 3 ln 3 ln 3 t 4300 = 100e t = 43 = e = ln 43 = t ln 3 The account balance will double after 10 years. = t = ln 43 ln 3 = Eponential Decay Eample 101. The half-life of radioactive radium 6 Ra is 1599 years. If there are 100 mg present, how much radioactive radium will remain after 00 years?

32 The half life can be found using the formula k = ln, where T is the half-life T k = ln 1599 = P (t) = 100e t ln ln = P (00) = 100e So there is about 91.7 mg left after 00 years.

33 9 Chapter 9: Systems of Equations and Matrices 9.1 Systems of Equations in Two Variables Eample 10. Is (, 4) a solution to 5 y = 3y = 8 We can check the ordered pair in each equation to see if it is true. 5( ) (4) = True but ( ) 3(4) = 16 8 is False So the ordered pair is not a solution Solving Systems of Equations Graphically Eample 103. Find the solution to the linear system f() The solution is the point of intersection of the graphs. The lines intersect at (3, ) 9.1. The Substitution Method Eample 104. Solve the system y = 5 3y = 3

34 Isolating in the first equation gives = 5 y, substituting into the second equation gives (5 y) 3y = 3 = 4y 5 = 3 = 4y = 8 = y = Substituting y back into the first equation gives = 5 () = 3. So the solution is (3, ) The Elimination Method Eample 105. Solve the system y = 5 3y = 3 Multiplying the first equation by 1 and adding gives ( y) = 5 ( 3y) = 3 0 4y = 8 Solving for y gives y =. Substituting y back into the first equation = 5 = = 3. So the solution is (3, ) Applications Eample 106. There are a total of 50 nickels and dimes in a jar. If the value of the coins is $3.00, how many nickels and dimes are there? Let N be the number of nickels and D be the number of dimes, then N D = 50. The value is.05n.10d = 3 = 5N 10D = 300. Solving the first equation for N gives N = 50 D and substituting into the second equation gives 5(50 D) 10D = 300 = 50 5D 10D = 300 = 5D = 50 = D = 10. So the number of nickels must be N = Solving Systems of Three Equations Eample 107. Solve y z = 6 4y 5z = 31 5 y z = 13 Multiplying equation 1 by 6 and adding it to equation gives 10y z = 19. Multiplying equation 1 by 5 and adding it to equation 3 gives 3y 3z = 3. Now multiply the new equation

35 3 by 1. This gives 3 y z = 10y z = 19 y z = 1 Multiply equation 3 by 10 and add to equation to get 9z = 9. Then interchange the new equation 3 with equation to get y z = y z = 1 9z = 9 Using back substitution gives 9z = 9 = z = 9. y 1 = 1 = y = and 1 = = = 3. So the solution is (3,, 1) 9..1 Mathematical Models Eample 108. Find the equation of the quadratic f() = a b c that contains the points f(0) = 5, f(1) = 4, and f( ) = 11 Evaluating the quadratic at the three points gives the three equations c = 5 a b c = 4 4a b c = 11 Putting c into the last two equations gives a b = 1 and 4a b = 16. Multiplying the first equation by and adding gives 6a = 18 = a = 3 and 3 b = 1 = b =. So the equation is f() = Matrices and Systems of Equations Eample 109. Write the system as an augmented matri. y z = 6 4y 5z = 31 5 y z = Gauss-Jordan Elimination Eample 110. Solve the system 3 4y = 3y = 4 The can be solved on a TI-Graphing Calculator using the RREF command or as shown below by hand.

36 [ R R 1 R 1 [ R 1 R R [ R R 1 R 1 [ This gives (8, 56) ] ] ] ] 9.4 Matri Operations Most matri operations can and should be done on a TI-Graphing Calculator. Please refer to your notes or the tutorials in MyLabs on how to use your graphing calculator. The Following Matrices will be used in this section [ ] [ ] [ ] [ A = B = C = D = E = ] Matri Addition and Subtraction Eample 111. Find A D and B C The can be solved on a TI-Graphing Calculator. The sum A D is undefined. The matrices have different sizes. B and C can be added [ ] [ ] [ ] B C = = Scalar Multiplication Eample 11. Find 3D The can be solved on a TI-Graphing Calculator or as shown below by hand D =

37 9.4.3 Products of Matrices Eample 113. Multiply AE and EA The can be solved on a TI-Graphing Calculator. [ ] [ ] 5 4 AE = The Product EA is undefined. = [ ] Matri Equations Eample 114. Write the system 3y = 5 3 y = 4 as a matri equation. [ [ ] [ 3 5 = 3 ] y 4] 9.5 Inverses of Matrices The Inverse of a Matri Eample 115. Find the inverse of A = [ ] 4 3 Use TI graphing calculator, or if you wish by hand Create an augments matri with the Identity and A and reduce. [ ] R 1 R R and 1 R 1 R 1 [ ] R R 1 R 1 and 1R R [ [ A 1 3 = 1 1 ] ]

38 9.5. Solving Systems of Equations Eample 116. Solve the system 4y = 6 3y = 4 as a matri equation. Use TI graphing calculator, or if you wish by hand [ ] [ [ 4 6 = 3 y] 4] Multiplying both sides by A 1 (Found in the previous eample) gives [ 3 ] [ ] [ [ 4 = y] 3 ] [6 ] [ [ ] 17 = y] 10 The solution is ( 17, 10)

39 11 Chapter 11:Sequences, Series, and Combinatorics 11.1 Sequences and Series Sequences Eample 117. Find the 3rd and 5th term of the sequence a n = n 7 n a 3 = 3 7 (3) = 3 a 5 = 5 7 (5) = = Finding the General Term Eample 118. Find the general term 3,, 7,... Each successive sequence member is 5 greater than the previous, this gives a n = 8 5n Sums and Series Eample 119. Find S 3 for the sequence 3,, 7,... S 3 = a 1 a a 3 = 3 7 = Sigma Notation Eample 10. Find the sum 6 n=4 6 n=4 1 n 1 n = = = Eample 11. Write in sigma notation = 1 n n=1

40 11. Arithmetic Sequences and Series Arithmetic Sequences Eample 1. Find the general term of the sequence, 6, 10, 14,... The first term is a 1 = and the common difference is d = 6 = 4. The general term is a n = a 1 (n 1) d = a n = (n 1) 4 = a n = 4n 11.. Sum of the First n Terms of an Arithmetic Sequence Eample 13. Find the sum of the first 100 terms of the sequence, 6, 10, 14,... a 1 =, d = 4, n = 100, and a 100 = 4(100) = 398 (See previous eample) This gives S n = n (a 1 a n ) = S 100 = 100 ( 398) = = Applications Eample 14. The sith term of an arithmetic sequence is a 6 = 54 and the thirtieth is a 30 =. Find the general term of the arithmetic sequence. Method I: There are 30 6 = 4 terms between the two given terms, so the common difference must be added 4 times. This gives the equation a 6 4d = a d =. Solving for d gives d = 7. Using the same technique to solve for the first terms gives the equation a 1 5d = a 6 a 1 5(7) = 54. Solving gives a 1 = 19. So the general term is a n = a 1 (n 1) d a n = 7n 1. Method II: Using the formula for the general term, a n = a 1 (n 1) d, and using the given terms yields the system of equations. { 54 = a 1 5d = a 1 9d Solving the linear system gives that a 1 = 19 and d = 7. The general term is a n = a 1 (n 1) d a n = 7n Geometric Sequences and Series Geometric Sequences Eample 15. Find the 5th term of a geometric sequence with a 1 = 7 and r = 1

41 a n = 7 ( ) n 1 ( ) = a 5 = 7 = Sum of the First n Terms of a Geometric Sequence Eample 16. Find the sum a 1 = 5,r = 1, and n = 11. This gives 3 ) 11) S 11 = 5(1 ( ( ) 1 = Infinite Geometric Series Eample 17. Find (the sum) 5 n=1 ( 1 ) n 1 3 a 1 = 5,r = 1 3 and r < 1 so the series converges. S = 5 1 ( 1 3) = = 15 4 = Applications Eample 18. Write the decimal as a fraction = = = This gives that a 1 = and r = s = = = =