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# Introductory maths analysis chapter 03 official

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### Introductory maths analysis chapter 03 official

1. 1. INTRODUCTORY MATHEMATICALINTRODUCTORY MATHEMATICAL ANALYSISANALYSISFor Business, Economics, and the Life and Social Sciences ©2007 Pearson Education Asia Chapter 3Chapter 3 Lines, Parabolas and SystemsLines, Parabolas and Systems
2. 2. ©2007 Pearson Education Asia INTRODUCTORY MATHEMATICAL ANALYSIS 0. Review of Algebra 1. Applications and More Algebra 2. Functions and Graphs 3. Lines, Parabolas, and Systems 4. Exponential and Logarithmic Functions 5. Mathematics of Finance 6. Matrix Algebra 7. Linear Programming 8. Introduction to Probability and Statistics
3. 3. ©2007 Pearson Education Asia 9. Additional Topics in Probability 10. Limits and Continuity 11. Differentiation 12. Additional Differentiation Topics 13. Curve Sketching 14. Integration 15. Methods and Applications of Integration 16. Continuous Random Variables 17. Multivariable Calculus INTRODUCTORY MATHEMATICAL ANALYSIS
4. 4. ©2007 Pearson Education Asia • To develop the notion of slope and different forms of equations of lines. • To develop the notion of demand and supply curves and to introduce linear functions. • To sketch parabolas arising from quadratic functions. • To solve systems of linear equations in both two and three variables by using the technique of elimination by addition or by substitution. • To use substitution to solve nonlinear systems. • To solve systems describing equilibrium and break-even points. Chapter 3: Lines, Parabolas and Systems Chapter ObjectivesChapter Objectives
5. 5. ©2007 Pearson Education Asia Lines Applications and Linear Functions Quadratic Functions Systems of Linear Equations Nonlinear Systems Applications of Systems of Equations 3.1) 3.2) 3.3) 3.4) 3.5) Chapter 3: Lines, Parabolas and Systems Chapter OutlineChapter Outline 3.6)
6. 6. ©2007 Pearson Education Asia Slope of a Line • The slope of the line is for two different points (x1, y1) and (x2, y2) is Chapter 3: Lines, Parabolas and Systems 3.1 Line3.1 Line       = − − = changehorizontal changevertical 12 12 xx yy m
7. 7. ©2007 Pearson Education Asia The line in the figure shows the relationship between the price p of a widget (in dollars) and the quantity q of widgets (in thousands) that consumers will buy at that price. Find and interpret the slope. Chapter 3: Lines, Parabolas and Systems 3.1 Lines Example 1 – Price-Quantity Relationship
8. 8. ©2007 Pearson Education Asia Solution: The slope is Chapter 3: Lines, Parabolas and Systems 3.1 Lines Example 1 – Price-Quantity Relationship 2 1 28 41 12 12 −= − − = − − = qq pp m Equation of line • A point-slope form of an equation of the line through (x1, y1) with slope m is ( )1212 12 12 xxmyy m xx yy −=− = − −
9. 9. ©2007 Pearson Education Asia Find an equation of the line passing through (−3, 8) and (4, −2). Solution: The line has slope Using a point-slope form with (−3, 8) gives Chapter 3: Lines, Parabolas and Systems 3.1 Lines Example 3 – Determining a Line from Two Points ( ) 7 10 34 82 −= −− −− =m ( )[ ] 026710 3010567 3 7 10 8 =−+ −−=− −−−=− yx xy xy
10. 10. ©2007 Pearson Education Asia Chapter 3: Lines, Parabolas and Systems 3.1 Lines Example 5 – Find the Slope and y-intercept of a Line • The slope-intercept form of an equation of the line with slope m and y-intercept b is . cmxy += Find the slope and y-intercept of the line with equation y = 5(3-2x). Solution: Rewrite the equation as The slope is −10 and the y-intercept is 15. ( ) 1510 1015 235 +−= −= −= xy xy xy
11. 11. ©2007 Pearson Education Asia a.Find a general linear form of the line whose slope-intercept form is Solution: By clearing the fractions, we have Chapter 3: Lines, Parabolas and Systems 3.1 Lines Example 7 – Converting Forms of Equations of Lines 4 3 2 +−= xy 01232 04 3 2 =−+ =−+ yx yx
12. 12. ©2007 Pearson Education Asia b. Find the slope-intercept form of the line having a general linear form Solution: We solve the given equation for y, Chapter 3: Lines, Parabolas and Systems 3.1 Lines Example 7 – Converting Forms of Equations of Lines 0243 =−+ yx 2 1 4 3 234 0243 +−= +−= =−+ xy xy yx
13. 13. ©2007 Pearson Education Asia Chapter 3: Lines, Parabolas and Systems 3.1 Lines Parallel and Perpendicular Lines • Parallel Lines are two lines that have the same slope. • Perpendicular Lines are two lines with slopes m1 and m2 perpendicular to each other only if 2 1 1 m m −=
14. 14. ©2007 Pearson Education Asia Chapter 3: Lines, Parabolas and Systems 3.1 Lines Example 9 – Parallel and Perpendicular Lines The figure shows two lines passing through (3, −2). One is parallel to the line y = 3x + 1, and the other is perpendicular to it. Find the equations of these lines.
15. 15. ©2007 Pearson Education Asia Chapter 3: Lines, Parabolas and Systems 3.1 Lines Example 9 – Parallel and Perpendicular Lines Solution: The line through (3, −2) that is parallel to y = 3x + 1 also has slope 3. For the line perpendicular to y = 3x + 1, ( ) ( ) 113 932 332 −= −=+ −=−− xy xy xy ( ) ( ) 1 3 1 1 3 1 2 3 3 1 2 −−= +−=+ −−=−− xy xy xy
16. 16. ©2007 Pearson Education Asia Chapter 3: Lines, Parabolas and Systems 3.2 Applications and Linear Functions3.2 Applications and Linear Functions Example 1 – Production Levels Suppose that a manufacturer uses 100 lb of material to produce products A and B, which require 4 lb and 2 lb of material per unit, respectively. Solution: If x and y denote the number of units produced of A and B, respectively, Solving for y gives 0,where10024 ≥=+ yxyx 502 +−= xy
17. 17. ©2007 Pearson Education Asia Chapter 3: Lines, Parabolas and Systems 3.2 Applications and Linear Functions Demand and Supply Curves • Demand and supply curves have the following trends:
18. 18. ©2007 Pearson Education Asia Chapter 3: Lines, Parabolas and Systems 3.2 Applications and Linear Functions Example 3 – Graphing Linear Functions Linear Functions • A function f is a linear function which can be written as ( ) 0where ≠+= abaxxf Graph and . Solution: ( ) 12 −= xxf ( ) 3 215 t tg − =
19. 19. ©2007 Pearson Education Asia Chapter 3: Lines, Parabolas and Systems 3.2 Applications and Linear Functions Example 5 – Determining a Linear Function If y = f(x) is a linear function such that f(−2) = 6 and f(1) = −3, find f(x). Solution: The slope is . Using a point-slope form: ( ) 3 21 63 12 12 −= −− −− = − − = xx yy m ( ) ( )[ ] ( ) xxf xy xy xxmyy 3 3 236 11 −= −= −−−=− −=−
20. 20. ©2007 Pearson Education Asia Chapter 3: Lines, Parabolas and Systems 3.3 Quadratic Functions3.3 Quadratic Functions Example 1 – Graphing a Quadratic Function Graph the quadratic function . Solution: The vertex is . • Quadratic function is written as where a, b and c are constants and ( ) 22 ++= bxaxxf 0≠a ( ) 1242 +−−= xxxf ( ) 2 12 4 2 −= − − −=− a b ( )( ) 2and6 260 1240 2 −= −+= +−−= x xx xx The points are
21. 21. ©2007 Pearson Education Asia Chapter 3: Lines, Parabolas and Systems 3.3 Quadratic Functions Example 3 – Graphing a Quadratic Function Graph the quadratic function . Solution: ( ) 762 +−= xxxg ( ) 3 12 6 2 =−=− a b 23 ±=x
22. 22. ©2007 Pearson Education Asia Chapter 3: Lines, Parabolas and Systems 3.3 Quadratic Functions Example 5 – Finding and Graphing an Inverse From determine the inverse function for a = 2, b = 2, and c = 3. Solution: ( ) cbxaxxfy ++== 2
23. 23. ©2007 Pearson Education Asia Chapter 3: Lines, Parabolas and Systems 3.4 Systems of Linear Equations3.4 Systems of Linear Equations Two-Variable Systems • There are three different linear systems: • Two methods to solve simultaneous equations: a) elimination by addition b) elimination by substitution Linear system (one solution) Linear system (no solution) Linear system (many solutions)
24. 24. ©2007 Pearson Education Asia Chapter 3: Lines, Parabolas and Systems 3.4 Systems of Linear Equations Example 1 – Elimination-by-Addition Method Use elimination by addition to solve the system. Solution: Make the y-component the same. Adding the two equations, we get . Use to find Thus,    =+ =− 323 1343 xy yx    =+ =− 12128 39129 yx yx 3=x ( ) 1 391239 −= =− y y    −= = 1 3 y x 3=x
25. 25. ©2007 Pearson Education Asia Chapter 3: Lines, Parabolas and Systems 3.4 Systems of Linear Equations Example 3 – A Linear System with Infinitely Many Solutions Solve Solution: Make the x-component the same. Adding the two equations, we get . The complete solution is     =+ =+ 1 2 5 2 1 25 yx yx    −=−+− =+ 25 25 yx yx 00 = ry rx = −= 52
26. 26. ©2007 Pearson Education Asia Chapter 3: Lines, Parabolas and Systems 3.4 Systems of Linear Equations Example 5 – Solving a Three-Variable Linear System Solve Solution: By substitution, we get Since y = -5 + z, we can find z = 3 and y = -2. Thus,      −=−− =++− =++ 63 122 32 zyx zyx zyx      −+= −=− =+ 63 5 1573 zyx zy zy      = −= = 1 2 3 x y z
27. 27. ©2007 Pearson Education Asia Chapter 3: Lines, Parabolas and Systems 3.4 Systems of Linear Equations Example 7 – Two-Parameter Family of Solutions Solve the system Solution: Multiply the 2nd equation by 1/2 and add to the 1st equation, Setting y = r and z = s, the solutions are    =++ =++ 8242 42 zyx zyx    = =++ 00 42 zyx sz ry srx = = −−= 24
28. 28. ©2007 Pearson Education Asia Chapter 3: Lines, Parabolas and Systems 3.5 Nonlinear Systems3.5 Nonlinear Systems Example 1 – Solving a Nonlinear System • A system of equations with at least one nonlinear equation is called a nonlinear system. Solve (1) (2) Solution: Substitute Eq (2) into (1),    =+− =−+− 013 0722 yx yxx ( ) ( )( ) 7or8 2or3 023 06 07132 2 2 =−= =−= =−+ =−+ =−++− yy xx xx xx xxx
29. 29. ©2007 Pearson Education Asia Chapter 3: Lines, Parabolas and Systems 3.6 Applications of Systems of Equations3.6 Applications of Systems of Equations Equilibrium • The point of equilibrium is where demand and supply curves intersect.
30. 30. ©2007 Pearson Education Asia Chapter 3: Lines, Parabolas and Systems 3.6 Applications of Systems of Equations Example 1 – Tax Effect on Equilibrium Let be the supply equation for a manufacturer’s product, and suppose the demand equation is . a. If a tax of \$1.50 per unit is to be imposed on the manufacturer, how will the original equilibrium price be affected if the demand remains the same? b. Determine the total revenue obtained by the manufacturer at the equilibrium point both before and after the tax. 50 100 8 += qp 65 100 7 +−= qp
31. 31. ©2007 Pearson Education Asia Chapter 3: Lines, Parabolas and Systems 3.6 Applications of Systems of Equations Example 1 – Tax Effect on Equilibrium Solution: a. By substitution, and After new tax, and 100 50 100 8 65 100 7 = +=+− q qq ( ) 5850100 100 8 =+=p ( ) 70.5850.51100 100 8 =+=p ( ) 90 65 100 7 50.51100 100 8 = +−=+ q q
32. 32. ©2007 Pearson Education Asia Chapter 3: Lines, Parabolas and Systems 3.6 Applications of Systems of Equations Example 1 – Tax Effect on Equilibrium Solution: b. Total revenue given by After tax, ( )( ) 580010058 === pqyTR ( )( ) 52839070.58 === pqyTR
33. 33. ©2007 Pearson Education Asia Chapter 3: Lines, Parabolas and Systems 3.6 Applications of Systems of Equations Break-Even Points • Profit (or loss) = total revenue(TR) – total cost(TC) • Total cost = variable cost + fixed cost • The break-even point is where TR = TC. FCVCTC yyy +=
34. 34. ©2007 Pearson Education Asia Chapter 3: Lines, Parabolas and Systems 3.6 Applications of Systems of Equations Example 3 – Break-Even Point, Profit, and Loss A manufacturer sells a product at \$8 per unit, selling all that is produced. Fixed cost is \$5000 and variable cost per unit is 22/9 (dollars). a. Find the total output and revenue at the break-even point. b. Find the profit when 1800 units are produced. c. Find the loss when 450 units are produced. d. Find the output required to obtain a profit of \$10,000.
35. 35. ©2007 Pearson Education Asia Chapter 3: Lines, Parabolas and Systems 3.6 Applications of Systems of Equations Example 3 – Break-Even Point, Profit, and Loss Solution: a. We have At break-even point, and b. The profit is \$5000. 5000 9 22 8 +=+= = qyyy qy FCVCTC TR 900 5000 9 22 8 = += = q qq yy TCTR ( ) 72009008 ==TRy ( ) ( ) 500050001800 9 22 18008 =      +−=− TCTR yy