IB MATH STUDIES We are so ecited that you have decided to embark upon an eciting journey through IB Math Studies. Make no mistake, the road ahead will be challenging and perhaps overwhelming at times. However, if we work together it can be both fun and rewarding eperience. Math Studies is designed to build confidence and encourage an appreciation of mathematics in students. This two-year course will eplore a wide range of mathematical topics concentrating on those that can be applied to other subjects. Objective: The objective of the entire summer assignment is to present the IB Math Studies teacher, Mrs.Wisniewski, a demonstration of each student s individual knowledge and thought process. If a student has questions or needs assistance, he/she should email jlwisniewski@cps.edu at any time. Epectations: This is a college level course; therefore, as with all IB and AP courses, it demands higher epectations than other college preparatory and standard level courses. Some of these epectations are: Consistent, eemplary attendance. A determined work ethic. Strong organizational skills. A strong sense of self-respect and respect for others Act with integrity and honesty, with a strong sense of fairness, justice, and respect for the dignity of the individual, groups and communities. Summer Assignment: The summer assignment for IB Math Studies contains two parts: (Part 1) Presumed Knowledge The IBO has set guidelines of mathematics students should understand prior to taking IB Math Studies. Each student should complete all questions in part 1 to the best of their ability as eplained in the directions preceding the problems. You may use a calculator, however remember No Work = No Credit. If you use a calculator to answer a question, you must write down the work that you are entering in the calculator to receive credit. In addition, this assignment is designed to demonstrate your individual knowledge. Therefore, you are epected to complete the work by yourself. (Part ) Recognizing Math The goal of IB Math Studies is to develop an appreciation for math in our everyday lives. Therefore, below are links to two articles from the NY Times. Please read both of these articles and summarize them. In your summary, reflect on what caught your attention, what you found interesting or what have you learned from them. All summaries must be typed and no longer than a page each. http://opinionator.blogs.nytimes.com/010/03/8/power-tools/ http://opinionator.blogs.nytimes.com/010/03/14/square-dancing/ Grading: This assignment will be collected for a quiz grade on the first day of school. All writing should be neat and legible. The summaries need to be typed (1 point, single space). Do not wait until the last minute to begin. You will need to review some of the mathematics concepts learned in Algebra and Geometry in order to complete this assignment. Feel free to use any books or websites to help you review the material. Have a wonderful summer! I know we are going to have a great school year! I am ecited about the upcoming school year. Mrs. Wisniewski
Coordinate Geometry In this part you will find eamples with full solutions first. Look at those eamples as you are working on questions from each section. SECTION I: Distance and Midpoint Formulae Given two points 1, y 1 and, y : Distance between the points = Midpoint of the points = 1 y y1 1 y1 y Eamples: a) Find the value(s) of b given that A(3, -) and B(b, 1) are 13 units apart. 13 13 ( b 3) 4 ( b 3) 4 b 3 3 b b 3 1 ( ) ( b 3) 3, b 5,1 b) Find the coordinates of B if M is the midpoint of AB, A is at (1, 3) and M is at (4, -). 1 3 y, 4, 1 3 y 4 1 8 3 y 4 7 y 7 SECTION II: Gradient, Equations of Lines and Graphing Vocabulary and Information: Gradient is another word for slope. Formula for slope: m y y1 The gradient of a horizontal line is zero. Its equation is of the form = a where a is a constant. The gradient of a vertical line is undefined. Its equation is of the form y = b where b is a constant. Parallel lines have the same gradient. Perpendicular lines have opposite reciprocal gradients. Three or more points are collinear if they lie on the same line. Two forms for writing equations of lines: Gradient-Intercept Form: y = m + b, where b is the y-intercept and m is the slope General Form: A + By = C, where A, B, and C are integers. The gradient of a line in this A form can be found by calculating. B Ais Intercepts: To find an -intercept, set y equal to zero and solve for. To find a y-intercept, set equal to zero and solve for y. 1
Eamples: a) Find t given that the line joining D(-1, -3) to C(1, t) is perpendicular to a line with gradient. Line DC has slope ½ (opposite reciprocal of ). t 3 1 11 t 3 1 t 3 1 t 4 b) Find the equation of the line, in both gradient-intercept form and general form, passing through (5, 3) and (1, 1). 31 1 1 1 1 m y b y 5 1 4 1 1 1 1 (1) b y 1 b y 1 General Form 1 1 Gradient-intercept form: y c) Graph the line y. 3 d) Graph the line 3y = 1 using ais-intercepts. y-intercept (0, ) Set = 0: 3y = 1 Set y = 0: = 1 slope = 1/3 y = -4 = 6 Point: (0, -4) Point (6, 0) SECTION III: Perpendicular Bisectors l Line l is the perpendicular bisector of segment AB. Since l bisects segment AB, we know that the midpoint of AB is on the perpendicular bisector. Also, points on the perpendicular bisector are A B equidistant from points A and B.
Eample: Find the equation of the perpendicular bisector of AB for A(-1, ) and B(3, 4). 1 3 4 Find the midpoint of AB:, 1,3 This is a point on the perpendicular bisector. 4 1 Find the slope of AB: m. Therefore, the slope of the perpendicular bisector is -. 3 1 Use the slope of the perpendicular bisector (-) and a point on the perpendicular bisector (1, 3) to write the equation. y b 3 (1) b 5 b y 5 SECTION IV: Epansion Rules FOIL: a bc d ac ad bc bd Quadratic Algebra Difference of Two Squares: a b a b a ab ab b a a b a b a b a ab b Perfect Squares Epansion: a b a ba b a ab b Further Epansion: a bc d e a bc a bd a be ac bc ad bd ae be b SECTION V: Factorization of Quadratic Epressions A quadratic epression in is an epression of the form a + b + c where is the variable and a, b and c represent constants with a 0. Factorization is the reverse process of epansion. Eample 1: Removal of common factors. (GCF Greatest Common Factor) 4 a) 6 ( 3) b) ( ) ( ) ( ) ( ) ( )( 4) Eample : Difference of Two Squares. b a b a b **The sum of two squares is prime (not factorable).** a) 4 5 ( 5)( 5) a = b) 3 3( 48 16) 3( 4)( 4)
Eample 3: Quadratic Trinomial Factorization (a + b + c when a = 1). Find two values that multiply to c and add to b. Use Big X 3 9 6 5 4 a) b) c) 3( 3 ) 1 4 4 13 3( )( 1) Eample 4: Factoring a + b + c (a 1) Use the Big X splitting the middle term and using GCF a) 3 + 13 + 4 Product = ac = 1 Find two factors of 1 that add to b = 13 Factors: 1 and 1 Rewrite the epression by splitting the middle term: 3 + 1 + 1 + 4 Group and factor: 3 1 (1 4) 3( 4) 1( 4) (3 1)( 4) SECTION VI: Solving Quadratic Equations A quadratic equation in is an equation of the form a + b + c = 0 where is the unknown and a, b, and c are constants with a 0. Eample 1: Solve for by isolating the quadratic variable or epression and taking the square root of both sides. 1 31 3 16 11 3 8 3 3 4 3 6 a) b) c) 11 d) 16 3 4 11 4 7, 1 Eample : Solving by using the zero-product property. When the product of two or more numbers is zero, then at least one of them must be zero. If ab = 0, then a = 0 or b = 0. Set the equation equal to zero by moving everything to one side. Factor and apply the zero product property. 1 11 3 8 5 3 ( 6) 0 5 3 8 0 5 3 0 1 0 5( 1) ( 11) a) ( 7)( 4) 0 b) (5 )( 1) 0 c) 13 0 0 d) 5 5 11 7 0 7, 4 0 4 5 0, 5 1 0 1 ( 5)( 4) 0 5, 4 0 0 ( 5)( 1) 5, 6 5 1
Eample 3: Solving by using the quadratic formula. b a) 4 b 4ac a 8 8 8 1 0 64 4(4)(1) (4) 64 16 8 8 48 8 8 4 8 3 Eample 4: Problem-solving with quadratics. a) The product of a number and the number increased by 4 is 96. Find the two possible answers for the number. ( 4) 96 ( 1)( 8) 0 4 96 0 1, 8 3 b) A rectangle has length 5 cm greater than its width. If it has an area of 84 cm, find the dimensions of the rectangle. + 5 ( 5) 84 5 84 0 ( 1)( 7) 0 1, 7 Dimensions: 7 by 1 c) A rectangular enclosure is made from 40 m of fencing. The area enclosed is 96 m. Find the dimensions of the enclosure. y 40 y 0 y y 96 y 0 (0 ) 96 0 0 0 ( 1)( 8) 1,8 y 8,1 96 0 96 Dimensions: 1 by 8
Practice Problems: Don t forget to show all work!! SECTION I: Distance and Midpoint 1. Find b given that P(b, ) is equidistant from A(5, 4) and B(, 5).. PQ is the diameter of a circle, center (5, -½ ). Find the coordinates of P given that Q is (-1, ). SECTION II: Gradient, Equations of lines, Graphing 1. Find t given that the line joining P(t, -) to Q(5, t) is perpendicular to a line with gradient -1/3.. Given the points A(1, 3), B(-1, 0), C(6, 4), and D(t, -1), find t if AC is parallel to DB. 3. Find the equation of a line going through the point (-1, 3) that is a) horizontal b) vertical 4. Find the equation of the line in general form which passes through the points (5, -1) and (-1, -). 5. Find k if (-1, 3) lies on the line with equation 5 y = k.
6. Write the equation in gradient-intercept form. y a) b) (-3, 4) 7. Draw the graph of the line given. a) y = + 1 b) y = 3 + c) 4 + 3y = 1 d) 9 y = 9 8. Graph the lines + y = 6 and y = 6 and find their point of intersection.
SECTION III: Perpendicular Bisectors 1. Find the equation of the perpendicular bisector of AB for A(3, 1) and B(-3, 6). (Hint: Find a point on the perpendicular bisector and the slope of the perpendicular bisector and write the equation of the line. Use the eample in section III to help you.). Segment PQ has endpoints P(7, -1) and Q(a, b). The segment is perpendicularly bisected by line l at the point (, ). (Use the graph to help you visualize this problem.) a. Find the values of a and b. b. Find the equation of the perpendicular bisector (line l). SECTION IV: Epand and simplify. 1. 1 (5 ). 1 3 3. 3 4y3 4y 4. 5 5. 3 4 5 6. 3
SECTION V: Use the method of your choice to factor the epression. If the epression is not factorable, write prime. 1. 10 5. 3 9 3. 4 4. 4 1 5. 4 8 60 6. 3 40 7. 3 8. 15 9. 3 6 7 10. 7 1 8 11. 6 5 1 1. 8 14 3
SECTION VI: Solve the quadratic equation using the given method. Solve using square roots. 1. 1 3 8. 1 1 8 3 Solve using the zero-product property. 3. 10 3 4. 3 8 3 5. 4 5 6. 10 7 3 7. 3 5 3 17 8. 1 3 1 Solve by using the quadratic formula. 9. 3 0
Use problem-solving strategies to solve. 10. When 4 is subtracted from the square of a number, the result is five times the original number. Find the number. 11. A triangle has base cm more than its altitude (height). If its area is 49.5 cm, find its altitude. 1. A rectangular pig pen is built against an eisting brick fence. 4 m of fencing was used to enclose 70 m. Find the dimensions of the pen. (Hint: Write an equation for area and an equation for perimeter and solve the system of equations.) 13. A rectangular swimming pool is 1 m long by 6 m wide. It is surrounded by pavement of uniform width, 7 the area of the pavement being the area of the pool. 8 a. If the pavement is m wide, show that the area of the pavement is 4 36 m. b. Hence, show that 4 36 63 0. c. How wide is the pavement? 6 1