MATH 4 part 2

6. To solve simple logarithmic equations, transform the equation in exponential form, then solve for the missing term.7. To solve exponential equations involving different bases, find the logarithm of both sides of the equation with the aid of a scientific calculator or a table of logarithms then solve for the missing term applying the properties of equality. What have you learned1. Which of the following is the inverse of y = ax? a. y = ax c. y = logax b. y = xa d. a = yx2. Which of the following is equivalent to log464 = 3? a. 34= 64 c. 644 = 3 b. 43 = 64 d. 163 = 43. Which of the following is equivalent to (27 ) 1 = 3? 3 a. log273 = 1 c. log 1 3 = 27 3 3 b. log327 = 1 d. log3 1 = 27 3 34. At what point does the graph of y = log2x intersect the x-axis? a. (0, 1) c. (-1, 0) b. (0, 0) d. (1, 0)5. The graphs of y = 4x and y = log4x are symmetric with respect to what line? a. y = x + 1 c. y = x b. y = 4x d. y = 4x6. What is the point common to the graphs of functions y = log3x and y = log10x? a. (0, 0) c. (0, 1) b. (1, 1) d. (1, 0)7. Write 3 log a + 5 log b – 4 log c as a single logarithm. a. log  a3 • b5  c. log  a3 + b5  c4  c 4 b. log (a3 + b5 – c4) d. none of the above8. Simplify log381 + log33 – log39 as a single number.

a. 27 c. 3b. 5 d. 29. Given that log102 = 0.3010, log103 = 0.4771, and log105 = 0.6990, what islog 107.5?a. 0.5229 c. 0.8751b. 0.0791 d. 1.447110. Determine the value of x if log5 (8x - 7) = 2a. 4.875 c. 3.125b. 4 d. 2.25

Answer KeyHow much do you know1. d 6. d2. a 7. a3. a 8. d4. c 9. b5. c 10. dTry this outLesson 1A. 1. 25 = 32 2. 34 = 81 3. 121 = 12 4. 105 = 100000 5.  1 −3 = 125 5 6. 43 = 64 7. 2-4 = 1 16 8. rp = t 9. as = p 10. qm = pB. 1. log3243 = 5 2. log91 = 0 3. log11121 = 2 4. log 2 16 = 4 81 3 5. log16 4 = 1 3 2 9 6. log6 1 = −2 36 7. loga 1 = −5 a5 8. log16 4 = 1 2

9. log125 25 = 2 3 10. log64 16 = 2 3C. 1. 2 2. 7 3. 3 4. 2 5. 0 6. 3 2 7. 1 3 8. -3 9. -4 10. 2 5Lesson 2A. 30 28 26 y = 3x 24 22 20 18 16 14 12 10 8 6 1. y = log3x 4 2 0 -2 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 -2 -4 -6

18 y = 4x 2. y = log4x 16 14 246 8 10 12 14 16 18 20 12 10 8 6 4 2 0-4 -2 -2 0 -4 30 y = 5x 25 3. y = log5x 20 5 10 15 20 25 30 15 10 5 0-5 0 -53027 y =  1 x2421  3 1815129630-6 -3-3 0 3 6 9 12 15 18 21 24 27 30-6 4. y = log 1 x 3

18 16 14 y=  1  x 4 12 10 8 6 4 2 0-4 -2 0 2 4 6 8 10 12 14 16 18 -2 -4 5. y = log 1 x 42826 y =  1  x2422  5 201816141210864 6. y= log 1 x2 50-4 -2-2 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28-4B. The graph of y = log3x has the same shape and size as the graph of y = log 1 x . The same is true for the graphs of y = log4x and y = log 1 x , and y 34 = log5x and y = log 1 x . The graphs of y = log3x, y = log4x and 5 y = log5x are increasing while the graphs of y = log 1 x , y = log 1 x , and 34

y = log1 x are decreasing. The six graphs have a common point (1, 0). 5Lesson 3 Domain Range x-intercept Asymptote Trend A. + Real Nos. Real Nos. (1, 0) y-axis increasing No Equation + Real Nos. Real Nos. (1, 0) y-axis increasing . + Real Nos. Real Nos. (1, 0) y-axis increasing 1 y = log3x + Real Nos. Real Nos. (1, 0) y-axis decreasing 2 y = log4x 3 y = log5x 4 y = log 1 x 35 y = log 1 x + Real Nos. Real Nos. (1, 0) y-axis decreasing 46 y = log1 x + Real Nos. Real Nos. (1, 0) y-axis decreasing 5B. Common Properties: Domain, range, x-intercepts, and asymptotes.Lesson 4A. 6. log 2 xy 3 1. log232 = 5 z 2. log39 = 2 3. 4 log55= 4 7. log (x+ 5) 33 4. 4log77 = 4 8. logb(2x2 – 3x – 5) 5. 5 log33 = 5 22 9. log a x6y3z4 w3B. 1. 1.3222 10. log x2 − 4(x + 2)3 2. 1.7993 3. 0.6232 6. 1.03 4. 3.3804 7. -0.1091 5. 0.4133 8. 1.5066 9. 1.2131 10. 1.8178C. 4. z 1. 2z 5. z + 3 2. 6z 3. -3z

Lesson 5 10. 2 11. 4 A. 12. 4 1. 1 13. 4 343 14. 2 2. 9 15. 20 2 16. 13 3. 1 17. 1 6 4. 78125 3 5. 1 18. 8 2 19. 3 6. 7 3 7 7. 12000 20. 3 8. 5 9. 30 B. 1. 1.16 2. 1.32 3. 1.29 4. 8.21 5. -3.53What have you learned 1. c 2. b 3. a 4. d 5. c 6. d 7. a 8. c 9. c 10. b

Module 4 Quadratic FunctionsWhat this module is about This module is about the application of quadratic equations and functions in everyday situations.As you go over the different problems you will apply your knowledge and skills related to quadraticequations and functions in solving problems. Frequently, only one solution of the equation isrelevant to the problem at hand. The root which does not satisfy the conditions of the problem mustbe rejected. It is always a good practice, then, to check the solutions to determine if one or bothmay be used. The lessons were presented in a very simple way so it will be easy for you tounderstand and be able to solve problems alone without difficulty. Treat the lesson with fun andtake time to go back if you think you are at a loss.What you are expected to learn This module is designed for you to: 1. recall the different steps in solving word problems 2. translate verbal statements into symbols 3. apply knowledge and skills related to quadratic equations and functions in solving problems. How much do you know A. Write the letter of the correct answer. 1. The roots of a quadratic equation are 8 and -8. What is the equation? i. a. x2 + 64 = 0 b. x2 – 64 = 0 c. x2 – 8 = 0 d. x2 + 8 = 0 2. What are the values of a, b, and c in the equation 6x = 2x2 + 1 i. a. a = -2 b = -6 c = 1 ii. b. a = 2 b = -6 iii. c. a = -2 b = 6 c. = -1 iv. d. a = 2 b = -6 c. = 1 1

3. Which of the following is a solution of ( x – 3 ) ( x – 4 ) = 20i. a = -5 b = 4 c = 8 d = 204. Which of the following has roots of -3 and 7?i. a. x2 + 4x – 21 = 0 b. x2 – 4x – 21 = 0 c. x2 – 4x + 21 = 0 d. x2+ 4x + 21 = 05. Find the product of the roots of 2x2 = x + 3i. a. 1 b. 3 c. -1 d. − 3 2 22B. Write an equation to show the functional relationship between the two quantities involved in the problems using the indicated variables.1. The area (A) of a square of side s is s2.2. The perimeter of a rectangle is equal to twice the length plus twice the width.3. 12x plus seven is equal to 3x reduced by 44. The circumference (C) of a circle is twice the product of π and the radius (r).5. The total distance (d) covered is equal to the product of the rate (r) and the time (t).C. Solve the following problems.1. The product of two consecutive odd integers is 143. Find the integers.2. One side of a square is 10 cm longer than its width. The area of the rectangle is 96 sq. cm. Find the dimensions of the rectangle.3. The sum of two numbers is 22 and the sum of their squares is 250. Find the numbers.4. The sum of two numbers is 24. What is the maximum possible product of the two numbers?What you will do Lesson 1 Problems Involving Quadratic Equation Number Problems Many students could not solve this type of problem easily because they fail to translatecorrectly the different expressions given by the word problem into a correct equation. In the 2

following examples, you will see that before you solve a word problem, you must first translate theword problem into an equation.Steps in Solving word problems involving quadratic equation and function: 1. Identify your unknowns. 2. Write your equation. 3. Operate on the numbers. 4. Convert the equation into the standard form. Remember to change signs. 5. Factor the equation into its two corresponding linear equations. By factoring, you reduce the quadratic equation into two equivalent linear equations.Example 1.Find two consecutive integers whose product is 195.Solution:1st step: Identify your unknowns. Let x = one odd number, so x + 2 = next consecutive odd number.2nd step: Write your equation. x ( x + 2 ) = 195 one odd consecutive their number odd number product3rd step: Operate on the numbers. Multiply the two quantities. Thus: x2 + 2x = 1954th step: Bring all terms to the left side of the equal sign and equate to zero. Remember to change signs. So x2 + 2x - 195 = 05th step: Factor the equation. Thus: ( x + 15 ) ( x – 13 ) = 0 3

6th step: Solve for x. x + 15 = 0 or (x – 13) = 0 x = -15 or x = 13 the consecutive numbers are : x = 13 and x + 2 = 15, 13 and 15 or x = -15 and x + 2 = -15 + 2 = -13 , -13 and -15Check: or x ( x + 2) = 195 x ( x + 2) = 195 (-13)(-15) = 195 (13)(15) = 195 195 = 195 195 = 195Example 2:The sum of two numbers is 19 and their product is 60. Find the numbers.Solution:1st step: Identify the unknown. Let x = one number So 19 - x = the other number And x(19 – x) = their product.2nd step: Write the equation. x ( 19 – x ) = 603rd step: Operate the numbers to remove the parentheses grouping symbol. 19x – x2 = 604th step: Convert the equation to its standard form. Thus: 19x - x2 - 60 = 0 or x2 - 19x + 60 = 0 4

5th step: Factor the equation.( x – 15 ) ( x - 4 ) = 06th step: Solve for x.x – 15 = 0 and x – 4 = 0 x = 15 x =4The numbers are therefore 15 and 4Check: 15 + 4 = 19 Their sum: 19 = 19 check.Their product: ( 15 ) ( 4 ) = 60 60 = 60 check.Example 3.The sum of two numbers is 22 and the sum of their squares is 250. Find the numbers.Solution:1st step: Identify your unknown Let x = one of the two numbers 22 – x = the other number2nd step: Write the equation x2 + ( 22 – x )2 = 250 3rd step: Operate on the numbers to remove the parentheses grouping symbol. x2 + ( 22 – x )2 = 250 x2 + 484 - 44x + x2 = 250 2x2 - 44x + 484 = 250 5

4th step: Convert the equation to its standard form.2x2 - 44x + 484 - 250 = 02x2 - 44x + 234 = 0 divide the equation by 2x2 - 22x + 117 = 05th step: Factor the quadratic equation to obtain the two equivalents linear equations. ( x - 13 ) ( x - 9 ) = 06th step: Solve for x. x - 13 = 0 and x - 9 = 0 x = 13 and x = 9if x = 13, 22 – 13 = 9.if x = 9, 22 - 9 = 13.The numbers are 9 and 13.Check: 9 + 13 = 22;92 + 132 = 81 + 169 = 250. Try this out!Analyze and solve. 1. Find 2 consecutive integers such that three times the square of the first is equal to seven more than five times the second. 2. Find 2 consecutive even integers whose product is 224.. 6

3. Find an integer such that the square of the integer is eighty one less than eighteen times the integer. 4. One number is three times another and their product is equal to their sum increased by fifty five. Find the numbers. 5. The sum of two numbers is 16 and the sum of their squares is 146. Find the numbers. 6. A number is 16 less than its multiplicative inverse. Find the number. 15 7. Find two numbers whose difference is 5 and the difference of their squares is 65. 8. One number is three more than a second number. The sum of their squares is 37 more than the product of the numbers. Find the two numbers. 9. The product of two consecutive integers is 47 more than the next consecutive integer. Find the two numbers. 10. The tens digit of a certain number is 7 less than the units digit. The sum of the squares of the two digits is 85. Find the numbers. Lesson 2 Problems Involving Quadratic Equation Geometry Problems Many government and classroom examinations contain problems involving measurements of lengths or widths or areas or volumes of squares or rectangles or triangles or circles or other geometric figures. These problems are therefore also called geometry problems.Example 1:The altitude of a triangle is 3 cm. less than the base. The area of the triangle is 35 squarecentimeters. What are its dimensions?Solution: 7

1st step: Identify the unknown. Let x cm = length of the base So ( x – 3 )cm = length of the altitude Area of a triangle = (base)(altitude) 22nd step: Write the equation.35 = x(x − 3) 23rd step: Operate the numbers to remove the parentheses grouping symbol.x ( x – 3 ) = 70 x2 - 3x = 704th step: Convert the equation to its standard form. x2 - 3x - 70 =05th step: Factor the equation. ( x – 10 ) ( x + 7 ) = 06th step: Solve for x. and x + 7 = 0 x = -7 x – 10 = 0 x = 10We can not accept -7 as a value for the base of the triangle. Therefore the base of the triangle is10 cm while the altitude is x - 7 = 7 cm.Example 2:The perimeter of a rectangle is 138 meters and the area is 1080 square meters. Find the lengthand the width of the rectangle.Solution:1st step: Identify the unknown. 8

Let x cm = length of the rectangle So ( 69 - x )cm = the width Area of a rectangle = length x width2nd step: Write the equation. 1080 = x ( 69 – x )3rd step: Operate the numbers to remove the parentheses grouping symbol. x ( 69 - x ) = 1080 69x - x2 = 10804th step: Convert the equation to its standard form. x2 - 69x + 1080 = 05th step: Factor the equation. ( x – 45) ( x - 24 ) = 06th step: Solve for x.x – 45 = 0 and x - 24 = 0 x = 45 x = 24 x = 45 m is the length of the rectangle69 – 45 = 24 m is the widthPerimeter = 2( 45 ) + 2(24) = 138 m Area = 45( 24 ) = 1080 square metersExample 3:If the hypotenuse of a right triangle measures 13 cm and one leg is 7 cm more than the other, whatare the lengths of the two legs?Solution: 9

1st step: Identify the unknown. Let x cm = length of shorter leg ( x + 7 )cm = length of other leg Formula: a2 + b2 = c22nd step: Write the equation. x2 + ( x + 7 )2 = 1333rd step: Operate the numbers to remove the parentheses grouping symbol. x2 + x2 + 14x + 49 = 169 2x2 + 14x + 49 = 1694th step: Convert the equation to its standard form. 2x2 + 14x + 49 - 169 = 0 2x2 + 14x - 120 = 0 simplify by dividing by 2 x2 + 7x - 60 = 05th step: Factor the equation. ( x + 12 ) ( x - 5 ) = 06th step: Solve for x.x + 12 = 0 and x - 5 = 0 x = -12 x =5 Disregard -12 as a solution because it has no real meaning. You cannot have a length ofnegative 12. x = 5 m is the length of of shorter legx + 7 = 12 m is the length of the other legCheck: a2 + b2 = c252 + ( 5 + 7 )2 = 132 10

25 + 144 = 169 169 = 169Try this out!Solve: 1. The length of a rectangular floor is twice the width. The area of the floor is 32 m2. What are the dimensions of the room? 2. If the perimeter of a rectangular garden is 76m and the area is 360 square meters, what are the dimensions of the garden? 3. The base of a triangle is 7 cm more than its height. If the area is 294 square centimeters, find the base and the height of the triangle. 4. A square pool was surrounded by 3 meters wide Bermuda grass. If the total area of the sidewalk and the square pool is 196 sq. m , how long is each side of the pool? What is the area of the Bermuda grass sidewalk? 5. A rectangular flower garden with dimensions 3m by 7m is surrounded by the walk of uniform width. If the area of the walk is 11 sq. m, what is the width in meters? Lesson 3 Problems Involving Quadratic Equation Motion Problems Motion problems deal with three quantities. They are: Distance Rate or speed Time All uniform motion problems are tied-up with the formula: Distance = Time x Rate or D = rtExample 1 11

A car travels 10 kilometers per hour faster than a truck. The car goes 600 kilometers in 5 hoursless time than it takes the truck to travel the same distance. Find the rate of each vehicle inkilometers per hour.Solution:1st step: Identify the unknown.Truck Rate Time DistanceCar X 600 600 X + 10 x 600 600 x + 102nd step: Write the equation. 600 = 600 + 5 x x + 103rd step: Operate the numbers to remove the parentheses grouping symbol. Multiply both sides by the least common denominator which is x( x + 10 ) 600( x + 10 ) = 600x + 5( x2 + 10x ) 600x + 6000 = 600x + 5x2 + 50x4th step: Convert the equation to its standard form. Thus: 5x2 + 600x - 600x + 50x - 6000 = 0 Simplify by dividing the equation by 5 x2 + 10x - 1200 = 05th step: Factor the equation. ( x + 40 ) ( x - 30 ) = 06th step: Solve for x. x + 40 = 0 and x – 30 = 0 12

x = -40 x = 30We must disregard the negative value. Thus, the rate of the truck is 30 kilometers per hour. Therate of the car is x + 10 = 40 kilometers per hour.Example 2:A man drives 500 km to a business convention. On the return trip, he increases his speed by 25km per hour and saves 1 hour of driving time. How fast did he go in each direction?Solution:1st step: Identify the unknown.Let r = speed in going to the convention r + 25 = speed of the return trip 500 = length of time to the convention r 500 = length of time in returning from the conventionr + 25 ( length of time in going ) = ( length of time in returning ) + 12nd step: Write the equation. 500 = 500 + 1 r r + 253rd step: Operate the numbers to remove the parentheses grouping symbol. Multiply both sides by the least common denominator which is r( r + 25 ) 500( r+ 25 ) = 500r + r( r + 25 ) 500r + 12, 500 = 500r + r2 + 25r 500r + 12,500 = 525r + r24th step: Convert the equation to its standard form. Thus: 0 = r2 + 525r - 500r - 12,5005th step: Factor the equation. 13

0 = ( r + 125 ) ( r - 100 )6th step: Solve for x.r + 125 = 0 or r – 100 = 0 r = -125 r = 100We must disregard the negative value. Thus, the rate of the man is 100 kilometers per hour ingoing to the convention and returns at 100 + 25 = 125 kilometers per hour.Example 2:A man can row 8 kilometers downstream and back in 6 hours. If the rate of the stream is 1kilometer per hour, what is the rate of the man rowing in still water?Solution:1st step: Identify the unknown. Let x = rate of the man in still waterDownstream DrTUpstream 8 X+1 8 x +1 8 X-1 8 x −12nd step: Write the equation. 8 + 8 =6 x +1 x −13rd step: Operate the numbers to remove the parentheses grouping symbol. Multiply both sides by the least common denominator which is ( x + 1 ) ( x - 1 )8( x – 1 ) + 8( x + 1 ) = 6( x + 1)(x – 1)8x – 8 + 8x + 8 = 6( x2 – 1 ) 14

16x = 6x2 - 64th step: Convert the equation to its standard form.Thus: 0 = 6x2 - 16x - 6Simplify by dividing the equation by 20 = 3x2 - 8x - 35th step: Factor the equation.0 = ( 3x + 1 ) ( x - 3 )6th step: Solve for x.3x + 1 = 0 or r – 3 = 0x = −1 r =3 3We must disregard the negative value. Thus, the rate of the man is 3 kilometers per hour.Try this outAnswer the following problems: 1. Two cyclist A and B traveled the same distance of 120 kilometers. A traveled 4 kilometers per hour faster than B and covered the distance in one hour less than B. Find the rate of each. 2. If Mr. Cruz will increase his usual rate by 10 kilometers per hour, it will take him one hour shorter to cover a distance of 200 kilometers. What is his usual speed in driving? 3. If you can row upstream to a landing 5 kilometers away and then row back to your starting point all in three hours and 20 minutes and if the river has an average current of 2 kilometers per hour, at what rate are you able to row in still water? 4. A horse travels 40 kilometers per hour faster than a mule. The horse goes 360 kilometers in three hours less time than the mule goes 360 kilometers. Find the speed of each animals? 15

5. An airplane flies 900 miles against a headwind of 25 miles per hour. The plane took 15 minutes longer for this flight than with a tailwind of 25 miles per hour. How fast could the plane fly in still air? Lesson 4 Problems Involving Quadratic Equation Work ProblemsWork problems are important not only to students but also to industrialists, engineers, productionexecutives and managers as well. This is because everyone often meets work problems in real life.Common work problems have the following characteristics 1. They ask you how long individual can finish a job alone, when working with others when his rate of working is given; and 2. They ask you how long a group of workers can finish a job, if rate of work of each member of the group is given.Example 1:Mario takes 5 days more to do a job than Jose. Together they can do it in 6 days. How long does ittake each alone to do the job?Solution:1st step: Identify the unknown. No. of hrs. to do job Amount of work done X 1JoseMario X+5 xTogether 1 6 x+5 1 62nd step: Write the equation. 16

1+ 1 = 1 x x+5 63rd step: Operate the numbers to remove the parentheses grouping symbol. Multiply both sides by the least common denominator which is 6x( x + 5 ) 6( x + 5 ) + 6x = x( x + 5 ) 6x + 30 + 6x = x2 + 5x4th step: Convert the equation to its standard form. Thus: 0 = x2 + 5x - 12x - 30 0 = x2 - 7x - 305th step: Factor the equation. 0 = ( x + 3 ) ( x - 10 )6th step: Solve for x. x+3 = 0 or x – 10 = 0 r = -3 r = 10The solutions of the fractional equation are 10 and -3. However, the solution of the problem is x =10 and x stands for the number of days it takes Jose to do the work alone x + 5 = 15 is the numberof days it takes Mario to do the work alone.Example 2:A tank has a supply pipe and an exhaust pipe. The exhaust pipe takes 5 minutes longer to emptythe tank than for the supply pipe to fill it. If both are open, it takes the supply pipe 30 minutes to fillthe tank. Find how long it takes the supply pipe to fill the tank when the exhaust pipe is closed.Solution:1st step: Identify the unknown.Let x = no. of minutes to fill in the tank 17

1 = part of the tank filled in one minute x x + 5 = no. of minutes to empty the tank 1 = part of the tank emptied in one minute x+5 It takes 30 minutes to fill the tank with both pipes open2nd step: Write the equation. 30 - 30 = 1 x x+53rd step: Operate the numbers to remove the parentheses grouping symbol. Multiply both sides by the least common denominator which is x( x + 5 ) 30( x+ 5 ) - 30r = x( x + 5 ) 30x + 150 - 30x = x2 + 5x4th step: Convert the equation to its standard form. Thus: 0 = x2 + 5x - 1505th step: Factor the equation.0 = ( x + 15 ) ( x - 10 )6th step: Solve for x.x + 15 = 0 or x– 10 = 0 x = -15 x = 10It takes the supply pipe 10 minutes to fill the tank when the exhaust pipe is closed.Try this outAnswer the following problems: 18

1. Charlie can construct 100 identical boxes in two days less than Cholo. If they work together they can finish the work in 5 5 days. How long will it take each of them to 6 finish 100 boxes? 2. It takes Peter 6 hours longer than Luis to do a certain job. Together they can do it in 4 hours. How long would it take each working alone to do the job? 3. Mang Tano and his son Mario can finish planting their crops in 4 days if they work together. Working alone, Mario will need 2 1 days more than the number of days that it 3 will take his father to finish the work alone. How many days will each of the father and son need to finish the work if each will work alone? 4. Mr. Cruz can paint a house in 2 days less than his son. When they work together they can do the job in 4 4 days. How long would it take each working alone to do job? 9 Lesson 5 Problems Involving Quadratic Function (Maximum or Minimum Point) Quadratic functions have practical applications. Many applications of quadraticfunctions are in simple maximization and minimization problems. The technique for solvingsuch problems is based on the fact that a quadratic function defined by f(x) = ax2 + bx + cattains its maximum value (for a 〈 0) or minimum value (for a 〉 0) at x = h = - b and the 2amaximum/minimum value is k = f(h) = 4ac − b2 . 4aExample 1:The sum of two numbers is 48. Find the maximum product and the value of the two numbers.Solution: 19

Let x = the first number 48 – x = the second number f(x) = the product of the two numbers y = f(x) = x(48 – x) = 48x – x2 or y = -x2 + 48x attains its maximum value at x = − b = − 24 = 12 2a 2(−1) The maximum product is 12(24 – 12) = 144Example 2:A rectangular garden is x meters long and (18 – x) meters wide. If y is the area, find themaximum area that the garden could have. What could be its dimensions?Solution: Let x = the length 18 – x = width y(area) = x(18 – x) = 18x – x2 or y = -x2 + 18x attains its maximum value at x = − b = −18 = 9 2a − 2 The maximum area is 9(18 – 9) = 18 square meters and the dimensions are length = 9mand the width = 9m.Example 3: A rectangular lot is bounded on the side by a riverand on the other three sides by a total of 80m fencing.Find the possible dimensions of the lot. 20

Solution: Step 1. The quantity to be maximized is the area of the lot. Step 2. Let x = be the width y = be the length If A is the area, then A = xy Step 3. Since the total length of fencing is 80 meters, 2x + y = 80 or y = 80 – 2x Substitute 80 – 2x for y in A = xy A = x(80 – 2x) = -2x2 + 80x Step 4. x = − b = − 80 = − 80 = 20 2a 2(−2) − 4 Area = -2x2 + 80x = -2(20)2 + 80(20) = -800 + 1600 = 800 Then maximum value of A is 800 and occurs when x = 20. Step 5. Since x = 20 and 2x + y = 80, y = 40. The dimensions of the largest possible rectangular lot are 40 meters by 20 meters.Example:A company charges P200 for each leather bag on order of 150 or less. The cost of each bag isreduced by P1 for each order in excess of 150. How many bags on order would result in amaximum revenue? what is the maximum revenue?Solution: Step 1. The quantity to be maximized is total revenue (R) in pesos from the leather bag to be sold. Step 2. Let x = be the number of bags ordered x - 150 = be number of bag ordered in excess of 150 200 – 1(x – 150) = 200 – x + 150 = 350 – x is the cost of each bag R = ( cost per bag ) ( number of bags sold ) = ( 350 - x )x 21

= 350x - x2 or -x2 + 350x Step 3. x = − b = − 350 = − 350 = 175 2a 2(−1) − 2 R = -x2 + 350x = -( 175 )2 + 350( 175 ) = 30,625 The maximum value of R is 30,625 and it occurs when x = 175 Step 5. A purchase of 175 bags maximizes the company’s revenue. The maximum revenue is P30,625Try this outAnswer the following problems: 1. What are the dimensions of the largest rectangular field that can be enclosed with 60 meters of wire? 2. Two numbers have a difference of 16. Find the numbers if their sum and product is a minimum. 3. The sum of two numbers is 40. If one is x, what is the other? What is the maximum product that the 2 numbers could have? 4. A rectangular lot is bordered on one side by a string and on the other three sies by 600 meters of fencing. Find the dimensions of the lot if its area is a maximum. 5. A theater seats 2000 people and charges P10 for a ticket. At this price, all the tickets can be sold. A survey indicates that if the ticket price is increased by a peso, the number of tickets sold will decrease by 100. What ticket price result in the greatest revenue? 22

Let’s summarize Steps in Solving word problems involving quadratic equation and function: 1. Identify your unknowns. 2. Write your equation. 3. Operate on the numbers. 4. Convert the equation into the standard form. Remember to change signs. 5. Factor the equation into its two corresponding linear equations. By factoring, you reduce the quadratic equation into two equivalent linear equations.What have you learnedA. Define a variable and write a quadratic function to describe each of the following. 1. The product of two numbers whose sum is 40. 2. The product of two numbers whose difference is 25. 3. The area of a rectangle whose perimeter is 20 centimeter. 4. The product of two consecutive integers is nine less than the square of the second. 5. Write a quadratic function to describe the area of a circle in terms of its radius.B. Answer each of the following problems completely. 1. Find two consecutive even integers whose product is 168. 2. The sum of two numbers is 11. The difference of their squares is 11. What are the numbers? 3. The perimeter of a garden is 66 meters and its area is 270 sq. meters. Find the dimensions of the garden. 4. What two integers having a sum equal to 66 will have a maximum product? 5. Two bikers started at the same corner, one going east, the other going north. One biker is traveling at 3 kph faster than the other. After one hour, the two bikers are 15 kilometers apart. Find the rate of each.Answer Key 23

How much do you knowA. 1. b B. 1. A = s2 C. 1. 11 and 13 or -13 and -112. d 2. P = 2l +2w 2. l = 16 cm. w = 6 cm3. c 3. 12x + 7 = 3x – 4 3. 9 and 134. b 4. c = 2π 4. 1445. d 5. d = rtLesson 1 Try it out1. x = 3 and x + 1 = 4 6. − 5 and 32. 14 and 16 35 7. 4 and 93. n = 9 8. -7 and -4 or 4 and 74. 5 and 15 9. 7 and 8 or -7 and -85. 5 and 11 10. n = 29Lesson 2 Try it out1. width = 4 meters and length = 8 meters2. length = 20meters and width = 18 meters3. base = 28 meters height = 21 meters4. The length of each side of the pool = 8 meters area of the Bermuda grass sidewalk = 132 square meters5. ½ m is the width of the walkLesson 3 Try it out1. B = 20kph A = 24 kph 2. The usual speed of Mr Cruz is 40 kph3. The rate in still water 4. Rate of mule = 14.5 kph Rate of horse = 56.6 kph 24

5. r = 425 kphLesson 4 Try it out 1. Charlie = 5 days Cholo = 7 days 2. Luis = 6 hours and Peter = 12 hours 3. Mario = 9 1 days Mang Tano = 7 days 4. son = 10 days Mr. Cruz = 8 days 3Lesson 5 Try it out 1. width = 15 meters length = 15 meters 2. – 9 and 7 3. 20 is the other number; 400 is the maximum product 4. length = 300m width = 150 meters 5. P20 ticket price would result to P21,000 greatest revenue.What have you learned?A. 1. y = x( 40 – x ) 2. y = x( x – 25 )3. A = l( 10 – l ) 4. P = x( x + 1 ) = ( x + 1 )2 - 9 5. A = πr2B. 1. -14 and -12 or 12 and 142. 6 and 5 3. length = 18 meters and width = 12 meters4. the maximum product is 10895. rate of slower biker = 9 kph rate of faster biker = 12 kph 25

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Module 4 Circular Functions What this module is about This module will teach you about graphical representation of trigonometricfunctions how this function behave when plotted in a rectangular coordinatesystem and what shape will it be form. You will also learn to define and findthe values of the six trigonometric functions of an acute angle which is in thestandard position. What are you expected to learn This module is designed for you to : 1. Describe the properties of sine and cosine function 2. Draw the graph of sine and cosine function 3. Define the six trigonometric functions of an angle in standard position whose terminal point is not on the unit circle 4. Find the values of six trigonometric functions of an angle , given some conditions. How much do you knowGiven the following functions, identify the amplitude of :1. y = 2 cos x2. y = ¾ sin x3. y = -2 sin 3/2 xGiven the following functions, determine the period of:4. y = 2 cos ½ x5. y = 4 sin 2/3 x6. y = sin 1/5 x7. What is the value of y = 4 sin ½ x, if x = 5π/3 ?

Given the figure, find the values of the radius and the six trigonometric functions. 12 -5 8. r = 9. Sin A = 10. Cos A = 11. Tan A = 12. Cot A = 13. Csc A = 14.Sec A=Find the values of the other five trigonometric function for /S if Tan S = -6/8,Sin S < 0 15. r = 16. Sin S = 17. Cos S = 18. Cot S = 19. Csc S = 20. Sec S =

What you will do Lesson 1 Describe the properties of Sine and Cosine functions Two properties of Sine and Cosine functions are amplitude and a period ofa function. The function in the form of y = a sin bx, the amplitude is a of thefunction which is the maximum point of the graph and the period is 2π/b wherethe graph repeat the cycle.Example : Determine the amplitude and the period of the given function: 1. y = 3 sin 2x Solution: a = 3; b = 2 a. amplitude is a = / a / = /3/ = 3 b. period is P = 2π/b = 2π /2 = π 2. y = ½ cos x Solution: a = ½; b = 1 a. amplitude is a = /a/ = /1/2 / = ½ b. period is P = 2π /b = 2π/1 = 2π 3. y = -3 sin 4x Solution: a = -3; b = 4 a. amplitude is a = /a / = /-3/ = 3 b. period is P = 2π/ b = 2π/ 4 = π/2 4. y = cos 4x Solution: a = 1; b = 4 a. amplitude is a = /a / = /1/ = 1 b. period is P = 2π/b = 2π/4 = π/2 5. y = -2/3 sin x Solution:

a = -2/3; b = 1 a. amplitude is a = /a/ = /-2/3/ = 2/3 b. period is P = 2π/b = 2π/1 = 2πTry this out A. Determine the amplitude of the following functions. 1. y = 3 sin x 2. y = 2 sin 1/2x 3. y = 3/2 cos 2x 4. y = -2 c0s 2x 5. y = cos 4 x 6. y = 2 sin 2x 7. y = ½ sin x 8. y = ¾ sin ½ x 9. y = -4 sin 3x 10. y = -2 sin 3/2 x B. Determine the period of the following functions. 1. y = 6 sin 2/3x 2. y = 5 sinx 3. y = 4 sin 1/2x 4. y = -1/2 cos 3/4x 5. y = 3 cos 1/2x 6. y = cos 3x 7. y = -3 sin 2/3x 8. y = -6 sin 2x 9. y = 2 sin 1/5x 10. y = sin 4x Lesson 2 Draw the graph of Sine and Cosine functions where 0< A <2π To graph the sine or cosine function as y = sin x, where x represents theabscissa or the x-coordinate of a point while y represents the ordinate or the y-coordinate of a point. Start by constructing table of values assigning values of xa set of real numbers or angles in degrees and then solve for y. Then markthese points on the rectangular coordinate system following the table of valuesand connect forming a smooth curve.Example 1: Construct table of values and draw the graph of y = sin x , 0 < x < 2π. (note: use your scientific calculator to lessen difficulty in computation.)

Solution: 2π a = 1; b = 1 0 a. amplitude is a = /a/ = /1/ = 1 b. period is P = 2π/b = 2π/1 = 2π Table of valuesx 0 π/4 π/2 3π/4 π 5π/4 3π/2 7π/4Y 0 √2/2 1 √2/2 0 -√2/2 -1 -√2/2The graph of Y = sin x0π π 3 π 2π 2 2Example 2: Construct table of values and draw the graph of y = cos x , 2π 0 < x < 2π. 1 Table of values x 0 π/4 π/2 3π/4 π 5π/4 3π/2 7π/4 Y 1 √2/2 0 √2/2 -1 -√2/2 0 √2/2 Graph of y = cos x0π π 3 π 2π 3 2

Example 3: Construct table of values and draw the graph of y = 3 sin x , where o < x < 2π Solution: a = 3; b = 1 a. amplitude is a = /a/ = /3/ = 3 b. period is P = 2π/b = 2π/1 = 2πTable of valuesx 0 π/4 π/2 3π/4 π 5π/4 3π/2 7π/4 2πY = 3sin x 0 2.12 3 2.12 0 -2.12 -3 -2.12 0Solutions in finding values of y:a. If x = 0 b. If x = π/4 c. If x = π/2 d. If x =3π/4 y = 3 sin π/2 y = 3 sin 3π/4y = 3 sin 0 y = 3 sin π/4 y = 3 (1) y = 3 sin (√2/2) y = 2.12y = 3 (0) y = 3 (√2/2) y=3y=0 y = 2.12e.if x =π f. x = 5π/4 g. if x = 3π/2 h. if x = 7π/4 y = 3 sin π y = 3 sin 5π/4 y = 3 sin 3π/2 y = 3 sin 7π/4 y = 3 (0) y = 3 (-√2/2) y = 3 (-1) y = 3 (-√2/2) y= 0 y = -2.12 y = -3 y = -2.12i.If x = 2π y = 3 sin 2π y = 3 (0) y=0Graph of y = 3 sin x0 π/4 π/2 3π/4 π 5π/4 3π/2 7π/4 2π Notice that the height of the graph shifted 3 units

compared to the graph of y = sin x in example 1 but their period remain the same.Example 4: y = 3 cos x , 0< x < 2πSolution: A = 3 ; b= 1 a. amplitude is a = /a/ = /3/ = 3 b. period is P = 2π/b = 2π/1 = 2πTable of valuesx 0 π/4 π/2 3π/4 π 5π/4 3π/2 7π/4 2πY =3 cosx 3 2.12 0 -2.12 -3 -2.12 0 2.12 3Solutions in finding values of y : a. If x = 0 b. If x = π/4 c. If x = π/2 d. If x = 3π/4 y = 3 cos π/2 y = 3 cos 3π/4 Y =3 cos0 y = 3 cos π/4 y = 3 (0) y = 3 (-√2/2) y =0 y = -2.12 Y = 3 (1) y = 3(√3/2) Y=3 y = 3 √2/2 e. If x = π f. If x = 5π/4 g. If x = 3π/2 h. If x = 7π/4 y = 3 cos π y = 3 cos 5π/4 y = 3 cos 3π/2 y = 3 cos 7π/4 y = 3 (-1) y = 3 (-√2/2) y = 3 (0) y = 3 (√3/2) y = -3 y = -2.12 y=0 y = 2.12 i. If x = 2π y = 3 cos 2π y = 3 (1) y=3 Graph of y = 3 cos x 2 1 0 π/4 π/2 3π/4 π 5π/4 3π/2 7π/4 2π -1 -2

-3 Looking at the graph the amplitude shifted 3 units as comparedto the graph of y = cos x but their period remained unchanged.Try this out 1. Construct the table of values and draw the graph of y = ½ cos x, where 0 < x < 2π Table of values Graph of y = ½ cos x x 0 π/4 π/2 3π/4 π 5π/4 3π/2 7π/4 2πy = ½ cos x π/4 π/2 3π/4 π 5π/4 3π/2 7π/4 2π2. Construct the table of values and draw the graph of y = 4 sin ½ x, where 0 ≤ x ≤ 2π. Table of values x 0 π/3 π/2 2π/3 π 4π/3 3π/2 5π/3 2π y = 4 sin 1/2x

Graph of y = 4 sin ½ xπ/3 π/2 2π/3 π 4π/3 3π/2 5π/3 2π3. Construct the table of values and draw the graph of y = 2/3 cos x, where 0 ≤ x ≤ 2π. Table of valuesx 0 π/3 π/2 2π/3 π 4π/3 3π/2 5π/3 2πy = 2 cos 1/2x Graphs of y = 2 cos 1/2 x.5.5 π/3 π/2 2π/3 π 4π/3 3π/2 5π/3 2π4. Construct the table of values and draw the graph of y = 2 cos 1/2x, where 0 < x < 2π.

Table of values x0 π/4 π/2 3π/4 π 5π/4 3π/2 7π/4 2πy = 2/3 cos x Graph of y = 2 cos ½ x0 π/3 π/2 2π/3 π 4π/3 3π/2 5π/3 2π5. Construct the table of values and draw the graph of y = 3/4 sin 1/2x, where 0 < x < 2π Table of values x 0 π/3 π/2 3π/3 π 4π/3 3π/2 5π/3 2πy = ¾ sin 1/2x Graph of y = ¾ sin ½ x0 π/3 π/2 2π/3 π 4π/3 3π/2 5π/3 2π

Lesson 3Define the six trigonometric function of an angle in standard position where theterminal point is not on the unit circle.Recall trigonometric functions of unit circle as:Sin A = yCos A = xTan A = y/xThe three other functions are:Secant of A : Sec A = 1/xCosecant of A : Csc A = 1/yCotangent of A : Csc A = x/y Unit circleThe trigonometric functions of /A if the terminal side of A is not within theunit circle , the function are defined as: Sin A = y/r Cos A = x/r Tan A = y/x Sec A = r/x Csc A = r/y Cot A = x/yExample !: Find the ratios of the functions of /A in standard position if coordinates of P(5,12) lies on its terminal side.Solution: x = 5 ; y = 12 r = y = 12 A x=5You need to find r :Using Pythagorean Theorem r = √ x2 + y2 = √ 52 + 122 = √ 25 + 144

= √ 169 r = 13The ratio of the functions are : Sin A = y/r = 12/13 Cos A = x/r = 5/13 Tan A = y/x = 12/5 Sec A = r/x = 13/5 Csc A = r/y = 13/12 Cot A = x/y = 5/12Example 2: If P(-4,5) find the values of radius and six trigonometric function for angle A . Solution: x = -4 ; y = 3 Figure y=3 r x = -4 You need to find r: By Pythagorean Theorem r = √ x2 + y2 = √ -42 + 32 = √ 16 + 9 = √ 25 r =5 The ratio of six functions are: Sin A =y/r = 3/5 Cos A = x/r = -4/5 Tan A = y/x = 5/-4 Cot A = x/y = -4/5 Csc A = r/y = 5/3 Sec A = r/x = 5/-4Example 3: If P(-1,-7) find the values of radius and six trigonometric function for angle A .

Solution: x = -1 ; y = -7 Figure: x =-1 y = -7 r Solve for r:By Pythagorean Theorem r = √ x2 + y2 = √ -12 + -72 = √ 1 + 49 = √ 50 = √ (25)(2) r = 5√ 2The ratios of the six functions are: Sin A = y/r = -7 5√ 2 = -7 (5√ 2) (5√ 2) (5√ 2) = -7√ 2 10 Cos A = y/r = -1 5√ 2 = -1 ((5√ 2) (5√ 2) (5√ 2) = -√ 2 10 Tan A = y/x = -7

-1 =7 Sec A = r/x = 5√ 2 -1 = -5√ 2 Csc A = r /y = 5√ 2 -7 Coy A = x/y = -1 -7 =1 7Example 4: If P(8,-10) find the values of radius and six trigonometric function for angle A .. Solution: x = 8 ; y = -10 Figure: X=8 r y = -10 Solve for r: Using Pythagorean Theorem r = √ x2 + y2 = √ 82 +( -10) 2 = √ 64 + 100 = √ 164 = √ (4)(41) r = 2√ 41 The ratios of six functions are: Sin A = y/r = -10 2√ 41

= -10 (2√ 41) rationalize (2√ 41) (2√ 41) = 10 (2√ 41) 4(41) = - 5√ 41 4 Cos A = x/r = 8 2√ 41 = 8 (2√ 41) ( 2√ 41)(2√ 41) rationalize = 8(2√ 41) 4(41) = 4√ 41 41 Tan A = y/x = 8 -10 = -4/5 Sec A = r/x = 2√ 41 8 = √ 41 4 Csc A = r/y = 2√ 41 -10 = -√ 41 5Try this out: Sketch the figure then find the value of r and six trigonometric functionsgiven are the coordinates of the terminal point:1. (-5, 7) Draw the figure Find: r=a. Sin A = d. Csc A =

b. Cos A = e. Sec A = f. Cot A = c. Tan A = Draw the figure.2. (-8, -15) Find : d. Csc A = r= e. Sec A = f. Cot A = a. Sin A = Draw the figure b. Cos A = d. Csc B = c. Tan A = e. Sec B = d. Cot B =3. (24, - 7) Find: d. Csc A = Draw the figure r= e. Sec A = Draw the figure f. Cot A = a. Sin B = d. Csc A = b. Cos B = e. Sec A = f. Cot A = c. Tan B =4. (2, 3) Find: r= a. Sin A = b. Cos A = c. Tan A =5. (-9, 40) Find: r= a. Sin A = b. Cos A = c. Tan A =