MATH 1

Answer: To find the coordinates of a point on the Cartesian plane, we draw a rectangle thathas the given point and the origin as two of its vertices. The x-coordinate is the distance ofthe vertical line passing through the given point from the y-axis while the y-coordinate isthe distance of the horizontal line passing though the given point from the x-axis. a) T(-4, 3) c) A(4, -4) b) E(3, 2) d) M(-3,-2) Every point on the Cartesian plane has a corresponding distance from the x and they axes. If a point has coordinates (x, y), the absolute value of x represents the distance ofthe point from the y-axis and the absolute value of y represents the distance of the pointfrom the x-axis.Example 6 Plot the following points on the Cartesian plane: a) A(-3, 1) c) C(-2, -1) b) B(3, 4) d) D(3, -2).Answers:Referring to the Cartesian plane that follows,a) A(-3, 1) means that point A is located 3 units to the left of the y- axis and 1 unit above the x-axis.b) B(3, 4) means than point B is located 3 units to the right of the y- axis and 4 units above the x-axis.c) C (-2, -1) means that point C is located 2 units to the left of the y- axis and 1 unit below the x-axis.d) D (3, -2) means that point D is located 3 units to the right of the y-axis and 2 units below the x-axis. 10

y-axis 5 4 • B(3,4) 3A (-3, 1) • 2 x-axis -5 -4 -3 -2 -1 1 • 01 2 3 45 -1 C (-2,-1) -2 • D(3, -2) -3 -4 -5 Did you know1. The Cartesian coordinate system, which uses the Cartesian plane in plotting points, is also known as the rectangular coordinate system.2. The Cartesian plane is named after the French mathematician Rene Descartes, one of the pioneer mathematicians who invented and used the system.3. The x-value of the coordinates (x, y) of a point is also known as the abscissa, while the y-value is also known as the ordinate.4. The Cartesian plane is divided into four quadrants by the x and the y axes. These quadrants are labeled counterclockwise as QI, QII, QIII, and QIV. As a general rule, we consider the axes not to belong to any of the quadrants. 11

The Four Quadrants y-axis QI QII 5 (+,+) 4 (–,+) 3 2 -5 -4 -3 -2 -1 1 QIII x-axis (–, –) 01 2 3 45 -1 -2 -3 QIV -4 (+, –) -5Example 7 Determine the quadrant where each point lies.a) (-4, 6) b) (7, 4) c) (0, 5)Answers: a) Since (-4, 6) has a negative abscissa and positive ordinate, it lies in QII. b) Since (7, 4) has a positive abscissa and positive ordinate, it lies in QI. c) The point (0, 5) does not belong to any quadrant. It lies along the positive y-axis.Self-check 2 A. Use the following figure to answer items 1 and 2. 12

y-axis 5 4 .A. 3 Q 2 .B .O .1 x-axis L -5 -4 -3 -2 -1 0 34 5 -11 2 E. -2 .T .P -3 -41. Identify the coordinates of the following points.a) B b) A c) E d) T2. Name the point that corresponds to the given coordinates.a) (-4, -3) b) (-3, 1) c) (0, 2) d) (3, -2)B. Identify the quadrant/s that satisfy the given condition.1. The x-coordinate is negative and the y-coordinate is positive.2. The x and y coordinates are both negative.3. The ordinate is positive.4. The abscissa is negative.5. The quadrant is above the x-axis and to the left of the y-axis.C. Plot the following points on the Cartesian plane. 5. E (0, -2)1. A (-1, -3) 2. B (4, 0) 3. C (3, -4) 4. D (1, 1)D. Define the following terms.1. origin 2. abscissa 3. ordinate 4. quadrant 5. y-axis Answer Key on page 20 13

Lesson 3 Finding the Distance and the Midpoint of Two Points In Lesson 2, we learned how to plot points on the Cartesian plane. We now focus onhow to find the distance between two given points and their midpoint.Did you know? Given two points A(x1, y1) and B(x 2 , y 2 ) , the distance d between A and B is given bythe formula: d = (x1 − x2 )2 + ( y1 − y2 )2 . This formula is commonly known as the distance formula. Note that the distance dbetween the points A and B is also the length of the line segment determined by A and B. y-axis 5 • B(x2,y2) 4 d3 • A (x1,y1) 2 1 x-axis-5 -4 -3 -2 -1 0 1 2 3 4 5 -1 -2 -3 d = length of the line segment determined by A and B -4 -5Example 8 Find the distance between each pair of points:a) (-1, 3) and (5, -5) b) (4, 0) and (-1, 5) 14

Answer: a) Let x1 = −1 , y1 = 3 , x 2 = 5 , and y 2 = −5 . By substituting these into the formula,we have d = (−1 − 5)2 + (3 − (−5))2 d = (−6)2 + (8)2 = 36 + 64 d = 100 = 10 units.b) Let x1 = 4 , y1 = 0 , x 2 = −1, and y 2 = 5 . By substituting these into the formula,we have d = (4 − (−1))2 + (0 − 5)2 d = (5)2 + (−5)2 = 25 + 25 d = 50 units.Example 9 What is the length of the line segment determined by A(-2, 3) and B(4, 1)?Answer: The length of the line segment determined by the points A and B is given byd. We let x1 = −2 , y1 = 3 , x 2 = 4 , and y 2 = 1. By substituting thesevalues into the formula, we have d = (−2 − 4)2 + (3 −1)2 d = (5)2 + (−5)2 = 25 + 25 d = 50 units.Did you know? The midpoint of a line segment is the point that divides the segment into two equalparts. It is also the middlemost point of the line segment. To find the coordinates of themidpoint M of the line segment determined by the points A(x1, y1) and B(x 2 , y 2 ) , we usethe midpoint formula given by: M =  x1 + x 2 , y1 + y 2  . 2 2 15

y-axis 5 • B(x2,y2) 4 •3 M• A (x1,y1) 2 1 x-axis-5 -4 -3 -2 -1 0 1 2 3 4 5 -1 -2 -3 M = the midpoint of A and B -4 -5Example 10 Find the coordinates of the midpoint of the line segment determined by the following pair of points:a) (-4, 3) and (6, 5) b) (3, 4) and (2, -6)Answer: a) We let x1 = −4 , y1 = 3 , x 2 = 6 , and y 2 = 5 . By substitution, the coordinates of the midpoint is given byM =  − 4 + 6 , 3 + 5  =  2 , 8  = (1,4) .  2 2  2 2 b) We let x1 = 3 , y1 = 4 , x 2 = 2 , and y 2 = −6 . By substitution,the coordinates of the midpoint is given byM =  3 + 2 , 4 + (−6)  =  5 , − 2  =  5 ,−1 .  2 2  2 2  2 Example 11. The midpoint of the line segment AB is M(-3, 4). If point A has coordinates (7, 2), find the coordinates of point B. 16

Answer: Since A has coordinates (7, 2), we let x1 = 7 , y1 = 2 . The coordinates of the midpoint M are (-3,4) and so x1 + x 2 = −3 and y1 + y 2 = 4 . 22By substituting x1 = 7 into x1 + x2 = −3 , we have 2 7 + x 2 = −3 2 2 7 + x 2  = 2(−3) 2 7 + x 2 = −6 x 2 = −13 .Also, by substituting y1 =2 into y1 + y2 = 4 , we have 2 2+ y2 = 4 2 2 2 + y 2  = 2(4) 2 2+ y2 = 8 y2 = 6 .Therefore, point B has coordinates (-13, 6).Self-check 31. Use the distance formula to find the distance between the given pair of points. a) (-4,2) and (1,-11) b) (-3,1) and (2,3)2. Find the length of the line segment determined by the given pair of points. a) A(-3,0) and B(7,1) b) P(2,-2) and Q(-1,2)3. Find the coordinates of the midpoint M of the line segment determined by the given pairof points. a) C(0,0) and D(-8,-4) b) M(-4,1) and N(6,4) 17

4. Let M(-2,4) be the midpoint of the line segment AB. a) If A has coordinates (-5,6), find the coordinates of B. b) If B has coordinates (7,5), find the coordinates of A. Answer Key on page 21 Let’s summarize The following concepts are the highlights of the three lessonscontained in this module: The absolute value of a number x, denoted by x , is its distance from the origin of the number line. The Cartesian plane consists of two perpendicular lines: the horizontal line or the x-axis and the vertical line or the y-axis.The intersection of the two axes is called the origin.The two axes separate the plane into four regions called quadrants.A point on the coordinate plane is represented by an ordered pair (x, y).These numbers are not interchangeable.The x-coordinate is also called the abscissa, while the y-coordinate isalso called the ordinate. Given two points A(x1, y1) and B(x 2 , y 2 ) , the distance d between A and B is given by d = (x1 − x 2 )2 + (y1 − y 2 )2 , the distance formula.The midpoint of a line segment is the point that divides the segmentinto two equal parts. To find the coordinates of the midpoint M of theline segment determined by the points A(x1, y1) and B(x 2 , y 2 ) , we usethe midpoint formula given by M =  x1 + x 2 , y1 + y 2  .2 218

What to do after (Posttest)Direction. Read each item and choose the letter of the correct answer.1. What is the absolute value of -7?a. 7 b. -7 c. 1 d. − 1 7 72. On the number line, what is the distance between 8 and -2?a. 16 b. 10 c. 6 d. 33. What is the value of 3 4 − (−3) − 2 3 − 4 ?a. 19 b. 23 c. 25 d. 274. What is the ordinate or the y-coordinate of the point (-3, 3)?a. -3 b. 3 c. − 1 d. 1 3 35. A point has negative abscissa and negative ordinate. Which quadrant does itlie?a. I b. II c. III d. IV6. A point lies 4 units to the right of the x-axis and 5 units below the y-axis. Whatare the coordinates of the point?a. (4, 5) b. (-4, 5) c. (-4, -5) d. (4, -5)7. The abscissa of a point on the y-axis is alwaysa. negative c. positiveb. zero d. undefined.8. What is the distance between the points (1, 3) and (-5, -5)?a. 14 c. 5b. 10 d. 89. On a map, Lisa’s house has coordinates (2, 0). If her school has coordinates (0, -2), what is the distance between Lisa’s house and her school?a. 12 c. 8b. 4 d. 810. What are the coordinates of the midpoint of (7, -6) and (-1, -2)?a. (3, -4) c. (8, -8)b. (3, 4) d. (6, -8) Answer Key on page 22 19

Answer KeyPretest page 3 1) b 2) b 3) c 4) a 5) d 6) c 7) b 8) b 9) c 10) bLesson 1 Self-Check 1 page 71 a) -5 b) 0 c) 32 a) d = | -5 - (-2) |= | -5 + 2 | = | -3 | = 3 unitsb) d = | -2 – 3 | = | -5 | = 5 unitsc) d = | -5 – 0 | = | -5 | = 5 units3 a) 5 − 9 − 4 = 5 − 5 = 5 − 5 = 0b) 2 7 − (− 3) − 4 = 2 7 + 3 − 4 = 210 − 4 = 2(10) − 4 = 20 − 4 = 16c) 10 − 2 + 18 − (− 3) = 8 + 18 + 3 = 8 + 21 = 8 + 21 = 29d) 6 − 2 + 21− (− 1) = 4 + 21+ 1 = 4 + 2 2 = 4 + 2(2) = 4 + 4 = 8e) 3 − 5 − 1 − 2 6 − 2 = 3 − 6 − 2 4 = 3(6) − 2(4) = 18 − 8 = 10f) − 4 − 6 − 3 5 − (− 1) + 7 − 4 = − 10 − 3 5 + 1 + 3 = 10 − 3(6) + 3 = 10 − 18 + 3 = −5Lesson 2 Self-Check 2 page 12A 1 a) B=(2,2) b) A=(-4,3) c) E=(-2,-2) d) T=(3,-2)2 a) P b) O c) Q d) TB 1) QII 2) QIII 3) QI and QII 4) QII and QIII 5) QII 20

C y-axis54321• D x-axis •B-5 -4 -3 -2 -1 0 1 2 3 4 5 -1 • E-2 •C• A -3 -4-5 D 1) origin – the point of intersection of the x-axis and the y-axis. 2) abscissa – the x-coordinate of a point. 3) ordinate – the y-coordinate of a point. 4) quadrant – one of the four divisions of the Cartesian plane. 5) y – axis – the vertical line of the Cartesian plane.Lesson 3 Self-Check 3 page 17 1 a) d = (−4 − 1)2 + (2 − (−11))2 d = (−5)2 + (13)2 = 25 + 169 d = 225 = 15 units b) d = (−3 − 2)2 + (1− 3)2 d = (−5)2 + (−2)2 = 25 + 4 d = 29 units 2 a) d = (−3 − 7)2 + (0 − 1)2 21

d = (−10)2 + (−1)2 = 100 + 1 d = 101 units b) d = (2 − (−1))2 + (−2 − 2)2 d = (3)2 + (−4)2 = 9 + 16 d = 25 = 5 units 3 a) M =  0 + (−8) , 0 + (−4)  =  −8 , −4  = (− 4,−2)  2 2   2 2  b) M =  − 4 + 6 , 1+ 4  =  2 , 5  = 1, 5   2 2  2 2  2 4 a) By substituting x1 = −5 into x1 + x2 = −2 , we have − 5 + x2 = −2 2 2⇒ 2 − 5 + x2  = 2(−2) ⇒ − 5 + x2 = −4 ⇒ x2 = 1. 2 Also, by substituting y1 = 6 into y1 + y2 = 4 , we have 6 + y2 =4 2 2⇒ 2 6 + y2  = 2(4) ⇒ 6 + y2 = 8 ⇒ y2 = −2 . 2 Therefore, point B has coordinates (1, -2). b) By substituting x2 = 7 into x1 + x2 = −2 , we have x1 + 7 = −2 2 2⇒ 2 x1 + 7  = 2(−2) ⇒ x1 + 7 = −4 ⇒ x1 = −11. 2 Also, by substituting y2 =5 into y1 + y2 = 4 , we have y1 + 5 =4 2 2⇒ 2 y1 + 5  = 2(4) ⇒ y1 + 5 = 8 ⇒ y1 = 3 . 2 Therefore, point A has coordinates (-11, 3).Posttest page 19 1) a 2) b 3) a 4) b 5) c 6) d 7) b 8) b 9) c 10) a END OF MODULE 22

BIBLIOGRAPHYFuller, G. (1977). College algebra. (4th ed). New York, NY: Van Nostrand Company.Leithold, L. (1989). College algebra. USA: Addison-Wesley Publishing Company, Inc.Mckeague, C.P. (1990). Intermediate slgebra: A text/workbook. (3rd ed). USA: Harcourt Brace Jovanovich, Inc. 23

Module 15 The Straight Line and The Absolute Value Function What this module is all about The straight line is the simplest geometric curve. Despite its simplicity, the straightline is a vital concept of mathematics and enters into our daily activities in numerousinteresting and useful ways. This module will teach you how find the equation of a line or write the equation of a graph of a linear function. This module will also strengthen your understanding of the connection between equations and the linear functions they represent. Also, you will learn how to deal with real world data that tend to be linear. Many types of calculations give us negative numbers. When we are interested in real quantities such as distance and location, there is a mathematical way to use these numbers without representing them as signed numbers. The concept of absolute value is used to describe such distances and locations. The last part of this module focuses on the concept of the absolute value of a number, the graphs of absolute value functions, and solutions of equations involving absolute values. This module has five lessons: Lesson 1 Point-Slope Equation of a Line Lesson 2 Two-Point Equation of a Line Lesson 3 Intercept Equation of a Line Lesson 4 Applications of Linear Functions/Equations Lesson 5 Absolute Value Functions 1

What you are expected to learnAfter going through this module, you are expected to:• obtain the equation of a line given the following:i) the slope and a pointii) any two pointsiii) intercepts• use linear equations in two variables to solve problems• define an absolute value equation y = |x|• construct a table of ordered pairs and draw the graphsy=x y = |x| + by=x+b y = |x| – by=x–b y = |x| + by = |x| y = |x| – b How to learn from this moduleFollow the steps suggested in the flowchart below. Take the Pretest Check your answers with the Answer Key placed at the later part of this Module Read each lesson and answer Self-check Check your answers with the Answer Key Take the Posttest 2

What to do before (Pretest)Multiple Choice: Read the problems carefully and then select the letter of the correct answer. Write your answers on a separate sheet of pad paper.1. What is the absolute value of the sum of –12 and 4?A. -8 B. -16 C. 8 D. 162. What is the equation of the line passing through (0, -5) and with slope of 1?A. y = -5x + 1 B. y = x – 5 C. y = -x + 5 D. y = 5x – 13. What is the equation of the graph? A. y = 2 x + 3 Y 3 X B. y = − 3 x + 3 2 C. y = 3 x + 3 2 D. y = − 2 x + 3 34. Which of the following graphs is represented by the table of the ordered pairs? x -1 1 0 -2 2 y -2 -2 -3 -1 -1 3

A. B. Y Y X XC. D. Y Y X X5. What is the table of ordered pairs of the equation y = |x| + 2?A. x 0 -1 2 1 -4 B. x -2 -1 0 1 2 y23436 y01234C. x -1 0 1 2 -3 D. x -3 -2 -1 0 2 y 1 2 3 4 -1 y541206. The graph of the equation x - y = 5 crosses the y-axis atA. -5 B. -1 C. 1 D. 57. What is the equation of the line that passes through the points (-2, 2) and (3,1)? 4

A. 5x + 5y = 8 B. 5x – 5y = 8 C. x + 5y = 8 D. 5x – y = 88. What is the equation of the graph at the right? Y A. 3x + 2 y = 1 B. 2x + 3y = 1 X C. 3x − 2 y = 1 D. 2x − 3y = 19. The graph of a line through (-2, -2) and has a slope of -2/3 is:A. B. Y Y XXC. D. Y Y XX 5

10. What is the slope and the y-intercept of the graph at the right?A. m = 1 ; y − int = 2 Y 4 XB. m = 1 ; y − int = 4 2C. m = 4; y − int = −4D. m = −4; y − int = −411. What is the equation of the line whose x-intercept is 1/3 and whose y-intercept is -1/4? A. 3x – 4y = 1 B. 3x + 4y = 8 C. 4x – 3y = 1 D. 4x + 3y = 112. What is the slope and the coordinates of a point on the graph of the equation y - 2 = 3(x - 5)? A. m=3; (2,5) B. m=3; (5,2) C. m=3; (-2,-5) D. m=-3; (-2,-5)13. A 9-liter container of freshly squeezed orange juice costs P35. A 12-liter container costs P45. How much would a 16-liter container cost?A. Php42.15 B. Php48.33 C. Php55.00 D. Php65.0014. Miss Torres offers tutorial services in mathematics. She charges a flat fee of P150, plus P60 per hour. Write an equation of the tutorial services fee in slope- intercept form.A. y = 6x + 150 C. y = 150x + 60B. y = 60x + 150 D. y = 150x – 6015. What is the value of |a - 3b| when a = -2 and b = 4?A. -14 B. 10 C. -10 D. 14 6

What you will doLesson 1 Point-Slope Equation of a Line In the previous activity, you learned about the slope of a line. You have learned howto describe the slopes of lines mathematically and how to apply them as rates of change.You also learned how to find the slope-intercept form of an equation of a line, that is,y = mx + b, with slope m and y-intercept b.Exploration Now you will explore ways to find the equation of the line when the slope and a pointon the line are given. Let's have a simple example.Example 1. Find an equation of the line through (2, 4) with slope 1/3.Solution: a) Draw a Cartesian plane and locate the point (2,4). Thus, Y (2,4) X 7

b) From the point (2,4), find another point on the line. Since the slope of the line is 1/3, move one unit upward and 3 units to the right. At what point did you stop?___________________Y (2,4) (x,y) XThe slope from (2,4) to (x, y) ism= y−4 x−2but m = 1 , so Multiplying both sides by (x-2) 3 1 = y−4 3 x−2(x-2)  1  =  y − 4  (x-2) Simplifying 3  x−2y - 4 = 1 (x - 2) 3 This equation is said to be in the point-slope form since this form of theequation was generated using the coordinates of a known point other than the y-intercept and the slope of the line. So, the point-slope form of an equation of a line through the point (x1, y1) withslope m is written as y - y1 = m(x - x1) This equation tells us that the point (x1,y1) is any point on the line and that theslope of the line is m. 8

You can now write an equation in point-slope form for the graph of anynonvertical line if you know the slope of the line and the coordinates of one point onthe line.Example 2. Write the point-slope form of an equation for each line: a) The line passes through (-2, 5) and has a slope of -3/4. b) The horizontal line passes through (3,1).Solution:a) Applying the point-slope form of a line, y - y1 = m(x - x1) point-slope form y - 5 = - 3 [x - (-2)] 4 y - 5 = - 3 (x + 2) equation of the line 4b) The slope of a horizontal line is 0 and using the point-slope form of a line, you get y - y1 = m(x - x1) point-slope form y - 1 = 0(x - 3) y-1=0 y=1 equation of the lineSelf-checkA. Write the equation of the line which passes through the point P and with the slope m. Draw the lines.1. P(-2, 1), m=3 2. P(0, 0), m = 5/63. P(5, -2), m = -3/2 4. P(-3, -4), m = 1/4 9

B. State the slope and the coordinates of a point through which the line represented by each equation passes.1) y - 2 = -3(x + 1) 2) y + 1 = - 4 (x - 2)3) 2(x + 5) = y - 1 5 4) y = -6Lesson 2 Two-Point Form of the Equation of a Line In the previous lesson you learned how the concepts of slope and a point on the lineare used to write the equation of a line. Now, you will see how to graph a linear equation ifyou know only two points on the line. So far, you have been using the equation of a line to find ordered pairs. But supposeyou know only the coordinates of two points. How could you find the equation of the linethat contains them? There is only one line that passes through two given points. Thus,given two points, one can determine a line following these steps:• To write the equation of a line through two given points, use the slope-intercept form of a line, y = mx + b,• Find m from the given coordinates,• Then, find b by solving the equation.Example 1. Write an equation of the line that passes through the points (-1, 3) and (2, -4).Solution:a) First, find the slope of the line. m = y2 − y1 slope formula x2 − x1 slope m = (− 4) − (− 3) = −7 = −7 2 − (−1) 2 +1 3b) Substitute this value of m into the slope-intercept form y = mx + b, 10

y = mx + b slope-intercept formy= −7x+b 3c) Now, find the y-intercept, b, by substituting the coordinates of either points for x and y in the equation. Thus, using the point (-1, 3), you get,3 = − 7 (-1) + b 33= 7 +b 3b = 3- 7 3b= 2 3d) Finally, substitute b = 2 in the equation obtained in part (b). thus, 3y= −7x+b 3y= −7x+ 2 This is the equation of the line 33 that passes through the points (-1, 3) and (2, -4). An alternative way of finding the equation of a line through two given points is to usethe point-slope form of the line. That is,y - y1 = m(x - x1) point-slope formbut m = y2 − y1 slope formula x2 − x1Consequently, the equation of the line through (x1, y1) and (x2, y2) isy - y1 = y2 − y1 (x - x1) x2 − x1 11

This equation is called the two-point form of the equation of a straight line.Example 2. Find the equation of the line determined by the points (2, -2) and (4, 5).Solution: Using the two-point form of a line, you get,y - y1 = y2 − y1 (x - x1) x2 − x1y - (-2) = 5 − (−2) (x - 2) 4−2y + 2 = 5 + 2 (x - 2) 2y + 2 = 7 (x - 2) 2y = 7 (x - 2) – 2 2y = 7 x - 7 (2) - 2 22y = 7x - 9 equation of the line. 2Self-checkSolve for the equation of the line given: a) (3, 3) and (2, 1) b) (5, 1) and (1, 4) c) (0, 0) and (3, -4) 12

d) passing through the origin with slope 0Lesson 3 Intercept Form of the Equation of a Line Recall the x and y-intercepts of a line. Suppose the x-intercept of a line is a and they-intercept is b, where a ≠ 0 and b ≠ 0. See Fig. 1 below: Y (0,b) (a,0) X Fig. 1Then the line passes through the points (a, 0) and (0, b). What is the slope of the linedescribed in Fig. 1? ____________________________________________Applying the point-slope form of a line with slope − b , you get ay - y1 = m(x - x1) point-slope formy - b = − b (x - 0) ay - b = −b x multiply both sides by a aay - ay = -bxbx + ay = ab divide both sides by abx + y =1ab 13

This equation is called the intercept form of a line because the intercepts are given inthe denominators.Example 1. Write the equation of the line whose x-intercept is 5 and whose y-intercept is -2.Solution: Using the intercept-form of a line with a = 5 and b = -2, you have x + y =1 intercept form of a line ab multiplying both sides by 10 x + y =1 5 −2 2x - 5y = 10 equation of the lineExample 2. Write the equation 4x + 7y = 28 in the intercept form.Solution: Divide both sides of 4x + 7y = 28 by 28. x + y =1 74Self-checkA. Write the equation of the line with x-intercepts a and y-intercept b in each of the following:1. a = -3, b = -8 2. a = − 3 , b = 1 53. a = 4 , b= 1 4. a = -1, b = -2 9 2B. Express each equation in the intercept form: 14

a. 6x - 5y = 30 b. 9y = 4x + 36 c. 2x = 5y + 9 d. 3x - 4y = 8Lesson 4 Applications of Linear Equations/Functions You have learned how line, with their slopes and y-intercepts, or any two points on it,can be matched with equations. We continue to look at the relationship between theseequations and the practical part everyday function they represent. Real-world data often have functional relationships that can be written as equations.These equations can be used to make predictions, with or without graphs. Let's have someexamples.Example 1. Ana baked 10 cakes in 5 hours. She worked for 2 ½ more hours and had baked a total of 15 cakes. How many cakes had she baked after 10 hours?Solution:a) Know what is asked. The number of cakes she had baked after 10 hours.b) Make a table to show the number of cakes she baked in certain number of hours. No. of hours (x) 5 7 ½ 10 15 20 No. of cakes she 10 baked (y)c) Solve the problem. Show the table as a rule and as an equation. Rule: The number of cakes she baked was 2 times the number of hours. Equation: y = 2x Check: y = 2x 15

y = 2(10) y = 20She had baked 20 cakes after 10 hours.Example 2. How much exercise is the right amount? Most health experts suggest that you should exercise to the point where your heart rate reaches a target level based on your age. Here's a rule that is suggested:To find your target heart rate or pulse . . . ..Subtract your age from 220.a) Write an equation for the relationship.b) Find your target rate using the equation.c) What is the target heart rate for a 5-year old child?Solution: a) Know what is asked.i) Your target heart rateii) The target heart rate of a 5-year old childb) Make a table to show the target heart rate for some people.Age(x) 15 30 45Target heart rate(y) 205 190 175c) Solve the problem.Rule: Subtract the age from 220.Equation: y = 220 - xTo find your age, say your 16 years old,y = 220 - 16y = 204 target heart rate of a 16-year oldSimilarly, the target heart rate of a 5-year old child is,y = 220 - 5y = 215 16

Self-check Following the procedures used in examples 1 and 2 of this lesson, solve the givenproblems below:1. An employee makes 7 pans of pizzas per hour. How many pans of pizzas can he make if he works 8 hours?2. A swimming pool was being filled with water. It was 1.22 m deep after 3 hours and 1.52 m deep after 5 hours. a. Write an equation for the relationship. b. How long will it take a pool to a depth of 2.13 mLesson 5 Absolute Value Function In this lesson, you will learn how to express distances between points mathematicallyand to solve equations involving distances.Take a look of the number line below:5 units 3 units -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7What are the numbers on the right of 0 on the number line? ______________On the left of 0? ________________________________________________The numbers to the right of 0 on the number line are positive and the numbers to the left of0 are negative. The distance from 0 to 3 is 3 while the distance from 0 to -5 is what? _____Distance is always positive (or zero). In mathematics, we use absolute value to describedistance on a number line. 17

The absolute value of a number is the distance between the number and zero on thenumber line. We write |x| to mean the absolute value of x.Consider the following examples:1. | 9 | = 9 2. | -9 | = 93. |-1.2 | = 1.2 4. |0| = 05. |-3| + 4 = 3 + 4 = 7 6. |5| - 2 = 5 - 2 = 37. |-5 + 3| = |-2| = 2 8. |-3 - 1| = |-4| = 49. |6 - 4| = |2| = 2 10. |-3 + 4| + |-6 - 1| = |1| + |-7| = 1 + 7 = 8From the examples above, the absolute value of zero or a positive number is what?______How about the absolute value of a negative number, what can you say about it? ________Notice that the absolute value of any real number is a non-negative number. Try thefollowing numbers below:Find the numbers:1. |-6| 2. |-12|3. |4| 4. |10|5. |6|The answers are:1. 6 2. 123. 4 4. 105. 6We can build a definition for the absolute value of a number, that is |x| = x if x ≥ 0 -x if x ≤ 0Having known the definition of the absolute value of a real number, you will see what thegraph of the function y = |x| and related functions look like. 18

Example 1. Find a table of values for y = x and sketch the graph.The table of ordered pairs of y = x Graph x 0 1 -2 4 Y y 0 1 -2 4 X y=xThe graph of y = x is in Quadrant I and Quadrant III.Now, graph the function y = |x|.y=|x| Graphx 0 1 -1 -3 Yy0113 X Notice that the Quadrant III \"arm\" of thegraph y = x is not flipped over intoQuadrant II. The function y = |x| is called the absolute value function. Thus,The absolute value function is a relationship that can be representedby the equation y = |x|, where x is any real number. To graph the absolute value function, simply find enough ordered pairs of thefunction, then plot them and see what the graph looks alike. In the next example, you will explore and try to graph other functions involvingabsolute value. 19

ExplorationExample 2. Graph y = x + b where b = 2 by finding a table of values. y=x+2 GraphX 0 1 -1 -3 YY 2 3 1 -1 y = x +2 X Fig.1Example 3. Find a table of values for y = |x + 2| and y = |x - 2|. Compare the graphs with the graph of y = x + 2.Solution: The table of values for y = |x + 2| and y = |x - 2| is: x -5 -4 -3 -2 -1 0 123y = |x+ 2| 321 01 2 345 765 43 2 121y = |x - 2|Plotting these ordered pairs, you gety = |x + 2| y = |x - 2| Y Y XX 20

Fig. 2 Fig. 3Notice again that the Quadrant III \"arm\" of the graph of y = x + 2 (see Fig. 1) is now flippedover into Quadrant II (see Fig. 2). Also the graph of y = |x - 2| was moved four units to theright. So they differ only by their vertices.Example 3. a) Copy and complete the table to find points for the graphs of y = |x| + 2 and y = |x| - 2 x -3 -2 -1 0 1 23 y = |x| + 2 5 3 y = |x| - 2 1 -1 b) Sketch the graphs in different Cartesian planes. c) How do the graphs differ from one another?Solution: a) The table of values for y = |x| +2 and y = |x| - 2: x -3 -2 -1 0 1 23 3 45 y = |x| + 2 5 4 3 2 -1 01 y = |x| - 2 1 0 -1 -2 b)Graphs are: y = |x| - 2 Y y = |x| + 2 Y XX Fig. 4 Fig. 5c) The graphs are similar. They only differ by their vertices. The vertex of the graph of y = |x| + 2 , which is (0, 2) was moved to (0, -2). 21

We can now summarize the graphs of the absolute value function y = |x|, and check howthese are similar to other equations you have graphed. The graphs of the absolute function y = |x| when x is any real number are V-shaped. The graphs of the functions y = |x + b| and y = |x - b| are also V-shaped but their vertices are moved b units to the right/left from the origin. The graphs of the functions y = |x| + b and y = |x| - b are also V-shaped and their vertices are moved b units up/down from the origin. To graph y = |x| + b, move the graph of y = |x| b units upward if b is positive and b units downward if b is negative.Self-checkA. Perform the indicated operation.1. | -2 | - | 3 4. | -12 |2. | 6 – 7 | 5. 4 + | -3 |3. | -3 + 2 | + | -6 – 5|B. Identify how to translate the graph of y = | x | to obtain the graph of each function.c. y = | x | - 5d. y = | x – 4 |e. y = -6 + | x |f. y = | x + 3 |g. y = | x | + 4 22

Let’s summarizeThe point-slope form of the equation of a line is y - y1 = m(x - x1), where m is theslope of the line and (x1, y1) is a any point on it.The two-point form of the equation of a line is y - y1 = y2 − y1 (x - x1) , where (x1, x2 − x1y1) and (x2, y2) are any two distinct points on the line.• The intercept equation of a line is x + y = 1, where a is the x-intercept and ab b is the y-intercept.• The absolute value function is a function defined by y = |x| where x is any real number.• The graphs of the absolute value function are V-shaped figures.What to do after (Posttest)Multiple Choice: Choose the letter of the correct answer.1. What is the absolute value of the sum of -2 and -6?A. -8 B. 4 C. 8 D. 122. What is the equation of the line passing through (0, 6) and whose slope is -2? A. y = -2x – 6 B. y = 2x + 6 C. y = -2x + 6 D. y = 2x – 6 23

3. What equation is represented by the graph? Y A. y = x + 3 B. y = -x + 3 X C. y = x – 3 D. y = -x – 34. Which of the following graphs is represented by y = | x | + 1?A. Y B. Y X XC. Y D. Y XX5. What is the equation of the line passing through ( -3, 5) and (-1, 7)?A. y = -x – 8 B. y = -x + 8 C. y = x – 8 D. y = x + 86. The graph of the equation x + y = 2 crosses the x-axis atA. 2 B. -2 C. 1 D. -17. Which of the following points is contained in the graph of the equation 3x - y = 6?A. (1, -9) B. (-1, 9) C. (-1, 3) D. (1, -3)8. The equation of a line through (1, -3) and having a slope of -1 isA. x + y = -2 B. x + y = 2 C. x + 2y = 1 D. 2x + y = 1 24

9. The graph of the equation y = 4x - 1 is B. Y A. Y X X Y XC. Y D. X10. What is the slope and the y-intercept of the equation 5x - 10y - 1 = 0?A. m = 2; y − int = 5 C. m = −2; y − int = 1B. m = 2; y − int = 1 5 5 D. m = 2; y − int = −511. Which equation would give the graph at the right?A. y = |x - 4| YB. y = |x| - 4C. y = |x + 4| XD. y = |x| + 4 25

12. Which of the following graphs is the graph of y = x ? 4A. Y C. Y X XB. Y D. Y X X13. When you buy an item from a credit car on a mail order catalog you pay a 10% sales tax. If you want the item set immediately by a delivery service, you must pay an additional P100 per order. The equation that shows this relationship is y = 0.10x + 100, where x is the cost of the item before tax and y is the cost of one item sent immediately. What is the cost of an item priced P875 and shipped immediately? A. Php1875.00 B. Php1062.25 C. Php962.50 D. Php975.0014. The maximum heart rate during exercise for a 20-year old is 200 beats per minute, and the maximum heart rate for a 70-year old is 150 beats per minute.(Nordic Track Brochure). What is the equation that shows the relationship between the maximum heart rate of a person and his/her age?A. x + y - 180 = 0 C. x - y - 180 = 0B. x - y + 180 = 0 D. x + y + 180 = 015. Which of the following points contained in the graph of y = |x| + 1?A. (0,-1) B. (1,0) C. (-1,2) D. (-1,1) 26

Answer Key 3. m = 2; (-5, 1) 4. m = 0; (0, -6)Pretest Lesson 2. Self-Check 1. c 1. 2x - y = 3 2. b 2. 3x + 4y = 6 3. a 3. 3x + 4y = 16 4. b 4. 3x + 4y = 0 5. a 5. y = 0 6. d 7. a Lesson 3. Self-Check 8. b 9. c 1. a. 8x + 3y = -24 10. c 11. a b. 5x - 3y = -3 12. b 13. b c. 9x + 8y = 4 14. c 15. d d. 2x + y = -2Lesson 1. Self-Check 2. a. x + y = 1 5 −6A. 1. 3x - y = 7 2. 5x - 6y = 0 b. x + y = 1 3. 3x + 2y = 11 −9 4 4. x - 4y = -13 c. x + y =1B. 1. m = -3 ; (-1, 2) 9 −9 2. m = -4/5; (2, -1) 25 d. x + y = 1 8 −2 3Lesson 4. Self-Check 123456 7 8 7 14 21 28 35 42 49 561. No. of hours (x) No. of pizzas (y)Rule: The number of pans of pizzas made is 7 times the number of hours.Equation: y = 7xCheck: If x = 8; y = 7(8) 27

y = 56 Therefore: the employee made 56 pans of pizzas in 8 hours.2. a. 2x + y = 11 b. 9 hoursLesson 5. Self-Check B. 1. Vertex (0, -5) 2. Vertex (4, 0)A. 1. -1 3. Vertex (0, -6) 2. 1 4. Vertex (-3, 0) 3. 12 5. Vertex (0, 4) 4. 6 5. 7Posttest1. c2. c3. c4. a5. d6. a7. d8. a9. c10. b11. b12. a13. b14. d15. c END OF MODULE 28

Module 16 The “Poly Products” What this module is all about This module is about multiplication of polynomials. It includes lessons onspecial products such as squaring binomials and finding the sum or differenceof two quantities. The processes are presented in meaningful way throughvarious illustrations for your better understanding of the concepts andprinciples. The following lessons are found in this module: Lesson 1 Multiplying Monomials Lesson 2 Multiplying A Polynomial by a Monomial Lesson 3 Multiplying Polynomials Lesson 4 Special Products: Difference Between Two Squares Lesson 5 Squaring a Binomial What you are expected to learn After working on this module, you are expected to: • multiply a monomial by polynomial using distributive property; • multiply a binomial by another binomial using the distributive property; and the FOIL method • find the special product of some polynomials:  square of a binomial  difference of two squares 1

How to learn from this moduleThis is your guide for the proper use of the module: 1. Read the items in the module carefully. 2. Follow the directions as you read the materials. 3. Answer all the questions that you encounter. As you go through the module, you will find help to answer these questions. Sometimes, the answers are found at the end of the module for immediate feedback. 4. To be successful in undertaking this module, you must be patient and industrious in doing the suggested tasks. 5. Take your time to study and learn. Happy learning! The following flowchart serves as your quick guide in using this module. Start Take the Pretest Check your paper and count your correct answers. Is your score Scan the items you 80% or above? missed. No Proceed to the nextStudy this module module/STOP.Take the Posttest 2

What to do before (Pretest)I. Multiple Choice. Choose the letter of the correct answer.1. The product of 3b and (2b2 + 3b – 3) isa. 6b2 + 9b – 9 c. 6b3 + 9b2 – 9bb. 6b3 + 9b – 9 d. 6b3 – 9b2 – 9b2. In the expressions (x + 2), (x – 5), what is the middle term of theproduct?a. 2x c. – 3xb. – 5x d. – 103. Find the first and last term of the product of these binomials, (a – 6) (a + 6).a. a , 36 c. a2 , 36b. a , 36 d. a2 , -364. In multiplying these two polynomials, (a – 3) and (a2 + 3a – 5), how many terms are there in the final product?a. two c. fourb. three d. five5. The square of the binomial (2x + 4) isa. 4x2 + 16 c. 4x + 16x2 + 16b. 4x2 + 16x d. 4x2 + 16x + 16II. Multiply :a. 2m ( 3m + 5 )( )b. 1 a2 6a2 + 9a −12 3c. ( l + 3 ) ( l + 6 )d. ( 5x + 1 ) ( x – 3 )e.  1 x + 1  1 x − 1   5 3  2 3  3

III. Find the special product of these factors a. ( y + 6 ) ( y – 6 ) b. ( w + 7) ( w - 7) c.  y + 3  y − 3   4  4  d. ( s2 – 3t )( s2 + 3t ) e. ( m – 8 ) 2 f. ( 2n – 1 ) 2 g.  a + 2 2  7IV. a. Is ( x + y ) 2 equal to x2 + y2? Explain your answer. b. What is the square of 4 more than a number? Give 3 numbers and check your answer. Answer Key on page 21 What you will do Read the following lessons carefully and do the suggested activitiespatiently.Lesson 1 Multiplying Monomials This lesson is a review of the past lessons on multiplication of monomialsand polynomials. As such, the laws of exponent are applied. 4

Exploration Look at the following examples: a. (x3) (x4) = (x.x.x) (x. x. x. x) = x3+4 = x7 b. (y2 )(y8 ) = (y ⋅ y)(y ⋅ y ⋅ y ⋅ y ⋅ y ⋅ y ⋅ y ⋅ y) = y2+8 = y10 c. (r3)(r5)(r2) = r3+5+2 = r3+5+2 = r10Remember: For any real number x, and for all whole numbers m and n, xm  xn = xm+n c. (5a2) (-2a3) = (5) (-2) a2  a3 = -10a2+3 = -10a5 ( )d.  1 x4  30x2 = 15x6 2  e. (3a) (a2) (-4a3) = (3)(1)(-4) a6 5

= -12a6How did you get the numerical coefficient of the product in example c, d, and e? Examine these examples further: a. (y4)2 = y4 y4 = y8 b. (x8)3 = x 8 x8 x8 = x24Remember: For any real number x, and for whole numbers m and n, (xm)n = xmn c. (3x5)3 = 33(x5)3 = 27x15 d. (4 r2 s3 t4)3 = 64 r6 s9 t12Remember: For any real number x and y, and for whole number n, (x y)n = xnynSelf-check 1I. Simplify each expression. 5. (2b)3 1. (y3)(y5) 6. (5a2b)2 2. (-y6)(y8) 7. (23)2 3. (a2)(a3)(a4) 8. (22x3)2 4. (6x3)(-4x4)II. a. Given a square with side 5x, what is its area? 6

b. Find the volume of a cube with sides 3a. 3a Answer Key on page 22Lesson 2 Multiplying A Polynomial by a Monomial In multiplying a monomial and a polynomial, we use the distributiveproperty as well as the rules in multiplying monomials which we have justreviewed in the preceding lesson. Study the following examples:a. 2x ( 4x + 8) = (2x)(4x) + (2x)(8) Distributive Property = 8x2 + 16xb. 2y (y2 – 3y + 4) = 2y (y2) + 2y (-3y) +2y(4) Multiplying Monomials = 2y3 – 6y2 +8yc. 3a (2a3 – 4a2 + 3a – 5) = 6a4 – 12a3 + 9a2 – 15ad. m2 ( m + 4) – 3 ( m3 – 2)= m3 + 4m2 – 3m3 + 6 Applying Distributive Property in= –2m3 + 4m2 + 6 the 2 terms of the Polynomial Combining Like Terms 7

Self-check 2A. In each blank, write the letter that corresponds to the given problem on theright._____ 1. a(2a – 3b) a. -6a2b3 +ab2-4ab_____ 2. –a( a2 – 3a + 4) b. 2a2 – 6ab_____ 3. a(2a – 1) – 3(a2 + 2) c. 5a3b3-15a4b2_____ 4. 5a2b(ab2 – 3a2b) d. –a3+3a2-4ab_____ 5. -ab(6ab2– b + 4) e. –a3+ 4a2 + 4ab f. –a2-a-6B. Multiply:1. 8x ( x – 3 )2. 5m2 (–2m + 1)3. 3y6 (–2y3 + 2y2 + y – 4)4. 1 m3 (24m2 – 20m + 64) 45. .01x(18x3 + 240x2 + 5C. 1. Find the expression that represents the area of the figure given below. 7m - 5 3m 2. What is the area of a rectangle whose sides are 5x and 4x + 8 ? 3. A portion of a rectangular room was carpeted as indicated by the shadedportion in the adjacent figure. With the given dimensions. represent the areathat was carpeted. 6a 2a-4 3a 20a+8 Answer Key on page 22 8