Math 11-CORE Gen Math-Q1-Week-3 PDF

4 MORE ON FUNCTIONS for General Mathematics Senior High School (CORE) Quarter 1 / Week 3 1  FOREWORD This Self-Learning Kit for General Mathematics is designed specifically for Grade 11 students in the Senior High School. Thus, a...

4 MORE ON FUNCTIONS for General Mathematics Senior High School (CORE) Quarter 1 / Week 3 1  FOREWORD This Self-Learning Kit for General Mathematics is designed specifically for Grade 11 students in the Senior High School. Thus, a modest background in grade school mathematics is important, written in a precise, readable, and conventional manner to facilitate students’ understanding of the subject. It is aligned with the BEC of the Department of Education following the prescribed MELCs (Most Essential Learning Competencies. It has the following features proven to be valuable aids to learning Mathematics even at home. What Happened This section contains pre-activities like review of the prior knowledge and a pretest on what the learners have learned in their previous discussions. What I Need to Know (Discussion) This section contains definition of terms, different examples of real-life situations as application of functions. It gives examples and the corresponding situations that clearly illustrate the applicability of a mathematical concept. What I have Learned (Evaluation/Post Test) The exercises contained in this section are guaranteed to build mathematical comprehension, skills, and competence. These serve as a diagnostic tool to identify the learners’ areas of strengths and difficulties. 2  OBJECTIVES: At the end of the lesson, the students are expected to: K.: represent rational functions and determine the domain and range, intercepts, zeroes and asymptotes of a function; S: solve problems involving rational functions equations, and inequalities; and A: develop perseverance in solving problems involving rational functions, equations, and inequalities. LESSON 1 REPRESENTING REAL-LIFE SITUATIONS USING RATIONAL FUNCTIONS I. WHAT HAPPENED PRE-TEST Mathinik Challenge 1 Complete Me! Complete the table of values of the following functions. Write your answers on your activity notebook/sheets. 1 1. 𝑦 = 𝑥−1 x -2 -1 0 2 3 4 y 𝑥+1 2. 𝑦 = 𝑥−1 y -2 -1 0 2 3 4 x Mathinik Challenge 2 Locate Me! Plot the functions in Mathinik Challenge 1 on the Cartesian plane. 1 𝑥+1 1. 𝑦 = 𝑥−1 2. 𝑦 = 𝑥−1 3  II. WHAT YOU NEED TO KNOW DISCUSSION Consider the simple scenario involving rational functions. 𝑑 Average speed (or velocity) can be computed by the formula 𝑠 =. Consider a 𝑡 100-meter track used for foot races. The speed of a runner can be computed by taking 100 the time it will take him to run the track and applying it to the formula 𝑠 = 𝑡 , since the distance is fixed at 100 meters. Example 1 1. Represent the speed of a runner as a function of the time it takes to run 100 meters in the track. Let 𝑥 represent the time it takes the runner to run 100 meters. Then the speed can be represented as a function 𝑠(𝑥) as follows; 100 𝑠 (𝑥 ) = 𝑥 2. Continuing the scenario, construct a table of values for the speed of a runner against different run times. The table of values for run times from 10 to 20 seconds is as follows: 𝑥 10 12 14 16 18 20 𝑠(𝑥) 10 8.33 7.14 6.25 5.56 5 From the table we can observe that the speed decreases with time. We can use a graph to determine if the points on the function follow a smooth curve or a straight line. 3. Graph the function. Plot and connect the points from the table of values on a Cartesian plane. Assign points on the Cartesian plane for each entry in the table of values above: 𝐴(10,10) 𝐵(12,8.33) 𝐶(14,7.14) 𝐷(16,6.25) 𝐸(18,5.56) 𝐹(20,5) By connecting the points we can see that they are collinear but rather follow a smooth curve. 4  For the 100-meter dash scenario, we have constructed a function of speed against time, and represented our function with a table of values and graph. Example 2 𝑥−1 Represent the rational function given by 𝑓(𝑥 ) = using a table of values and plot a 𝑥+1 graph of the function by connecting points. Solution. Since we are now considering functions in general, we can find function values across more values of 𝑥. 1. Let us construct a table of values for some x-values from -10 to 10. 2. Plot the points on a Cartesian plane and connect the points. Observe that the function will be undefined at 𝑥 = −1. This means that there cannot be a line connecting point 𝐸 and point 𝐹 as this implies that there is a point in the graph of a function 5 where 𝑥 = −1. We will cover this aspect of graphs of rational functions in a future lesson, so far now we just present a partial graph for the function above as follows: Remove the segment connecting E and F as the graph does not pass through points with an x- value of -1. III. WHAT HAVE I LEARNED POST TEST: CHALLENGE YOURSELF. ANSWER ME! Do what is asked. Write your answers on your activity notebook/sheets. Represent the following rational functions through table of values and graph. 2𝑥−4 1. 𝑓 (𝑥 ) = 𝑥 𝑥 -4 -2 -1 0 1 2 4 6 𝑦 1 2. 𝑓 (𝑥 ) = 𝑥 𝑥 -3 -2 -1 0 1 2 3 4 𝑦 6  𝑥 3. 𝑓 (𝑥 ) = 2𝑥−5 𝑥 -3 -2 -1 0 1 2 3 4 𝑦 2 4. 𝑓 (𝑥 ) = 𝑥2 𝑥 -3 -2 -1 0 1 2 3 𝑦 LESSON 2 FINDING THE DOMAIN AND RANGE OF A RATIONAL FUNCTION I. WHAT HAPPENED PRE-TEST Mathinik Challenge 1 Why Not? Determine the restrictions of the variable 𝑥 in the denominator. (value of x that will make the expression undefined). The denominator should not be equal to zero. Do it in your activity notebook/sheets. 2𝑥−3 1 𝑥 2 −1 𝑥+1 1. 2. 2𝑥−1 3. 𝑥 2 +1 4. 5𝑥−2 𝑥 Mathinik Challenge 2 Undefined! For each of the rational function below find the: a. domain (possible value/s of variable 𝑥 can take) b. range (possible value/s of variable 𝑥 can take) 1 1 1 1. 𝑓 (𝑥) = 2. 𝑓 (𝑥) = 𝑥 2 3. 𝑓 (𝑥) = 𝑥−1 𝑥 𝑥 4. 𝑦 = 2𝑥−5 7 II. WHAT YOU NEED TO KNOW DISCUSSION Definition The domain of a function is the set of all values that the variable 𝑥 can take (Verzosa 2016). The range of the function is the set of all values that 𝑓(𝑥) or variable 𝑦 will take (Verzosa 2016). To find the domain of a rational function is to determine the restriction of the variable in the denominator. You may equate the denominator to zero then solve for the value of variable 𝑥 to find the restriction in the denominator (Verzosa 2016). For this time, to find the range of a rational function is to graph the rational function and determine the possible values that the variable y will restrict. Note: Always express given function in its simplest form. You may also equate the denominator to zero then solve for the value of variable 𝑥 to find the restriction in the denominator. Examples. Finding the domain and range of a rational function. 3 1. 𝑓 (𝑥 ) = 𝑥+1 Since the function is in simplest form, we may now equate the denominator to zero then, find the value of variable 𝑥. 𝑥+1=0 𝑥 = −1 -1 is the restriction Domain: {𝑥 ∈ ℝ, 𝑥 ≠ −1} 3 This can be read as the domain of the rational function 𝑓(𝑥 ) = are set of real 𝑥+1 number except for negative one. This can also read as Domain: 𝑥 𝑏𝑒𝑙𝑜𝑛𝑔𝑠 𝑡𝑜 𝑟𝑒𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 𝑒𝑥𝑐𝑒𝑝𝑡 𝑓𝑜𝑟 − 1. Graph of the function. 8 Restriction 𝑦 = 0 Range: {𝑦 ∈ ℝ, 𝑦 ≠ 0} 2𝑥−3 2. 𝑓 (𝑥 ) = 𝑥 Equate the denominator to zero. 𝑥=0 0 is the restriction Domain: {𝑥 ∈ ℝ, 𝑥 ≠ 0} Graph: Restriction is 𝑦 = 2 Range: {𝑦 ∈ ℝ, 𝑦 ≠ 2} 𝑥+3 3. 𝑓 (𝑥 ) = 2𝑥−1 Equate the denominator to zero. 2𝑥 − 1 = 0 2𝑥 = 1 2𝑥 1 =2 2 9  1 1 𝑥= 2 2 is the restriction 1 Domain: {𝑥 ∈ ℝ, 𝑥 ≠ 2} Graph. 1 Restriction 𝑦 = 2 1 Range: {𝑦 ∈ ℝ, 𝑦 ≠ 2} 𝑥 4. 𝑓 (𝑥 ) = 𝑥2 −1 Equate the denominator to zero. 𝑥2 − 1 = 0 𝑥2 = 1 √𝑥 2 = √1 𝑥 = ±1 ±1 is the restriction Domain: {𝑥 ∈ ℝ, 𝑥 ≠ ±1} Graph. 10  No restriction. Range: {𝑦 ∈ ℝ} 2𝑥+1 5. 𝑓 (𝑥 ) = 4𝑥 2 −1 Notice that the given function is not in simplest form, so simplifying it we have 2𝑥+1 2𝑥+1 𝑓 (𝑥 ) = 2 =4𝑥 −1 (2𝑥−1)(2𝑥+1) 1 = 2𝑥−1 Equate the denominator to zero. 2𝑥 − 1 = 0 2𝑥 = 1 1 𝑥= 2 1 1 The domain does not include − 2 and because from the original functions the 2 1 restriction of 𝑥 includes ± 2. 1 Domain: {𝑥 ∈ ℝ, 𝑥 ≠ ± 2} Graph. 11 Restriction 𝑦 = 0 Range: {𝑦 ∈ ℝ}, 𝑦 ≠ 0 III. WHAT HAVE I LEARNED POST TEST CHALLENGE YOURSELF. ANSWER ME! Find the domain and range of the following rational functions. Write your answer on your activity notebook/sheets. 𝑥+1 1. 𝑓 (𝑥 ) = 𝑥+3 1 2. 𝑓 (𝑥 ) = 𝑥 2 3. 𝑓 (𝑥 ) = 𝑥−2 𝑥+1 4. 𝑓 (𝑥 ) = 𝑥−2 12  LESSON 1 INTERCEPTS, ZEROES, AND ASYMPTOTES OF RATIONAL FUNCTIONS I. WHAT HAPPENED Quick Recall: Properties of Functions 1. a) The domain of a function is a set of all values that the variable x can take. 2. b) The range of a function is a set of all values that f(x) can take. 3. c) The zeroes of a function are the values of x which make the function zero. The real numbered zeroes are also x-intercepts of the graph of the function. 4. d) The y-intercept is the function value when x = 0. 5. e) The horizontal line is a horizontal asymptote of the function if gets closer to as increases or decreases without bound 6. f) The vertical line 𝑥 = 𝑎 is a vertical asymptote of a function 𝑓 if the graph of 𝑓 either increases or decreases without bound as the 𝑥 − 𝑣𝑎𝑙𝑢𝑒𝑠 approach from the right or left. II. WHAT YOU NEED TO KNOW Discussion How to find the Do the following: domain Find all x values which do not make the denominator 0. y-intercept If 0 is not in the domain, then there is no y- intercept. If 0 is in the domain, the y- intercept is the value of the function at x = 0. x-intercept Find the values of x where the numerator is zero, but the denominator is non-zero. Vertical asymptotes Find the values of x where the denominator of the reduced rational function is zero. Horizontal asymptote Use the degree of the polynomial in the numerator and denominator, as indicated above. range Graph the function. (However, there may be rational functions where more advanced techniques like calculus are needed) 13  𝑥−2 Example 1. Consider the function 𝑓 (𝑥 ) = 𝑥+2 (a) Find its zeroes or x-intercept, (b) y-intercept, (c) vertical asymptote and (d) horizontal asymptote. Solution. Recall that the 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑠 of a rational function are the values of 𝑥 that will make the function zero. A rational function will be zero if its numerator is zero. Therefore, the zeroes of a rational function are the zeroes of its numerator. The numerator 𝒙 – 𝟐 will be zero at 𝒙 = 𝟐. (a)Therefore 𝒙 = 𝟐 is a zero of 𝒇(𝒙). Since it is a real zero, it is also an 𝒙 − 𝒊𝒏𝒕𝒆𝒓𝒄𝒆𝒑𝒕. 2 (b)The 𝒚 − 𝒊𝒏𝒕𝒆𝒓𝒄𝒆𝒑𝒕 of a function is equal to 𝒇(𝟎). In this case, 𝒇(𝟎) = − 2 = −1. (c) In sketching the graph of 𝒇(𝒙), let us look at what happens to the graph near the values of 𝑥 which make the denominator undefined. Recall that in the previous lesson, we simply skipped connecting the points at integer values. Let us see what happens when x takes on values that brings the denominator closer to zero. The denominator is zero when 𝒙 = – 𝟐. Let us look at the values of 𝒙 close to – 𝟐 on its left side (i.e. x < –2, denoted –2 negative) and values of x close to –2 on its right side (i.e. x > –2, denoted –2+). i. Table of values for x approaching −𝟐−. x -3 -2.5 -2.1 -2.01 -2.001 -2.0001 As x approaches −2− f(x) 5 9 41 401 4001 40001 f(x) increases without bound ii.Table of values for x approaching −𝟐+ x -1 -1.5 -1.9 -1.99 -1.999 -1.9999 As x approaches −2+ f(x) -3 -7 -39 -399 -3999 -39999 f(x) increases without bound Plotting the points corresponding to these values on the Cartesian plane, we obtain 14  Note: the axes do not have the same scale Observe that as 𝑥 approaches –2 from the left and from the right, 𝑓(𝑥) gets closer and closer to the line 𝑥 = – 2, indicated in the figure with a dashed line. We call this line a vertical asymptote, formally defined as follows: (c) The vertical asymptote of the function is 𝑥 = 2 We will also look how the function behaves as 𝑥 increases or decreases without bound. We first construct a table of values for 𝑓(𝑥) as 𝑥 increases without bound, or in symbols. as 𝑥 → −∞. iii. Table of values for 𝒇( 𝒙) 𝒂𝒔 𝒙 → +∞ x 5 10 100 1 000 10 000 As x → +∞ f(x) 0.43 0.67 0.96 0.9960 0.99960 f(x) approaches 1− Next, construct a table of values for 𝒇(𝒙) as 𝒙 decreases without bound, or in symbols, as 𝑥 → −∞. iv. Table of values for 𝒇( 𝒙) 𝒂𝒔 𝒙 → −∞. x -5 -10 -100 -1 000 -10 000 As x → −∞ f(x) 2.33 1.41 1.041 1.00401 1.0004001 f(x) approaches 1+ Plotting the points saccording to these on the Cartesian Plane: 15  Note: the axes do not have the same scale Observe that as 𝑥 increases or decreases without bound, 𝑓(𝑥) gets closer and closer to1. (d) The line 𝒚 = 𝟏 indicated in the figure with a dashed line is called the horizontal asymptote Note: A rational function may or may not cross its horizontal asymptote. If the function does not cross the horizontal asymptote 𝒚 = 𝒃, then b is not part of the range of the rational function. III. WHAT HAVE I LEARNED POST TEST 1. Find the asymptotes and intercepts of the function. 𝑥−1 𝑓(𝑥) = (𝑥−2)(𝑥+3) 2. Identify the y- intercepts, x-intercept or zeroes, vertical asymptote, and horizontal 3x 2 - 8x - 3 asymptote of 𝑓(𝑥 ) = 2x 2 + 7x - 4 SOLVING PROBLEMS INVOLVING LESSON 4 RATIONAL FUNCTIONS, EQUATIONS, AND INEQUALITIES I. WHAT HAPPENED Quick Recall: Recall the steps in solving word problems in mathematics. 1. Read the problem carefully. 2. Identify and list the facts. 3. Figure out exactly what the problem is asking for. 4. Find or develop a formula. 5. Solve the problem. 16  6. Verify the answer. Interval and Set Notation An inequality may have infinitely many solutions. The set of all solutions can be expressed using set notation or interval notation. These notations are presented in the table below. PRE-TEST Find the solutions for each rational equation/inequality below. Make sure to check for extraneous solutions. Write your answer on your activity notebook/sheets. 3 2 a) 𝑥+1 = 𝑥−3 2𝑥 5 b) + 2𝑥 = 2 𝑥+1 𝑥+1 c) ≤2 𝑥+3 (𝑥+3)(𝑥−2) d) ≥0 (𝑥+2)(𝑥−1) II. WHAT YOU NEED TO KNOW DISCUSSION PROCEDURES FOR SOLVING RATIONAL EQUATIONS To solve rational equations: (a) Eliminate denominators by multiplying each term of the equation by the least common denominator. (b) Note that eliminating denominators may introduce extraneous solutions. Check the solutions of the transformed equations with the original equation. 17  2 3 1 Example 1. Solve for 𝑥: 𝑥 − 2𝑥 = 5 Solution: The LCD of all the denominators is 10𝑥. Multiply both sides of the equation by 10𝑥 and solve the resulting equation. 𝑥 1 8 Example 2. Solve for 𝑥: − 𝑥−2 = 𝑥2 −4 𝑥+2 Solution: Factor each denominator in the fraction. The LCD is (𝑥 − 2)(𝑥 + 2). Multiply the LCD to both sides of the equation to remove the denominators. Upon reaching this step, we can use strategies for solving polynomial functions. Check for extraneous solutions by substituting the answers back into the original the original equations. Since x =2 will make the original equation undefined, it is an extraneous solution. Since x =5 satisfies the original equation, it is the only solution. APPLICATION pplying the concepts we learned in solving rational equations to solve word problems. Rational equations can be applied to problems with variables in the denominator. 18 Example 3. Solve the problem. In an inter-barangay basketball league, the team from Barangay Nabago has won 12 out of 25 games, a winning percentage of 48%. How many games should they win in a row to improve their win percentage to 60%? Solution: Let 𝑥 represent the number of games that they need to win to raise their percentage to 60%. The team has already won 12 out of their 25 games. If they win x games in a row to increase their percentage to 60%, then they would have played 12 + 𝑥 games out of 25 + 𝑥 games. It can be represented by this equation: 12 + 𝑥 = 0.6 25 + 𝑥 The equation is a rational equation. Solve the equation using the techniques previously mentioned. Since 25 + 𝑥 is the only denominator, we multiply it to both sides of the equation. We then solve the resulting equation: Since 𝑥 represents the number of games, this number should be an integer. Therefore, Barangay Nabago needs to win 8 games to raise their winning percentage to 60%. (Applications involving Work) Steve and Janet are going to paint the fence that surrounds their house today. Steve can paint the fence in 12 hours. Janet can paint the fence alone in 9 hours. How long will it take them to paint the fence together? Solution: 𝟏 𝟏 𝒕 𝒕 𝒕 + 𝒕 = 𝟏 𝒐𝒓 + =𝟏 𝟏𝟐 𝟗 𝟏𝟐 𝟗 Let’s solve the equation for t: 𝑡 𝑡 + =1 12 9 19  𝑡 𝑡 36( 12 + 9) = 1(36) 3t +4t = 36 7t = 36 𝑡 = 36/7 or 5.15 Processing of the answer. To find an equation that represents the given scenario. Let’s determine how much of the fence Steve and Janet paint respectively. 1 Steve takes 12 hours to paint the fence, each hour he paints 12 hour. Since it takes them t hours to paint the fence together. Steve paints 1/12t of the fence. Since it takes Janet 9 hours to paint the fence, each hour she paints 1/9 of the fence. Since it takes them t hours to paint the fence together, Janet paints 1/9t of the fence. Since the 1/12 t of the fence that Steve paints and the 1/9t of the fence that Janet paints together constitute one complete fence, we obtain the equation 𝟏 𝟏 𝒕 𝒕 𝒕 + 𝒕 = 𝟏 𝒐𝒓 + =𝟏 𝟏𝟐 𝟗 𝟏𝟐 𝟗 PROCEDURES FOR SOLVING RATIONAL INEQUALITIES To solve rational inequalities: (a) Rewrite the inequality as a single fraction on one side of the inequality symbol and 0 on the other side. (b) Determine over what intervals the fraction takes on positive and negative values. (i) Locate the x-values for which the rational expression is zero or undefined (factoring the numerator and denominator is a useful strategy). (ii) Make the numbers found in (i) on the number line. Use shaded circle to indicate that the value is included in the solution set, and a hollow circle to indicate that the value is excluded. These numbers partition the number line into intervals. (iii) Select a test point within the interior of each interval in (ii). The Sign of the rational expression at this test point is also the sign of the ration expression at each interior point in the aforementioned interval. (iv) Summarize the intervals containing the solutions. 2𝑥 Example 1: Solve the inequality 𝑥+1 ≥ 1. Solution: (a) Rewrite the inequality as a single fraction on one side, and 0 on the other side. 20 (b) The value 𝑥 = 1 is included in the solution since it makes the fraction equal to zero, while 𝑥 = −1 makes the fraction undefined. Make these on a number line. Use a shaded circle for x=1 (a solution) and an unshaded circle for 𝑥 = −1 (not a solution). (c) Choose convenient test points in the interval determined by -1 and 1 to determine the sign 𝑥−1 of 𝑥+1 in these intervals. Construct a table of signs as shown below. (d) Since we are looking for the intervals where the fraction is positive or zero, we determine solution intervals to be 𝑥 < −1 and 𝑥 ≥ 1. Plot these intervals on the number line. 3 1 Example 2. Solve: < 𝑥−2 𝑥 Solution: (a) Rewrite as an inequality with zero on one side. (b) The fraction will be zero for 𝑥 = −1 and undefined for 0 and 2. Plot on a number line. Use hollow circle since these values are not part of the solutions. 21  (c) Construct a table of signs to determine the sign of the fraction in each interval determined by -1,0, and 2. (d) Summarize the intervals satisfying the inequality. Plot these intervals on the number line. III. WHAT HAVE I LEARNED POST TEST Answer the following problems in your notebook. Show your solution. 1. A taxi is rented by a group of x people for P2000 and the cost is shared equally. If there is one less person in the group, then each of the remaining people has to pay P100 more. What is 𝑥? 2. A group of teenagers hiked 6 kilometers from Barangay Poblacion to Barangay Calango at a rate of x kph. For the return trip their rate was 1 kph faster. It took them 5 hours for the entire round trip. What is 𝑥? 3. Four divided by the sum of a number and two is greater than 2. Find all such numbers. 22  SYNOPSIS This Self Learning Kit (SLK) is about Rational Functions. Here the students will learn how to determine and analyze the intercepts, zeroes, and asymptotes of a given rational function. This helps the learners develop skills in graphing different types of rational functions. Going over the discussion and exercises, makes one appreciate the importance of this function. Find enjoyment in learning this SLK and go over the discussion and examples if you have not yet mastered a concept ABOUT THE AUTHOR RICKLEOBEN V. BAYKING. He is a graduate of St. Paul University Dumaguete with a bachelor’s degree of Secondary Education specializing in Mathematics in the year 2016. He spent his first year of teaching in the same university as a Senior High School teacher. Currently, he is serving his fourth year of teaching in the Department of Education at Jimalalud National High School as a full-time faculty teacher in the Senior High School Department. MERCYDITHA D. ENOLPE received her undergraduate degree and master’s degree from Negros Oriental State University (NORSU). She is currently pursuing her Ph.D. Math Education degree of that same University. Mrs. Enolpe is presently the Teacher In-Charge-SHS Department of Jose Marie Locsin MHS, Zamboanguita District. At the same time, handling mathematics subjects of the school. She has contributed significantly in the crafting of the Daily Lesson Log for teachers in the Division of Neg. Or, specifically General Mathematics subject. She is also the District Planning Coordinator-SHS of the District of Zamboanguita. 23  24 Mathinik Challenge 2 Post Test Pre-test Mathinik Challenge1 1. x -2 -1 0 2 3 4 y 1 1 -1 1 1 1 − − 3 2 2 3 2. y -2 -1 0 2 3 4 x 1 0 -1 3 2 5 3 3 Post-test 1. 𝑥 - - - 0 1 2 4 6 4 2 1 𝑦 3 4 6 und - 0 1 4 2 3 2. 𝑥 -4 -2 -1 0 1 2 4 6 𝑦 1 1 -1 und 1 1 1 1 − − 3 2 2 4 6 B. 3. 𝑥 -3 -2 -1 0 1 2 3 4 𝑦 3 2 1 0 1 -2 3 4 11 9 8 3 3 4. 𝑥 -3 -2 -1 0 1 2 3 𝑦 2 1 2 und 2 1 2 9 2 2 9 ANSWER KEY  25 MATHEMATICS. Quezon City: Commission on Higher Education. Verzosa, D., Ph.D. 2016. Teaching Guide for Senior High School GENERAL VICARISH PUBLICATIONS AND TRADING, INC Tan, F., Labilyn Lasic, L., and Cruz, R., 2017. General Mathematics. Manila: BLR). 54-65. Department of Education-Bureau of Learning Resources (DepEd- Crisologo, L., et al. 2013. General Mathematics Teacher’s Guide. Pp. Logarithmic Functions. 2013. Retrieved from https://www.slideshare.net Logarithmic Functions. 35-42. Diwa Learning Systems, Inc. Albay, E., et al. 2016. General Mathematics: Graphs and Properties of REFERENCES Lesson 3 Lesson 2 Lesson 4 Post-test Pre-test 1 1. y-intercept is 6 x-intercept is 1 Mathenik Challenge1 Vertical asymptote x =2 and x= -3 Horizontal asymptote at y = 0 1 2 1) 0 2) 2 3) none 4) 5 2. to determine the y-intercept and x- Mathenik Challenge 2 intercept: 1) domain: {𝑥 ∈ ℝ; 𝑥 ≠ 0}; The numerator and denominator of f(x) range: {𝑦 ∈ ℝ; 𝑦 ≠ 0} can be factored as follows: 2) domain: {𝑥 ∈ ℝ; 𝑥 ≠ 0} 3x 2 - 8x - 3 (3𝑥 + 1)(𝑥 − 3) 𝑓(𝑥) = = range: {𝑦 ∈ ℝ; 𝑦 ≠ 0} 2x 2 + 7x - 4 (2x-1)(x+4) 3) domain: {𝑥 ∈ ℝ; 𝑥 ≠ 1} 0−0−3 3 range: {𝑦 ∈ ℝ; 𝑦 ≠ 0} y-intercept: 𝑓(0) = 0+0−4 = 4 2 4) domain: {𝑥 ∈ ℝ; 𝑥 ≠ 5} 1 1 range: {𝑦 ∈ ℝ; 𝑦 ≠ 2} x-intercept or zeroes: 𝑥 = − 3 𝑎𝑛𝑑 𝑥 = 3 1 Post-test Vertical asymptote: 𝑥 = 2 𝑎𝑛𝑑 𝑥 = −4 1.domain: {𝑥 ∈ ℝ; 𝑥 ≠ 3} range: {𝑦 ∈ ℝ; 𝑦 ≠ 1} horizontal asymptotes: the polynomials in 2. domain: {𝑥 ∈ ℝ; 𝑥 ≠ 0} the numerator and denominator have Range: {𝑦 ∈ ℝ; 𝑦 ≠ 0} equal degree. The horizontal asymptote is 3. domain: {𝑥 ∈ ℝ; 𝑥 ≠ 2} 3 the ratio of the leading coefficients 𝑦 = 2 Range: {𝑦 ∈ ℝ; 𝑦 ≠ 0} 4. domain: range: {𝑥 ∈ ℝ; 𝑥 ≠ 2} range: {𝑦 ∈ ℝ; 𝑦 ≠ 1} 26  DEPARTMENT OF EDUCATION SCHOOLS DIVISION OF NEGROS ORIENTAL SENEN PRISCILLO P. PAULIN, CESO V Schools Division Superintendent JOELYZA M. ARCILLA, EdD Assistant Schools Division Superintendent MARCELO K. PALISPIS, EdD Assistant Schools Division Superintendent NILITA L. RAGAY, EdD OIC - Assistant Schools Division Superintendent CID Chief ROSELA R. ABIERA Education Program Supervisor – (LRMS) ELISA L. BAGUIO, EdD Division Education Program Supervisor – MATHEMATICS MARICEL S. RASID Librarian II (LRMDS) ELMAR L. CABRERA PDO II (LRMDS) NAME OF WRITER RICKLEOBEN V. BAYKING MERCYDITHA D. ENOLPE NAME OF ILLUSTRATOR/LAY-OUT ARTIST/TYPESETTER _________________________________ ALPHA QA TEAM MERCYDITHA D. ENOLPE RONALD TOLENTINO DIDITH T. YAP BETA QA TEAM ELIZABETH A. ALAP-AP EPIFANIA Q. CUEVAS NIDA BARBARA S. SUASIN VRENDIE P. SYGACO MELBA S. TUMARONG HANNAHLY I. UMALI DISCLAIMER The information, activities and assessments used in this material are designed to provide accessible learning modality to the teachers and learners of the Division of Negros Oriental. The contents of this module are carefully researched, chosen, and evaluated to comply with the set learning competencies. The writers and evaluator were clearly instructed to give credits to information and illustrations used to substantiate this material. All content is subject to copyright and may not be reproduced in any form without expressed written consent from the division.