Unit 3: Relations and Functions

Unit 3: Relations and Functions 5-1: Binar Relations Binar Relation: - a set ordered pairs (coordinates) that include two variables (elements). (, ) = horizontal = vertical Domain: - all the -values (first elements) of a relation. Range: - all the -values (second elements) of a relation. Different Was to Describe a Relation. a. Table Form b. Arrow Diagram c. Ordered Pairs 3 9 4 1 1 0 0 1 1 4 3 9 3 0 1 1 4 0 9 1 3 ( 3, 9), (, 4), ( 1, 1), (0, 0), (1, 1), (, 4), (3, 9) d. Graph e. Equation 10 9 8 7 5 4 3 1 0-3 - -1-1 0 1 3 - = f. Words The second number is equal to the square of the first number. Domain { 3,, 1, 0, 1,, 3} Range {0, 1, 4, 9} Page 38. Coprighted b Gabriel Tang, B.Ed., B.Sc.

Relations and Functions Eample 1: Use the graph on the right to epress the relations as a. a set of ordered pairs. b. an arrow diagram. c. in words. d. an equation. e. Find the Domain and Range. 3 1 0-3 - -1 0 1 3 a. ( 3, 3), (, ), ( 1, 1), (0, 0), (1, 1), (, ), (3, 3) b. -1 - -3 3 3 1 1 0 0 1 1 3 3 c. The second number is the negative of the first number. d. = e. Domain { 3,, 1, 0, 1,, 3} Range { 3,, 1, 0, 1,, 3} 5-1 Homework Assignments Regular: pg. 14 to 15 #1 to 0 (ecept 19d, 0d, and 0e) AP: pg. 14 to 15 # 1 to 1 Coprighted b Gabriel Tang, B.Ed., B.Sc. Page 39.

5-: Linear Relation and Line of Best Fit Linear Relation: - a set ordered pairs that ehibit a straight line when plotted on a graph. Scatter Plot: - a graph that onl has ordered pairs showed. Line of Best Fit: - a line that will best describe the general relation of the ordered pairs on the graph. There are two tpes of data a. Discrete Data: - a graph with a series of separated ordered pairs or broken lines. Eample: Cost of Parking is $.00 ever hour with a Dail Maimum of $10.00. Time (hr) Cost ($) 0 0 1 4 3 4 8 5 10 10 7 10 8 10 Cost ($) 1 11 10 9 8 7 5 4 3 1 0 Parking Cost 0 1 3 4 5 7 8 Tim e (hour) b. Continuous Data: - a graph with an unbroken line that connects a series of ordered pairs. Eample: Distance versus Time of a car with a constant speed of 80 km/h. Time (hr) Distance (km) 0 0 1 80 10 3 40 4 30 5 400 480 7 50 8 40 Distance (km) Distance versus Time 700 00 500 400 300 00 100 0 0 1 3 4 5 7 8 Time (hour) Page 40. Coprighted b Gabriel Tang, B.Ed., B.Sc.

Relations and Functions Eample 1: Graph = + 1 for the domains { 3,, 1, 0, 1,, 3} and the real number set, R. 3 ( 3) + 1 = 7 ( ) + 1 = 5 1 ( 1) + 1 = 3 0 (0) + 1 = 1 1 (1) + 1 = 1 () + 1 = 3 3 (3) + 1 = 5 Domain { 3,, 1, 0, 1,, 3} Range { 5, 3, 1, 1, 3, 5, 7} 7 5 4 3 1 - -3-4 -5 - -7-8 0-5 -4-3 - -1-1 0 1 3 4 5-5 -4-3 - -1-1 0 1 3 4 5 Eample : Use a graphing calculator to graph the equation 3 + = 0 (cop it with aes properl labelled) and obtain a table of values from = 3 to =. 8-7 Domain: R -8 Range: R ( means belongs to ) 7 5 4 3 1 0 - -3-4 -5 - First, ou will have to solve for. 3 + = 0 = 3 Check WINDOW setting. WINDOW = = 3 3 OR ZOOM 3 = + 3 Select Option Then, using the graphing calculator, enter the equation. Y= GRAPH You MUST label at least two Points on the graph! To obtain a Table of Values using the graphing calculator: (, 0) (0, 3) Coprighted b Gabriel Tang, B.Ed., B.Sc. Page 41.

nd Set Tbl Start = 3 TBLSET WINDOW nd TABLE GRAPH You ma Scroll UP or DOWN to view more values. Eample 3: A vehicle is going at 110 km/h. a. Set up a table of values from t = 0 hours to t = 5 hours for ever hour and the distances travelled. b. Plot the distance versus time graph from the table of values above and obtain an equation relating distance and time. c. Eplain whether the graph should be continuous or discrete. a. b. Time (hr) Distance (km) 0 0 1 110 0 3 330 4 440 5 550 c. Data should be continuous because ou ma have t, time, between the whole number intervals (eample.5 hours is allowed). Distance (km) 00 500 400 300 00 100 Distance versus Time for a vehicle going at 110 km/h d = st d = 110 t 0 0 1 3 4 5 Time (hour) Page 4. Coprighted b Gabriel Tang, B.Ed., B.Sc.

Relations and Functions Eample 4: A school tracked 10 students attendance records and their final marks in the class of Applied Math 10. Student Number Number of Absences Final Marks (%) 1 3 75 7 3 8 51 4 1 88 5 80 4 78 7 10 4 8 7 55 9 3 70 10 5 5 a. Using the graphing calculator, decide on an appropriate WINDOW setting. Plot the final marks versus number of absences. Cop this graph with aes properl labelled. b. Using the graphing calculator, obtain the line of best fit and its correlation coefficient. c. Using the equation of the line of best fit; predict what final mark a student will likel get with 9 absences. d. Eplain whether the final graph should be discrete or continuous. a. To Enter a Table of values, ou have to use the STAT menu. STAT Choose Option 1 Enter Scores When finished entering, nd QUIT MODE To set WINDOW with : [ min, ma, scl ] and : [ min, ma, scl ] : [ 0, 10, 1 ] and : [ 40, 100, 10 ] WINDOW Coprighted b Gabriel Tang, B.Ed., B.Sc. Page 43.

To Graph from Data in STAT Lists, ou must turn the STAT PLOT On and be sure there is NO equation in the Y= Screen. nd Select On STAT PLOT Y= Choose Option 1 GRAPH Final Marks vs. Number of Absences 100 90 Final Mark (%) 80 70 0 50 40 0 1 3 4 5 7 8 9 10 Cop to grid lines with intervals and aes properl labelled Number of Absences b. Obtaining Equations with Correlation Coefficient: nd CATALOG 0 and Move Down using Again Select DiagnosticOn Note: After DiagnosticOn is selected; it will remain ON even when the calculator is turned Off. However, resetting the calculator will turn the Diagnostic Off (factor setting). Page 44. Coprighted b Gabriel Tang, B.Ed., B.Sc.

Relations and Functions STAT Select CALC, use Choose Option 4 Linear Regression Therefore, the Line of Best Fit Equation is: = 4.84 + 90.8 Again f = 4.84 a + 90.8 where f = final mark a = number of absences Correlation Coefficient is 0.95, which means a good fit, close to 1 or 1, and as increases, decreases. To Draw the Line of best Fit on the Graph: (Need to Cop the Equation to the Y= Screen) Y= VARS Choose Option 5 Notice that STAT PLOT is ON Select EQ, use Choose Option 1 The Linear Regression Equation is now copied onto the Y= Screen. Coprighted b Gabriel Tang, B.Ed., B.Sc. Page 45.

GRAPH 100 90 Final Marks vs. Number of Absences Final Mark (%) 80 70 0 Cop graph to actual grid with best fit line, intervals and aes properl labelled 50 40 0 1 3 4 5 7 8 9 10 Number of Absences c. At 9 Absences, a = 9, f =? or Using Graphing Calculator f = 4.84 a + 90.8 f = 4.84 (9) + 90.8 f = 47% TRACE Select Equation of the Best Fit Line to Trace b Press 9 for = 9 d. The data is discrete because ou cannot have a decimal as the number of absences. 5- Homework Assignments Regular: pg. 19 to 0 #1 to d, 7, 8a to 8d, 9 to 31 AP: pg. 19 to 0 # 1 to d, 7 to 31 Page 4. Coprighted b Gabriel Tang, B.Ed., B.Sc.

Relations and Functions 5-3: Non-Linear Relations Non-Linear Relation: - a set ordered pairs when graphs, ehibit a curve as the line of best fit instead of a straight line. - usuall when the and/or variable(s) have an eponent other than 0 or 1. Eample 1: Provide the table of values and a graph for the following equations. For each graph, state the domain and range. a. = + 3 b. = 3 3 ( 3) + 3 = 1 ( ) + 3 = 7 1 ( 1) + 3 = 4 0 (0) + 3 = 3 1 (1) + 3 = 4 () + 3 = 7 3 (3) + 3 = 1 : [ 5, 5, 1] and : [ 0, 0, ] Domain R Range: 3 3 ( 3) 3 = 54 ( ) 3 = 1 1 ( 1) 3 = 0 (0) 3 = 0 1 (1) 3 = () 3 = 1 3 (3) 3 = 54 : [ 5, 5, 1] and : [ 0, 0, 10] Domain R Range: R c. = 3 d. 3 3 ( 3) = no solution 3 ( ) = no solution 1 3 ( 1) = no solution 0 3 (0) = 0 1 3 (1) = 3 3 () = 4.4 3 3 (3) = 5.19 : [ 5, 5, 1] and : [ 10, 10, 1] Domain 0 Range: 0 = 3 = ( 3) = 3 ( ) 1 = ( 1) 0 = undefined (0) 1 = (1) = 3 () 3 = (3) : [ 5, 5, 1] and : [ 10, 10, 1] Domain 0 Range: 0 Coprighted b Gabriel Tang, B.Ed., B.Sc. Page 47.

e. = 3 f. = 3 1 (3) 3 1 = 7 (3) 1 = 9 (3) 1 = 3 1 0 (3) 0 = 1 1 (3) 1 = 3 (3) = 9 3 (3) 3 = 7 : [ 5, 5, 1] and : [ 30, 30, 5] Domain R Range: > 0 3 3 = 5 = 4 1 1 = 3 0 0 = 1 1 = 1 = 0 3 1 = 1 : [ 5, 5, 1] and : [ 7, 7, 1] Domain R Range: 0 Eample : A $1000 investment was left in an account that pa 5%/a (a = annum = ear). The table below shows the balance at the end of each ear, assuming no withdrawal is made in antime. Year Balance 0 $1000.00 1 $1050.00 $110.50 3 $1157.3 4 $115.51 5 $17.8 a. Graph the balance versus the number of ears on a graphing calculator. Show the WINDOW settings and label the aes properl. b. Eplain the pattern in words. c. Write out a possible equation. Verif this equation with our graphing calculator. Did the graph of our equation give a curve that goes through most of the points? d. Eplain whether the data is discrete or continuous. e. Use the equation to predict the balance at the end of the 10 ears. f. Using the graphing calculator, predict the minimum number of ears for the initial investment of $1000 to double. a. Balance vs Years of Investment Page 48. Balance ($) 00 000 1800 100 1400 100 1000 0 4 8 10 1 14 1 Years of Investment : [0, 1, ] and : [1000, 00, 00] Coprighted b Gabriel Tang, B.Ed., B.Sc.

Relations and Functions b. For ever ear it is invested, the balance increases b 5% of the previous ear s balance. c. A = $1000 (1.05) n where A = Balance and n = number of ears invested. Balance vs Years of Investment 00 000 Balance ($) 1800 100 1400 100 1000 0 4 8 10 1 14 1 Years of Investment d. The data is discrete because interest is paid out at the end of the ear. e. Using the equation, A = $1000 (1.05) n A = $1000 (1.05) 10 = $18.89 f. B drawing another line, Y 1 = 000 and Run the Intersect Function, Y= GRAPH nd CALC TRACE Choose Option 5 It will take 15 ears (round-up) Coprighted b Gabriel Tang, B.Ed., B.Sc. Page 49.

(AP) Eample 3: For the relation ( 3) + ( + 5) =1 a. Solve for b. Graph the resulting equations. c. Describe the shape, domain, and range of the graph. d. If =, what could be the value(s) for? e. If = 4, what could be the value(s) for? a. b. 3 + + 5 = 1 ( ) ( ) ( + 5) = 1 ( 3) + 5 = ± 1 ( 3) = ± 1 ( 3) ( ) 5 ( ) 5 Y 1 = 1 ( 3) Y = 1 ( 3) ( ) ( ) 5 c. CIRCLE (the graph appears d. Using the TRACE function for both equations, to be an ellipse because the = 1.13 and 8.87 calculator screen is a rectangle; each interval is longer than the equivalent interval). Domain: 1 7 Range: 9 1 e. Using the INTERSECT function for equations Y 1 and Y 3 twice, = 0.87 and.87 5-3 Homework Assignments Regular: pg. 5 to 7 #1 to 10, 1c and 13 (ecept 8e) AP: pg. 5 to 7 # 1 to 13 Page 50. Coprighted b Gabriel Tang, B.Ed., B.Sc.

Relations and Functions 5-4: General Relations Independent (Manipulated) Variable: - a variable that ou change in a situation to cause an effect. - label on the -ais (horizontal ais). Dependent (Responding) Variable: - a variable that ou measure because of the changes ou caused with the manipulated variable. - label on the -ais (vertical ais). Title Dependent Variable Independent Variable For an graph, there should be proper labelling on: a. Title (usuall vs. ) b. The name of the variables on the ais and their units. c. Proper intervals scaling. d. If there are two overlapping graphs on the same grid, a clear legend must be indicated. Creating a Scenario to Match a Graph Eample 1: Jane is driving to the shopping mall from her house. Using the graph, write a scenario that would describe her travel. Speed (km/h) 0 A Jane goes to the mall C A B C Jane Left her house and drove at 7 km/h (plaground zone). She stopped at the stop sign. Jane drove at 50 km/h (residential zone). D E D She entered the parking lot of the mall and was looking for a parking spot (10 km/h). B Time (min) 5 E Jane parked and stopped. Common Shapes of Different Graphs (Fill in the descriptions below.) As, at a stead rate. As, sharpl at first, but eventuall levels off As, slowl at first, but faster later. Coprighted b Gabriel Tang, B.Ed., B.Sc. Page 51.

As, at a stead rate.. As, slowl first, then faster later. As, sharpl first, then levels off. As, in a As, oscillates stepwise manner. back and forth. Creating a Graph to illustrate a Scenario As, remains constant. Eample : Imagine a mass attached to a spring. It occupies a position at rest above a level surface. If the mass is pulled down and then released, it will move up and down. Sketch the graph to represent the relationship between the height of the mass above the surface and the time after its release. Height from the surface MASS Rest Position Pull Down and Release Time after release 5-4 Homework Assignments Regular: pg. 9 to 31 #1 to 11a, 11c, 1 AP: pg. 9 to 31 # 1 to 11a, 11c, 1 Page 5. Coprighted b Gabriel Tang, B.Ed., B.Sc.

Relations and Functions 5-5: Functions Relation: - an equation that eplains how one variable (input ) can turn into another variable (output ). Function: - a special relation that must satisf the following two conditions: a. The graph is continuous (no break unless stated in the domain and range). b. For each input, there is onl one unique output. (Vertical Line Test If a vertical line moves from left to right of the graph and it did not cross the graph at two different points, then we can sa the graph passed the vertical line test). Eample 1: State whether each of the graphs below is a function. Provide reasons. Function: Continuous and pass Vertical Line Function Notation: - a wa to epress an equation to denote that it is a function (satisfies requirements of continuit and vertical line test). - instead of writing, we can write f(). - we can sa Function f of or f as a function of. - a number to be put in to replace in f() for the purpose of substitution. Eample : Given f() = 3 +, set up a Table of Values for = 3 to = 3. Graph f () a. Find f( 8). b. Find when f() = 5 c. Find f(3n 1). f() 3 3( 3) + = 7 3( ) + = 4 1 3( 1) + = 1 0 3(0) + = 1 3(1) + = 5 3() + = 8 3 3(3) + = 11 a. f() = 3 + f( 8) = 3 ( 8) + f ( 8) = NOT a Function: Does NOT pass Vertical Line Test b. f() = 3 + f() = 5 5 = 3 + 5 = 3 54 = 3 = 18 Function (at each branch): Continuous and pass Vertical Line Test NOT a Function: Does NOT pass Vertical Line Test c. f() = 3 + f(3n 1) = 3 (3n 1) + f(3n 1) = 9n 3 + f (3n 1) = 9n 1 5-5 Homework Assignments Regular: pg. 34 to 35 #1 to 18, 0 to 5, 31 AP: pg. 34 to 35 # 1 to 18, 0 to 5, 9, 31 Coprighted b Gabriel Tang, B.Ed., B.Sc. Page 53.

5-: Applications of Linear Functions Direct Variation: - a variable that varies directl (b a constant rate of change) with another variable. ( is directl proportional to ) or = k where k = constant of variation (constant of proportionalit rate of change) Eample 1: Gasoline at one time costs $0.70 per Litre. a. Write out the cost of gasoline as a function of volume bought. b. What is the constant of variation? c. Set up a table of values from V = 0 L to V = 50 L with a scale of 5 L. d. Graph the function. e. Find the cost of 3 L of gasoline. f. How much gasoline can ou bu with $.43? a. C(V) = 0.7 V b. Constant of Variation = $0.70/L (unit price of gasoline) c. d. V in L C(V) in $ 0 0.7 (0) = 14.00 5 0.7 (5) = 17.50 30 0.7 (30) = 1.00 35 0.7 (35) = 4.50 40 0.7 (40) = 8.00 45 0.7 (45) = 31.50 50 0.7 (50) = 35.00 Cost ($) Cost of Gasoline vs Volume Bought 40.00 30.00 0.00 10.00 0.00 0 5 10 15 0 5 30 35 40 45 50 55 0 Volume (L) Eample : The amount of fuel used b a vehicle varies directl with the distance travelled. On a particular trip, 4.35 L of gasoline is used for a distance of 51.5 km. a. Calculate the constant of variation. b. Find the function of volume of gasoline used in terms of distance travelled. c. What is the distance travelled if 35 L of gasoline is used? d. How much would it cost to fuel up the car when the price of gasoline was $5.90 / 100 L if the entire trip was 31 km? a. V(d) = kd b. V(d) = kd c. V = 35 L, d =? d. d = 31 km, V =? 4.35 = k (51.5) 35 = 0.08 d V = 0.08 (31) 4.35L 35 = k V(d) = 0.08d = d 51.5km 0.08 V = 51.74 L $5.90 = 100L 51. 74L k = 0.08 L/km d = 4.8 km = $34.10 Page 54. Coprighted b Gabriel Tang, B.Ed., B.Sc.

Relations and Functions Partial Variation: - a variable that varies partiall (b a constant rate of change and a fied amount) with another variable. = k + b where k = constant of variation (constant of proportionalit rate of change) b = fied amount (initial amount when = 0) Eample 3: The cost of a school dance organized b the student council $5.50 per person and $300 to hire the DJ. a. What is the constant of variation and the fied amount? b. Epress the cost as a function of the number of people attending. c. Set up a table of values from n = 0 to n = 100 people with a scale of 0 people. d. Graph the function. e. Eplain whether the graph should be discrete or continuous. f. State the Domain and Range. g. How man people attended if the cost was $115? a. Constant of Variation = $5.50 / person (cost per person) b. C(n) = 5.50n +300 Fied Amount = $300 (fied cost) c. d. n C(n) in $ 0 5.50 (0) +300 = 300 0 5.50 (0) +300 = 410 40 5.50 (40) +300 = 50 0 5.50 (0) +300 = 30 80 5.50 (80) +300 = 740 100 5.50 (100) +300 = 850 Cost ($) $1,000 $800 $00 $400 $00 $0 n (people) e. The graph should be discrete because ou can not have a decimal number to describe the number of people attending. f. Domain: n W Range: C = 5.50n +300 where n W or {300, 305.5, 311, 31.5, 3 } g. C(n) = $115, n =? C(n) = 5.50n +300 115 = 5.50n +300 115 300 = 5.50n 85 = 5.50n 85 n = 5.50 n = 150 people Cost of a School Dance 0 0 40 0 80 100 10 Coprighted b Gabriel Tang, B.Ed., B.Sc. Page 55.

Eample 4: A tai ride costs ou $18.00 for 5 km travelled, and $34.80 if ou travelled 1 km. a. Epress the above information in a table. b. Enter the table into the STAT menu of our graphing calculator. c. Using the Appropriate WINDOW settings, graph the data. d. Run LINEAR REGRESSION to obtain an equation. e. Epress the equation as a function of Cost in terms of distance travelled. f. What is the constant of variation and the fied amount? g. How much would it cost for a 0 km ride to the airport? h. If the ride costs $3.50, what is the distance travelled? a. b. d in km C(d) in $ 5 18.00 1 34.80 STAT c. : [ 0, 15, 1 ] and : [ 0, 40, 5 ] d. e. C(d) =.4d + f. Constant of Variation = $.40/km (cost per km travelled) Fied Amount = $.00 (Flat Rate) g. d = 0 km, C =? h. C = $3.50, d =? C(d) =.4d + C(d) =.4d + C(d) =.4(0) + 3.50 =.4d + C(d) = $54.00 3.50 =.4d.5 =.4d.5.4 = d 5- Homework Assignments Regular: pg. 38 to 40 #1 to 1, 4 AP: pg. 38 to 40 # 1 to d = 11.04 km Page 5. Coprighted b Gabriel Tang, B.Ed., B.Sc.