Скачать презентацию TYPES OF RELATIONS LINEAR RELATIONS the graph Скачать презентацию TYPES OF RELATIONS LINEAR RELATIONS the graph

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TYPES OF RELATIONS TYPES OF RELATIONS

LINEAR RELATIONS: the graph is a straight line the rate of change “a” is LINEAR RELATIONS: the graph is a straight line the rate of change “a” is constant (the rate of change between any 2 points on the line will be the same)

RELATION 1 – CONSTANT FUNCTION Consider the amount of fuel purchased and the price RELATION 1 – CONSTANT FUNCTION Consider the amount of fuel purchased and the price per litre. (Gas is 98. 9 cents per litre) Table of Values Amount (L) Price per Litre (¢) 1 98. 9 5 10 15 … x

RELATION 1 Consider the amount of fuel purchased and the price per litre. (Gas RELATION 1 Consider the amount of fuel purchased and the price per litre. (Gas is 98. 9 cents per litre) Table of Values Amount (L) Price per Litre (¢) 1 5 10 15 … x 98. 9 … 98. 9

GRAPH. . STRAIGHT LINE PARALLELTO X-AXIS Price of Fuel Price per Litre (¢) 100 GRAPH. . STRAIGHT LINE PARALLELTO X-AXIS Price of Fuel Price per Litre (¢) 100 75 50 25 0 5 10 15 Amount (L)

RULE The rule is: cost per litre = 98. 9 ¢ The rule is RULE The rule is: cost per litre = 98. 9 ¢ The rule is y = 98. 9 ¢ Table of Values No matter what the x value is. . the y value always stays the SAME

The cost per litre is constant The rate of change is zero (a The cost per litre is constant The rate of change is zero (a = 0) We call this a ZERO VARIATION relation The rule is: y = b (constant value) y = 98. 9

DO. . WORKBOOK PAGE 107 -108 ACTIVITY 2 COST OF BUS RIDE DO. . DO. . WORKBOOK PAGE 107 -108 ACTIVITY 2 COST OF BUS RIDE DO. . A, B, C, D, E, F PAGE 116 #8 PAGE 143 #2 TEXTBOOK #1 PAGE 153 #4 BRING TEXTBOOK #2 FROM NOW ON

RELATION 2 – DIRECT VARIATION The cost of apples is $2. 50 per kg. RELATION 2 – DIRECT VARIATION The cost of apples is $2. 50 per kg. Independent variable: number of kg you buy Dependent variable: cost of buying apples

RELATION 2 – DIRECT VARIATION The cost of apples is $2. 50 per kg. RELATION 2 – DIRECT VARIATION The cost of apples is $2. 50 per kg. Independent variable: # of kg of apples Dependent variable: cost ($)

TABLE OF VALUES Amount (kg) Cost ($) 0 1 2 5 … x TABLE OF VALUES Amount (kg) Cost ($) 0 1 2 5 … x

TABLE OF VALUES Number of kg purchased 0 1 2 5 Cost ($) 0 TABLE OF VALUES Number of kg purchased 0 1 2 5 Cost ($) 0 2. 50 5. 00 12. 50 … x

TABLE OF VALUES Number of kg purchased 0 1 2 5 Cost ($) 0 TABLE OF VALUES Number of kg purchased 0 1 2 5 Cost ($) 0 2. 50 5. 00 12. 50 … x 2. 50 x RULE Cost = 2. 50 times the amount of kg purchased y = 2. 50 x dep r. o. c indep Rule for direct y = a x Rule for this question: y=2. 50 x

GRAPH Cost of Apples Cost ($) 10. 00 7. 50 5. 00 2. 50 GRAPH Cost of Apples Cost ($) 10. 00 7. 50 5. 00 2. 50 0 1 2 3 Amount (kg)

DIRECT VARIATION There is no initial fee (start-up cost) Values are proportional (you can DIRECT VARIATION There is no initial fee (start-up cost) Values are proportional (you can cross multiply and always get the same answer) The rule is y = ax (“a” is the rate of change) y is dep var, x is indep. var The graph is a straight line that passes through the origin

WORK ON ZERO AND DIRECT RELATIONS Workbook Page 116 #8, 9 Text #2 Page WORK ON ZERO AND DIRECT RELATIONS Workbook Page 116 #8, 9 Text #2 Page 101 #1 Page 102 #4 a-h, #5 Page 103 #6, 7 Page 104 #9 (find the rate of change first then write it in the rule y = ax

QUIZ ON RATE OF CHANGE…TAKE OUT A PIECE OF PAPER. . PUT YOUR NAME QUIZ ON RATE OF CHANGE…TAKE OUT A PIECE OF PAPER. . PUT YOUR NAME ON IT… Calculate the rate of change…write coordinates, label, write the formula, solve for the r. o. c 1(2, 10) (6, 30) a =5 2 - (-2, 4) (5, -2) a= -6/7 3. - ( 1, 8) a= o 4 - (26, 12) (102, 50) a= 38/76 or 1/2 5 - (24, 4) a= -4/12 or -1/3 (8, 8) (12, 8)