MATH GRADE 8 UNIT 7 FUNCTIONS ANSWERS FOR EXERCISES

Transcription

1 MATH GRADE 8 UNIT 7 FUNCTIONS FOR EXERCISES

2 LESSON 2: WHAT A GRAPH CAN TELL YOU 1. A Increasing and nonlinear 2. D 4 to 3. a. x (input) (output) Not defined b. Answers will var. Possible answer: There are no definite maxima or minima points. From negative infinit to x = the left side of the graph is decreasing. The graph approaches x = but never reaches a value at x =. The graph decreases slowl at first but then decreases rapidl as it approaches x = 1. The inverse is true for the right side of the graph. From x = to positive infinit the right side of the graph is decreasing. The graph decreases rapidl at first but then decreases slowl as it approaches x = a. Answers will var. Possible answer: Federer was the best tennis plaer at the time. The graph shows that at age 24 Federer surpassed Nadal and Djokovic in the number of Grand Slam titles won. b. Starting at age 25 Nadal s graph slows down to a constant rate (not winning an more Grand Slam titles) while Federer s graph continues to rise at a stead rate. c. When plaers win Grand Slam titles their graph increases. When plaers do not win more Grand Slam titles their graph remains constant. The graph would never decrease because titles would not be taken awa. Challenge Problem 5. Answers will var. Ask a classmate to check our work. Copright 215 Pearson Education, Inc. 34

3 LESSON 3: LINEAR VERSUS NONLINEAR GRAPHS 1. B (1, 3), (2, 6), (3, 9) 2. D Line D 3. a. The truck passes the bike after 9 sec. b. Using the points (2, 4) and (9, 14) ou can determine the speed of the bike b finding the slope of the line. The slope is 1, or about ft/sec = = c. The speed of the new bike traveling at 1 mph needs to be converted into feet per second. 1 mi 1 hr 5,28 ft = 14.6 or about ft/sec 1 hr 3,6 sec 1 mi This speed is slightl faster than that of the previous bike, which has a speed of ft/sec. d. The third bike starts farther ahead of the other two, starting at 2 ft. But it travels significantl slower, at onl 1 ft/sec. e. At around 5 sec the slope of the truck s curve is about the same as the slope of the bike s line. This fact indicates that the are at about the same speed. Copright 215 Pearson Education, Inc. 35

4 LESSON 3: LINEAR VERSUS NONLINEAR GRAPHS 4. Since the lines are parallel ou know the have the same slope. The slope of the second line (described b the two points) can be calculated as = 6 = So, the slope of the first line must be the same. Thus, ou get the equation = 3 2 x + 4 for the first line and a = x for the second line. Challenge Problem 5. a. High tide is usuall around 4 m and low tide is roughl 1 m. b. There is a little more than 12 hr between each peak on the graph approximatel 12.5 hr. c. Height (m) a.m p.m. Time (hr) a.m Copright 215 Pearson Education, Inc. 36

5 LESSON 4: WHAT IS A FUNCTION? 1. C (1, 3), (1, 4), (1, 5) 2. D a. 3 C is the same as 86 F. f = 9 (3) f = f = 86 b. Yes, f = 9 c + 32 is a function. The equation represents a function because it is 5 a line and for an given input (c-value) there is exactl one associated output (f-value). c. Temperature Conversion from Degrees Celsius to Degrees Fahrenheit Temperature in Fahrenheit (F ) f ( x 9 ) = + 5 x x Temperature in Celsius (C ) (continues) Copright 215 Pearson Education, Inc. 37

6 LESSON 4: WHAT IS A FUNCTION? (continued) d. The constant rate of change is 9 ; this rate means that for ever increase of 9 F 5 there is an increase of 5 C. e. Yes, the dots can be connected because the graph is continuous. You could measure a temperature of 5.5º and that would still make sense The formula for converting from Fahrenheit to Celsius is c = (f 32). 9 To convert a temperature of 1 F to degrees Celsius: c = 5 (1 32) 9 c = 5 (68) 9 c = 37.7 Challenge Problem 5. Answers will var. Share our findings with a classmate. The equations involved are shown in the table. From Kelvin (K) To Kelvin (K) Celsius (C) C = K K = C Fahrenheit (F) F = K K = ( F ) 5 9 Copright 215 Pearson Education, Inc. 38

7 LESSON 6: WHAT IS A LINEAR FUNCTION? 1. B 4 2. C = 3x a. At t = Talisha starts 3 ft closer to the finish line than Erin. 7 4 = 3 b. Answers will var. Possible answer: Erin starts 7 ft from the finish line. She runs at 5 ft/sec the entire race. She finishes the race after 14 sec. Talisha starts onl 4 ft awa from the finish line. She runs at 2 ft/sec for the entire race. Talisha finishes the race after 2 sec. Even though Erin starts significantl behind Talisha, Erin passes Talisha after 1 sec and wins the race! c. Answers will var. Possible answer: The bos stor is quite similar to the girls stor. Marshall starts 4 ft from the finish, runs at 4 ft/sec for the entire race, and completes the race after 1 sec. Jacob starts the race 22 ft from the finish, runs at 1 ft/sec, and finishes the race after 22 sec. Marshall passes Jacob after 6 sec and wins the race. d. Here are all four runners speeds: Erin: 5 ft/sec Talisha: 2 ft/sec Marshall: 4 ft/sec Jacob: 1 ft/sec Erin is the fastest, followed b Marshall, then Talisha, and finall Jacob. e. Erin: d = 5t + 7 Talisha: d = 2t + 4 Marshall: d = 4t + 4 Jacob: d = 1t + 22 (continues) Copright 215 Pearson Education, Inc. 39

8 LESSON 6: WHAT IS A LINEAR FUNCTION? (continued) f. The girls graphs intersect at (1, 2). The bos graphs intersect at (6, 16). Girls: d = 5t+ 7 ( d) = ( 2t+ 4) = 3t + 3 t = 1 Bos: d = 4t+ 4 ( d) = ( 1t+ 22) = 3t + 18 t = 6 d = 5t+ 7 ( ) d = d = 2 d = 4t+ 4 ( ) d = d = 16 Challenge Problem 4. a Taxi Fares Cost ($) x Distance (mi) b. Answers will var. Possible answer: The first section indicates that for an distance between mi and 2 mi, ou will pa $4.. After that, the fare goes up prett steepl, at a rate of $.35/.2 mi. The slope is 7. For example, if ou go 6 mi our total fare is now $ If ou bring 3 friends, the entire graph shifts up verticall b $3., since the charge is $1. per each extra rider, no matter the distance. Copright 215 Pearson Education, Inc. 4

9 LESSON 7: COMPARING LINEAR FUNCTIONS 1. B = 4x C No, because the slope is the same and the intercepts are different. 3. a. Da Tickets Number of Tickets 2, 25, 3, 35, 4, 45, Income ($) 1,6, 2,, 2,4, 2,8, 3,2, 3,6, b. Income = $8(number of tickets) = 8x c. 4, Da Tickets 3,5 Income (in $1,s) 3, 2,5 2, 1,5 1, x Number of Tickets (in 1,s) d. Full Festival Tickets Number of Tickets 2, 25, 3, 35, 4, 45, Income ($) 4,, 5,, 6,, 7,, 8,, 9,, (continues) Copright 215 Pearson Education, Inc. 41

10 LESSON 7: COMPARING LINEAR FUNCTIONS (continued) e. Income = $2(number of tickets) = 2x f. Da Tickets and Full Festival Tickets Income (in $1,s) 4,5 4, 3,5 3, 2,5 2, 1,5 = 2x Full festival tickets = 8x Da tickets 1, x Number of Tickets (in 1,s) g. Da ticket rate of change: $8 per ticket Full festival ticket rate of change: $2 per ticket Challenge Problem 4. Since ou make the most profit from da tickets ou would want to sell a total of 6, da tickets, which means 2, people per da. Ideall the rest of the space will be filled with people who have full festival tickets. The maximum capacit is 5, people, so if ou sell 3, full festival tickets the festival will be filled each da. Total earnings from ticket sales would be $1,8,. Total = ($8 6,) + ($2 3,) = $4,8, + $6,, = $1,8, Copright 215 Pearson Education, Inc. 42

11 LESSON 8: GRAPHS AND FORMULAS 1. A 1 2. A The graph decreases from x = to x = From these two points ou can determine that the rate of change is 3 4. You know the -intercept, since it is given. a. The value will reach at the point (16, ). b. The equation for this linear relationship can be written = 3 4 x a. The range of the function is the -values interval of [ 11, 49]. b. f(x) = 6x 11 Substitute 6 and 26 for x and. 26 = 6(6) Contradiction! The point (6, 26) does not fit this function. c. g(x) = 6x 1 5. Using the times given ou can determine that the rate of the water level dropping is 1.5 ft/hr. The equation for this function can be written: f(t) = t t = number of hours after 6 a.m. The equation for the process of draining the tank is: = t 1.5t = 12 t = 8 hr Since the tank starts being drained at 6 a.m. it will be completel drained at 2 p.m. Challenge Problem 6. a. From 1995 ou can create an equation that models the amount of cheese based on time (in ears), where t = means f(t) = 7,, + 36,,t b. The ear 213 is 18 ears after So substitute 18 for t. f(18) = 7,, + 36,,(18) = 7,, + 648,, = 1,348,, kg of cheese If this growth model still holds in 213, the Dutch production of cheese for 213 would be 1,348,, kg. Copright 215 Pearson Education, Inc. 43

12 LESSON 9: MODEL SITUATIONS 1. A Season pass 2. D = 1x a. f (t) = 1t, or d = 1t b. Time t is positive in this situation. The domain is t. c. f (3) = 1(3) = 3 mi d. Answers will var. e. Distance (mi) d t Time (hr) 4. a. g(d) =.1d, or t = 1 1 d b. Distance d is positive in this situation. The domain is d. c. g(3) =.1(3) = 3 hr d. Answers will var. e. Time (hr) t Distance (mi) 8 d Copright 215 Pearson Education, Inc. 44

13 LESSON 9: MODEL SITUATIONS 5. a. The total area of the path is 3x x. The rough sketch shows how the area was calculated. x 2 5x 7x 5x x 2 7x x 2 b. A(x) = 3x x Substitute the value 1 for x: A(1) = 3(1) (1) A(1) = 27 c. Substitute the value 1.5 for x: A(1.5) = 3(1.5) (1.5) A(1.5) = Copright 215 Pearson Education, Inc. 45

14 LESSON 9: MODEL SITUATIONS 6. a. Jack charges $5 to come to our house and then $5 per hour. Marta charges $65 to come to our house and then $4 per hour. Marta has a higher initial cost but a lower hourl rate. b. Both equations are in terms of t (in hours) and are for the total cost of the plumber. The equation for Jack: J(t) = 5 + 5t The equation for Marta: M(t) = t c. A good wa to analze these two functions is to visualize them b graphing them on the same coordinate sstem. Plumber Cost 2 Jack Marta 15 Cost ($) x Hours on the Job d. The graph shows that the costs are exactl the same at 1.5 hr. So, if the visit will take less than 1.5 hr ou should hire Jack, but if the visit will take longer than 1.5 hr ou should hire Marta. Copright 215 Pearson Education, Inc. 46

15 LESSON 11: MORE MODELING 1. a. f (x) = 1,5x + 19, or = 19, 1,5x b. The -intercept is (, 19,). c. Domain of f : t 3 d. g(3) = 14,5 2,5(3) = 14,5 7,5 = 7, ft e. Domain of g : 3 t < 6 f. 6 hr Copright 215 Pearson Education, Inc. 47 g. 7, 19, 6 = 12, 6 = 2, ft/hr h. h(x) = 2,x + 19, or = 19, 2,x i. h(4) = 19, 2,(4) = 19, 8, = 11, ft 2. a. Miami: (x is measured miles) M(x) = 2.5 for x [, 1 6 ] M(x) = [.4 6(x 1 6 )] for x > 1 6 New York: N(x) = 2.5 for x [,.2] N(x) = [5(x.2)] for x >.2 b. Taxi Fares Cost ($) (continues) New York Miami x Distance (mi)

16 LESSON 11: MORE MODELING (continued) c. If ou are going less than 1 6 mi there is no difference. If ou are going between 1 6 and 1 5 mi then New York is slightl cheaper. If ou are going more than 1 5 mi then Miami is cheaper. As ou go farther and farther Miami is significantl cheaper. 3. a f(x) = 4x + 15 g(x) = 7x x b. The two lines do not intersect on this domain. c. The range of f(x) is the set of -values [ 5, 35]. The range of g(x) is the set of -values [ 38, 32]. Copright 215 Pearson Education, Inc. 48

17 LESSON 11: MORE MODELING Challenge Problem 4. f(x) and g(x) intersect at ( 6.193, 8.135) f(x) = 3.8x x 2 g(x) =.7x Copright 215 Pearson Education, Inc. 49

18 LESSON 13: USING FUNCTIONS TO PREDICT 1. D 2 sec 2. D You ma need more than three coordinate pairs. 3. Both graphs have the same start and end points on this domain. The difference is that the first graph is slightl curved while the second is a straight line. 4. From the graph it looks like the car has gone almost 55 m after 2 sec of braking. 5. a. f(x) is linear while g(x) is curved. On the domain from before, [, 3.2], g(x) is higher, but after x = 3.2 then f(x) is higher. Distance (m) Stopping Distances f(x) = 25x g(x) = 1.5x 2 + 3x x Time (sec) b. g(x) will eventuall stop gaining distance, which would show when the car is stopped. In fact, ou can see at the edge of this graph that the car will be stopped at 1 sec. c. The range of f(x) is the set of -values from [, 25]. The range of g(x) is the set of -values from [, 15]. Copright 215 Pearson Education, Inc. 5

19 LESSON 13: USING FUNCTIONS TO PREDICT Challenge Problem 6. a. Herring 1.5 Millions of Tons 1..5 f(x) = 1.4.1x x Years after 1989 b. 25 is 16 ears after On the graph, x = 16 actuall shows a negative value (.2 million) for the herring population, meaning that the fish has gone extinct! c. From the real data ou can see that the herring numbers began to increase again around 1996; the herring did not go extinct as the graph from part a showed. Copright 215 Pearson Education, Inc. 51