IAT SOLUTIONS - C_7.pdf

Chapter 7, continuedCopyright © by McDougal Littell, a division of Houghton Mifflin Company.20.(0, 6)(0, 3)1yy 5 3(0.25) x 1 315( 1,4 )3( 1,4 )1y 5 3(0.25) x x21.y3(0, 1)21(1 x 2 2y 5 3 1 21( 1,3 )(2, 3)(7(3,3 )(1 x xy 5 3Domain: all real numbers Domain: all real numbersRange: y > 3 Range: y > 222.f(x)x 23.g(x)1( )31, 24(0, 23)1f(x) 5 23 453( 2, 24 )(1, 23)f(x) 5 231( 4)( )x 2 1x(24, 1)g(x) 5 6g(x) 5 6(25, 4)1( 2)1( 2)x1x 1 5 2 2(0, 6)1(1, 3)Domain: all real numbers Domain: all real numbersRange: y < 0 Range: y > 2224.h(x)(21, 4)(0, 2)121h(x) 5 4(0, 4)(1, 2)1( 2)( )x1 x 1 1 xh(x) 5 4 2Domain: all real numbersRange: y > 025. y 5 ab x 2 h 1 k → y 5 3(0.4) x 2 2 2 1a. If a changes to 4, there will be a vertical stretch.b. If b changes to 0.2, the graph will be steeper becausethe decay factor is smaller.c. If h changes to 5, there will be a horizontal translation(of 3 units right)d. If k changes to 3, there will be a vertical translation(of 4 units up).26. The decay factor was written incorrectly. It should be1 2 percent decrease 5 1 2 0.02 5 0.98.y 5 (inital amount)(decay factor) ty 5 500(0.98) t27. D; The graph y 5 1 }1 22 x 2 2 1 3 has an asymptote at theline y 5 3.28. You need a base between 0.25 and 0.5 and a verticaltranslation up between 0 and 3 units. Sample answer:y 5 (0.3) x 1 129. Yes; (4) 2x is just another way of saying 1 }1 42 x or(0.25) x , so f (x) 5 5(4) 2x and g(x) 5 5(0.25) x representthe same function.(xProblem Solving30. a. When I 5 200 and t 5 1.5: A 5 I(0.71) tA 5 200(0.71) 15 ø 119.65There is about 120 milligrams of ibuprofen remainingin the bloodstream.b. When I 5 325 and t 5 3.5: A 5 I(0.71) tA 5 325(0.71) 3.5 ø 98.01There is about 98 milligrams of ibuprofen remainingin the bloodstream.c. When I 5 400 and t 5 5: A 5 I(0.71) tA 5 400(0.71) 5 ø 72.17There is about 72 milligrams of ibuprofen remainingin the bloodstream.31. a. When r 5 0.25 and a 5 200: y 5 a(1 2 r) ty 5 200(1 2 0.25) t → y 5 200(0.75) tWhen t 5 3: y 5 200(0.75) 3 ø 84.38The value of the bike after 3 years is about $84.38.c. Using the graph, you can estimate that the value of thebike will be $100 after about 2.5 years.Value (dollars)y20017515012510075502500 1 2 3 4 5 6 7 8 tYears32. Ratio from year 1–2: 1832 }1906 ø 0.96Ratio from year 2–3: 1762 }1832 ø 0.96Ratio from year 3–4: 1692 }1762 ø 0.96Ratio from year 4–5: }16271692 ø 0.96The ratio of depreciation remains a constant 0.96 or 96%.d 5 a(0.96) 1 5 1906 → a 5 }19060.96 ø 1985An equation is d 5 1985(0.96) t .33. a. yValue (dollars)28,00024,00020,00016,00012,0008000400000 1 2 3 4 5 6 7 8 tYearsUsing the graph, you can estimate that the value of thecar will be $10,000 after about 5 years.Algebra 2Worked-Out Solution Key425