AFM Unit 2A Review Sheet

Transcription

1 AFM Unit A Review Sheet Multiple Choice Identif the choice that best completes the statement or answers the question. 1. Write the ordered pairs for the relation. Find the domain and range. O x {(, 5), ( 1, ), (0, 1), (1, ), (, 5)}; domain: {, 1, 0, 1, }; range: {1,, 5} {(5, ), (, 1), (1, 0), (, 1), (5, )}; domain: {, 1, 0, 1, }; range: {1,, 5} c. {(, 5), ( 1, ), (0, 1), (1, ), (, 5)}; domain: {1,, 5}; range: {, 1, 0, 1, } d. {(5, ), (, 1), (1, 0), (, 1), (5, )}; domain: {1,, 5}; range: {, 1, 0, 1, }. Find the domain and range of the relation and determine whether it is a function. O x Domain: all real numbers; range: all real numbers; es, it is a function Domain: x > 0; range: > 0; es, it is a function. c. Domain: positive integers; range: positive integers; no, it is not a function. d. Domain: x 0; range: 0; no, it is not a function. 3. Use the vertical-line test to determine which graph represents a function.

2 c. O x O x d. O x O x Determine whether varies directl with x. If so, find the constant of variation k.. 6 = 5x es; 5 6 es; 6 5 c. es; 5 d. no 5. The range of a car is the distance R in miles that a car can travel on a full tank of gas. The range varies directl with the capacit of the gas tank C in gallons. Find the constant of variation for a car whose range is 31 mi with a gas tank that holds gal. Write an equation to model the relationship between the range and the capacit of the gas tank mi/gal; R = 151 C c. 31 mi/gal; R = 31 C 15 1 mi/gal; C = 151 R d. 750 mi/gal; RC = A leak valve on the water meter overcharges the residents for one gallon of water in ever months. The overcharged amount w varies directl with time t.

3 Find the equation that models this direct variation. How man months it will take for the residents to be overcharged for 8 gallons of water? ; 0 months c. ; months ; 0 months d. ; months Find the value of for a given value of x, if varies directl with x. 7. If = 166 when x = 83, what is when x = 3? c. 6 d If =.8 when x =., what is when x =.05? c d The distance traveled at a constant speed is directl proportional to the time of travel. If Olivia traveled 11 miles in 3.5 hours, how man miles will Olivia travel in 8.9 hours at the same constant speed? 99.6 mi 8.8 mi c mi d. 1. mi 10. Use a graphing calculator to find the relative minimum, relative maximum, and zeros of. If necessar, round to the nearest hundredth. relative minimum: ( 6., 0.36), relative maximum: (37.79, 3.69), zeros: x = 5,, relative minimum: (0.36, 6.), relative maximum: ( 3.69, 37.79), zeros: x = 5,, c. relative minimum: (0.36, 6.), relative maximum: ( 3.69, 37.79), zeros: x = 5, d. relative minimum: ( 6., 0.36), relative maximum: (37.79, 3.69), zeros: x = 5, 11. Suppose that x and var inversel, and x = 7 when = 11. Write the function that models the inverse variation. = 1.57x c. d. 1. Suppose that x and var inversel and that variation and find when x = 10. ; 1 0 c. when x = 3. Write a function that models the inverse ; 1 6 d. ; 1 0 ; Suppose that varies directl with x and inversel with z = 5 when x = 35, and z = 7. Write the equation that models the relationship. Then find when x = 1 and z =.

4 5 3 1 c. 7 3 d Suppose that varies jointl with w and x and inversel with z and = 360 when w = 8, x = 5 and z = 5. Write the equation that models the relationship. Then find when w =, x = and z = 3. c d Find the domain and range of the relation. Age of Person Books Read domain: {9, 9, 36} range: {17, 37, } domain: {9, 9, 36} range: {37, 37, } c. domain: {9, 36, 65} range: {37, 37, } d. domain: {9, 36, 65} range: {17, 37, }

5 AFM Unit A Review Sheet Answer Section MULTIPLE CHOICE 1. ANS: A PTS: 1 DIF: L1 REF: -1 Relations and Functions OBJ: -1.1 Graphing Relations NAT: NAEP Ab NAEP Ae CAT5.LV1/.5 IT.LV17/18.AM S9.TSK3.PRA S10.TSK3.PRA TV.LV1/.10 TV.LV1/.1 TV.LV1/.16 TV.LVALG.56 STA: NC.03a TOP: -1 Example KEY: ordered pair domain range relation MSC: NAEP Ab NAEP Ae CAT5.LV1/.5 IT.LV17/18.AM S9.TSK3.PRA S10.TSK3.PRA TV.LV1/.10 TV.LV1/.1 TV.LV1/.16 TV.LVALG.56. ANS: B PTS: 1 DIF: L REF: -1 Relations and Functions OBJ: -1. Identifing Functions NAT: NAEP Ab NAEP Ae CAT5.LV1/.5 IT.LV17/18.AM S9.TSK3.PRA S10.TSK3.PRA TV.LV1/.10 TV.LV1/.1 TV.LV1/.16 TV.LVALG.56 STA: NC.03a TOP: -1 Example 5 KEY: domain range relation MSC: NAEP Ab NAEP Ae CAT5.LV1/.5 IT.LV17/18.AM S9.TSK3.PRA S10.TSK3.PRA TV.LV1/.10 TV.LV1/.1 TV.LV1/.16 TV.LVALG ANS: C PTS: 1 DIF: L1 REF: -1 Relations and Functions OBJ: -1. Identifing Functions NAT: NAEP Ab NAEP Ae CAT5.LV1/.5 IT.LV17/18.AM S9.TSK3.PRA S10.TSK3.PRA TV.LV1/.10 TV.LV1/.1 TV.LV1/.16 TV.LVALG.56 STA: NC.03a TOP: -1 Example 5 KEY: graphing vertical-line test MSC: NAEP Ab NAEP Ae CAT5.LV1/.5 IT.LV17/18.AM S9.TSK3.PRA S10.TSK3.PRA TV.LV1/.10 TV.LV1/.1 TV.LV1/.16 TV.LVALG.56. ANS: A PTS: 1 DIF: L1 REF: -3 Direct Variation OBJ: -3.1 Writing and Interpreting a Direct Variation NAT: NAEP Aa NAEP Ab CAT5.LV1/.50 CAT5.LV1/.5 IT.LV17/18.AM S9.TSK3.PRA S10.TSK3.PRA TV.LV1/.10 TV.LV1/.16 TV.LV1/.5 TV.LVALG.5 STA: NC 1.05 TOP: -3 Example KEY: constant of variation MSC: NAEP Aa NAEP Ab CAT5.LV1/.50 CAT5.LV1/.5 IT.LV17/18.AM S9.TSK3.PRA S10.TSK3.PRA TV.LV1/.10 TV.LV1/.16 TV.LV1/.5 TV.LVALG.5 5. ANS: A PTS: 1 DIF: L1 REF: -3 Direct Variation OBJ: -3.1 Writing and Interpreting a Direct Variation NAT: NAEP Aa NAEP Ab CAT5.LV1/.50 CAT5.LV1/.5 IT.LV17/18.AM S9.TSK3.PRA S10.TSK3.PRA TV.LV1/.10 TV.LV1/.16 TV.LV1/.5 TV.LVALG.5 STA: NC 1.05 TOP: -3 Example 3 KEY: constant of variation direct variation multi-part question word problem problem solving MSC: NAEP Aa NAEP Ab CAT5.LV1/.50 CAT5.LV1/.5 IT.LV17/18.AM S9.TSK3.PRA S10.TSK3.PRA TV.LV1/.10 TV.LV1/.16 TV.LV1/.5 TV.LVALG.5 6. ANS: A PTS: 1 DIF: L REF: -3 Direct Variation OBJ: -3.1 Writing and Interpreting a Direct Variation NAT: NAEP Aa NAEP Ab CAT5.LV1/.50 CAT5.LV1/.5 IT.LV17/18.AM S9.TSK3.PRA S10.TSK3.PRA TV.LV1/.10 TV.LV1/.16 TV.LV1/.5 TV.LVALG.5 STA: NC 1.05 TOP: -3 Example 3 KEY: constant of variation direct variation multi-part question problem solving word problem MSC: NAEP Aa NAEP Ab CAT5.LV1/.50 CAT5.LV1/.5 IT.LV17/18.AM S9.TSK3.PRA

6 S10.TSK3.PRA TV.LV1/.10 TV.LV1/.16 TV.LV1/.5 TV.LVALG.5 7. ANS: D PTS: 1 DIF: L1 REF: -3 Direct Variation OBJ: -3.1 Writing and Interpreting a Direct Variation NAT: NAEP Aa NAEP Ab CAT5.LV1/.50 CAT5.LV1/.5 IT.LV17/18.AM S9.TSK3.PRA S10.TSK3.PRA TV.LV1/.10 TV.LV1/.16 TV.LV1/.5 TV.LVALG.5 STA: NC 1.05 TOP: -3 Example KEY: direct variation MSC: NAEP Aa NAEP Ab CAT5.LV1/.50 CAT5.LV1/.5 IT.LV17/18.AM S9.TSK3.PRA S10.TSK3.PRA TV.LV1/.10 TV.LV1/.16 TV.LV1/.5 TV.LVALG.5 8. ANS: A PTS: 1 DIF: L1 REF: -3 Direct Variation OBJ: -3.1 Writing and Interpreting a Direct Variation NAT: NAEP Aa NAEP Ab CAT5.LV1/.50 CAT5.LV1/.5 IT.LV17/18.AM S9.TSK3.PRA S10.TSK3.PRA TV.LV1/.10 TV.LV1/.16 TV.LV1/.5 TV.LVALG.5 STA: NC 1.05 TOP: -3 Example KEY: direct variation MSC: NAEP Aa NAEP Ab CAT5.LV1/.50 CAT5.LV1/.5 IT.LV17/18.AM S9.TSK3.PRA S10.TSK3.PRA TV.LV1/.10 TV.LV1/.16 TV.LV1/.5 TV.LVALG.5 9. ANS: B PTS: 1 DIF: L1 REF: -3 Direct Variation OBJ: -3.1 Writing and Interpreting a Direct Variation NAT: NAEP Aa NAEP Ab CAT5.LV1/.50 CAT5.LV1/.5 IT.LV17/18.AM S9.TSK3.PRA S10.TSK3.PRA TV.LV1/.10 TV.LV1/.16 TV.LV1/.5 TV.LVALG.5 STA: NC 1.05 TOP: -3 Example KEY: direct variation proportion problem solving word problem MSC: NAEP Aa NAEP Ab CAT5.LV1/.50 CAT5.LV1/.5 IT.LV17/18.AM S9.TSK3.PRA S10.TSK3.PRA TV.LV1/.10 TV.LV1/.16 TV.LV1/.5 TV.LVALG ANS: B PTS: 1 DIF: L REF: 6- Polnomials and Linear Factors OBJ: 6-.1 The Factored Form of a Polnomial NAT: NAEP A3b NAEP A3c CAT5.LV1/.50 CAT5.LV1/.5 CAT5.LV1/.56 IT.LV17/18.AM IT.LV17/18.CP S9.TSK3.GM S9.TSK3.PRA S10.TSK3.GM S10.TSK3.PRA TV.LV1/.10 TV.LV1/.13 TV.LV1/.5 TV.LVALG.53 STA: NC 1.03 TOP: 6- Example 3 KEY: factoring a polnomial graphing calculator polnomial function problem solving x-intercept relative maximum relative minimum zeros of a polnomial function MSC: NAEP A3b NAEP A3c CAT5.LV1/.50 CAT5.LV1/.5 CAT5.LV1/.56 IT.LV17/18.AM IT.LV17/18.CP S9.TSK3.GM S9.TSK3.PRA S10.TSK3.GM S10.TSK3.PRA TV.LV1/.10 TV.LV1/.13 TV.LV1/.5 TV.LVALG ANS: B PTS: 1 DIF: L1 REF: 9-1 Inverse Variation OBJ: Using Inverse Variation NAT: NAEP Ae CAT5.LV1/.50 CAT5.LV1/.53 CAT5.LV1/.56 IT.LV17/18.AM S10.TSK3.GM S10.TSK3.PRA TV.LV1/.1 TV.LV1/.15 TV.LV1/.17 TV.LV1/.5 STA: NC 1.05 TOP: 9-1 Example 1 KEY: rational function inverse variation MSC: NAEP Ae CAT5.LV1/.50 CAT5.LV1/.53 CAT5.LV1/.56 IT.LV17/18.AM S10.TSK3.GM S10.TSK3.PRA TV.LV1/.1 TV.LV1/.15 TV.LV1/.17 TV.LV1/.5 1. ANS: A PTS: 1 DIF: L1 REF: 9-1 Inverse Variation OBJ: Using Inverse Variation NAT: NAEP Ae CAT5.LV1/.50 CAT5.LV1/.53 CAT5.LV1/.56 IT.LV17/18.AM S10.TSK3.GM S10.TSK3.PRA TV.LV1/.1 TV.LV1/.15 TV.LV1/.17 TV.LV1/.5

7 STA: NC 1.05 TOP: 9-1 Example 3 KEY: rational function inverse variation MSC: NAEP Ae CAT5.LV1/.50 CAT5.LV1/.53 CAT5.LV1/.56 IT.LV17/18.AM S10.TSK3.GM S10.TSK3.PRA TV.LV1/.1 TV.LV1/.15 TV.LV1/.17 TV.LV1/ ANS: D PTS: 1 DIF: L1 REF: 9-1 Inverse Variation OBJ: 9-1. Using Combined Variation NAT: NAEP Ae CAT5.LV1/.50 CAT5.LV1/.53 CAT5.LV1/.56 IT.LV17/18.AM S10.TSK3.GM S10.TSK3.PRA TV.LV1/.1 TV.LV1/.15 TV.LV1/.17 TV.LV1/.5 STA: NC 1.05 TOP: 9-1 Example 5 KEY: direct variation combined variation MSC: NAEP Ae CAT5.LV1/.50 CAT5.LV1/.53 CAT5.LV1/.56 IT.LV17/18.AM S10.TSK3.GM S10.TSK3.PRA TV.LV1/.1 TV.LV1/.15 TV.LV1/.17 TV.LV1/.5 1. ANS: A PTS: 1 DIF: L REF: 9-1 Inverse Variation OBJ: 9-1. Using Combined Variation NAT: NAEP Ae CAT5.LV1/.50 CAT5.LV1/.53 CAT5.LV1/.56 IT.LV17/18.AM S10.TSK3.GM S10.TSK3.PRA TV.LV1/.1 TV.LV1/.15 TV.LV1/.17 TV.LV1/.5 STA: NC 1.05 TOP: 9-1 Example 5 KEY: direct variation combined variation joint variation MSC: NAEP Ae CAT5.LV1/.50 CAT5.LV1/.53 CAT5.LV1/.56 IT.LV17/18.AM S10.TSK3.GM S10.TSK3.PRA TV.LV1/.1 TV.LV1/.15 TV.LV1/.17 TV.LV1/ ANS: D PTS: 1 DIF: L1 REF: 5- Relations and Functions OBJ: 5-.1 Identifing Relations and Functions NAT: NAEP A1g CAT5.LV19.53 CAT5.LV19.5 IT.LV15.DI IT.LV15.PS S9.TSK1.PRA S10.TSK1.PRA TV.LV19.1 TV.LV19.16 TV.LVALG.56 STA: NC.01 NC.01a TOP: 5- Example 1 KEY: domain range MSC: NAEP A1g CAT5.LV19.53 CAT5.LV19.5 IT.LV15.DI IT.LV15.PS S9.TSK1.PRA S10.TSK1.PRA TV.LV19.1 TV.LV19.16 TV.LVALG.56