Unit 4 Practice Problem ANSWERS

Unit Practice Problem ANSWERS SECTION.1A 1) Parabola ) a. Root, Zeros b. Ais of smmetr c. Substitute = 0 into the equation to find the value of. -int 6) 7 6 1 - - - - -1-1 1 - - - - -6-7 - ) ) Maimum ) Minimum ) -int or zero Ais of smmetr verte -int 7) ais of smmetr: = 1 -intercept: (0, ) verte: (1, 7) domain: All Real Numbers range: 7 opens: down Vertical stretch 7 6 1 9) ais of smmetr: = 1 -intercept: (0, 1) verte: ( 1, ) domain: All Real Numbers range: opens: up Vertical stretch ais of smmetr: = 1 -intercept: (0, ) verte: (1, ) domain: All Real Numbers range: opens: up No vertical stretch nor compression - - - - -1-1 1 - - - - -6-7 - ais of smmetr: = 1 -intercept: (0, ) verte: (1,.) domain: All Real Numbers range:. opens: up Vertical compression ais of smmetr: = 1 -intercept: (0, 0) verte: (1, 1) domain: All Real Numbers range: 1 opens: down No vertical stretch nor compression 1 APPENDICES

Unit Practice Problem ANSWERS ) SECTION.1A (continued) ais of smmetr: = 1 -intercept: (0, ) verte: ( 1, 6) domain: All Real Numbers range: 6 opens: down Vertical stretch 11) a. (0, ), 1 =, down b. (0, 1), =, up 1) a. (0,7), =, up 1) a. b. (0, 1), =, down b. feet 7 6 1 - - - - -1-1 1 - - - - -6-7 - c. 0. seconds, at the verte (maimum) 1) d. 1 second if the juggler catches it; at approimatel 1.91 seconds if it hits the ground, this is an -intercept 1) a. Height of fireworks (ft) 1) a. Height of frogsa (ft) e. (0, 6); The height of the jugglers hand is 6 feet when he lets go of the ring. b. 7 ft c. 7 seconds, at the verte d. 1 sec, the positive -int b. 1 feet, at the verte c. seconds d. seconds Time (seconds) Time (seconds) 16) a. provides information whether the graph opens up of down provides information whether the graph is a vertical stretch or vertical compression of the parent graph. b. Use b to find the ais of smmetr; =. Then use this -value to find the -value of the verte. c. c is the -intercept where = 0. d. = e. Once ou know the coordinate from, ou can substitute the value into the original function and solve for 17) a. -intercept: (0, ) ais of smmetr: = 0 verte: (0, ) opens: up Vertical compression domain: All real numbers range: b. -intercept: (0, ) ais of smmetr: = verte: (, 7) opens: up Vertical: same domain: All real numbers range: 7 c. -intercept: (0, ) ais of smmetr: = 1 verte: (1, 1) opens: down Vertical stretch domain: All real numbers range: 1 Appendi B ~ Selected Answers ~ Unit 19

Unit Practice Problem ANSWERS 1) -int Ais of smmetr verte -int -int ) a. Answers will var. E: Start b using =, then substitute the -value into the original function to solve for and find the verte. From there, use a table to find other points on the parabola. Use the c- value of the equation to graph the -intercept at (0, 1). b. Answers will var. E: Set each of the factors equal to zero to find the two -intercepts. Then, graph the points (, 0) and (, 0) and find the -coordinate halfwa between them to use as the -coordinate of the verte. Then, substitute that -value to find the -value of the verte. Because there is no negative in front of either factor, the parabola opens up. c. Answers will var. E: With factored form ou start with the -intercepts, then find the verte. With standard form ou start b finding the verte using. You use the verte to graph with both forms. SECTION.1B ) a. -intercepts are ( 9,0) and (,0). The -intercepts represent the number $ decreases the compan should have in order to increase profit. b. $ decreases. $16 off $0, so $. Income at this price will be $,900 c. ) a. 0000 000 0000 1000 000 000 - - 1 0 d. Domain: 9 Range: 0,900 1 1 6 - -1 1 b. seconds c. 1. seconds d. 11.0 meters e. Domain: all real numbers Range: 11.0 ) f. * The domain represents time, and because time is negative it must be greater than or equal to 0. The object hits the ground after seconds, so an time after seconds implies the object is below the ground which is not possible. * The range represents the height of the object. The object starts at the ground and goes to a maimum height of 11 meters so the height will alwas remain between 0 and 11. ) 6) -intercept(s): (, 0), (6, 0) Ais of Smmetr: = Minimum at (, ) No Ma Domain: All real numbers Range: -intercept(s): ( 1, 0), (1, 0) Ais of Smmetr: =0 Minimum at (0, ) No Ma Domain: All real numbers Range: 0 APPENDICES

Unit Practice Problem ANSWERS 7) ) 9) SECTION.1B (continued) -intercept(s): (,0)(7,0) Maimum at (,1.) Ais of smmetr: = Domain: All real numbers Range: 1. - -1 1 6 7 - - -1-0 -intercept(s): (0,0)(,0) Ais of Smmetr: =. Minimum at (., 1.7) Domain: All real numbers Range: 1.7 1-7 -6 - - - - -1 1 -intercept(s): (,0)( 1,0) Minimum at (, ) Ais of smmetr: = Domain: All real numbers Range: - ) 11) 1) 6 - - -1 1 - - -6 - -intercept(s): (,0)( 1,0) Maimum at (1,) Ais of smmetr: = 1 Domain: All real numbers Range: -7-6 - - - - -1 1 - -intercept(s): (,0)(1,0) 1 6 - - - -1 1 6 - - -6 6 - -6 - Maimum at (,9) Ais of smmetr: = Domain: All real numbers Range: 9 -intercept(s): (,0)( 1,0) Minimum at (1, ) Ais of smmetr: = 1 Domain: All real numbers Range: 1) -int(s): ( 6,0), (, 0) Ais of smmetr: = Verte: (, ) Opens: Up 1) -int(s): (,0)(,0) Ais of smmetr: = Verte: (, 1) Opens: Down 1) -int(s): (1,0)( 7,0) Ais of smmetr: = Verte: (, ) Opens: Up 16) -int(s): (,0)(,0) Ais of smmetr: = 1 Verte: ( 1,) Opens: Down 17) a. The a value determines whether the parabola opens up or down and whether the function is verticall compressed or stretched. b. Set each factor equal to zero and solve. These values are the -intercepts. c. The -coordinate of the verte can be found b going half wa between the -intercepts. d. After finding the -coordinate of the verte, substitute it into the function for to solve for. e. The appropriate values are near the verte. 1) = ( 1)( ) Domain: all real numbers Range: Minimum at (, ) 19) = ( 1)( + 6) Domain: all real numbers Range: 1. Minimum at (., 1.) Appendi B ~ Selected Answers ~ Unit 1

Unit Practice Problem ANSWERS SECTION.1C 1) -int -int ) 7) Verte: (, 1) Maimum at (, 1) Opens: down Domain: All real numbers Range: 1 Ais of smmetr -int - ) verte - ) a. a gives the direction of the opening, opening up or down, and if there is a vertical stretch or compression of the parent graph. Verte: ( 1, ) Minimum at ( 1, ) Opens: Up Domain: All real numbers Range: - - b. You have to change the sign on h, and leave k as is; the verte is located at (h, k) c. Answers will var. E: ou can use - numbers on one side of the verte, then reflect those points over the ais of smmetr. 6) 9) Verte: (, ) Minimum at (, ) Opens: Up Domain: All real numbers Range: ) a. a =, h = 1, k = b. a= 0., h=, k = 1 ) Verte: (, 1) Maimum at (, 1) Opens: down Domain: All real numbers Range: 1 7) Verte: (, ) Minimum at (, ) Opens: Up Domain: All real numbers Verte: (, 1) Range: Minimum at (, 1) Opens: Up Domain: All real numbers - Range: 1 - APPENDICES

Unit Practice Problem ANSWERS ) SECTION.1C (continued) Form Equation Significant Features Verte Verte = ( h) + Verte opens up or down. Vertical stretch or compression of graph in relation to parent graph =. Intercept = ( )( ) -intercepts, opens up or down, Vertical stretch or compression of graph in reference to the parent graph =. Standard = + +! Opens up or down, Vertical stretch or compression of graph in relation to the parent graph =. The verte is (h, k). Remember to set the quantit inside the parentheses equal to zero to identif the value of h. The -coordinate is found half-wa between the -intercepts. Substitute that number in for in the original equation to find the -coordinate of the verte. = gives the -coordinate of the verte. Substitute this number into the original equation and solve for to get the -coordinate. 11) Answers will var. Possible choices include: Verte form because I get the verte right awa. Intercept form because I get the -intercepts right awa. Standard form because I am used to it and I can use the = to find the -intercepts. 1) a. 16 1-6 1 - b. (, 1) c. = feet, so at his highest point in the air he is feet awa (horizontall) from where he started on the diving board; = 1 feet, his highest point is 1 feet in the air. 1) d. (9.916, 0) The diver is 9.916 feet out from the diving board horizontall and is entering the water (height above the water is 0). 1) a.. seconds b. feet c. Domain: 0 < <. Range: 0 < < 1) d. Domain has to be positive because it represents time, and the ball is no longer in the air after. seconds, Range has to be positive because the ball can t go below ground level and doesn t reach a height above feet. 1) Up, =, (, ) 1) Down, =, (, 1) 16) Up, =, (, 0) 17) Up, = 1, ( 1, 1) 1) Down, =, (, 1) 19) Down, = 1, (1, 7) Appendi B ~ Selected Answers ~ Unit

Unit Practice Problem ANSWERS SECTION.A 1) + 1 1) + +1 ) + + ) + 6 1) + 0 0 6) + 16 ) + 1) + 7) + 7 ) 6 + 16) 6 + 1 6 ) 1 1 ) 1 17) + 19 + 60 9) + 7 6 6) a 1) + 1 + 0) + 0 7) b + 6b 19) + 7 + 1) + 1 ) k + k 0) + 9 + ) + 9) + 1 9 1) + +1 ) 6 + + 60 ) + ) + 1 ) 1 11) +1 + ) 9 1) 6n n n ) 6 16 1) = + 6 1 ) = ) = + ) = + ) = 1 6) = + SECTION.B ) a. -int: (, 0), (, 0) b. = + c. (0, ) d. 1 1) a. -int: ( 7, 0), (, 0) b. = +1 c. (0, 1) d. 7) = +1 + ) = + 0 9) a. -int: (, 0), ( 1, 0) b. = + + c. (0, ) d. graph 16 1-6 - - - - -1-0 - 11) a. -int: (1, 0), (, 0) b. = + c. (0, ) d. 1) a. 0,000 people b. P(0) = 0 {1,00,000 people} c. t 0 1 0 0 P(t) 0 1. 600 1. 0 1600 - - 6 - -6 - - - - -1-0 - APPENDICES

Unit Practice Problem ANSWERS SECTION.B (continued) 1) a. at 7 seconds b. 6.6 meters c. 0 1 6 7 6.6. 9 9. 6.6 9. 0 1) a. = + b. 16) a. = 9 6 b. 1) = 6 + ) = + 1 ) = ) = 16 + 67 ) = 6) =.. + 6. 7) Equivalent ) No 9) a. (1, 16) b. = 1 c. (0, 1) d. 1 0 1 9 0 7 1 1 16 6 1 1 7 0 9 ) a. (, 7) 11) C 1) A 1) D 1) B SECTION.C b. = + 1 + c. (0, ) d. 1 1 9 7 9 1 1) a. = 0. + 6. +. b. 6 meters c.. meters 16) a. = + 1 16) b. 17) a. = + b. 1) a. meters b. seconds c. 1.6 meters d. 19) C Appendi B ~ Selected Answers ~ Unit

Unit Practice Problem ANSWERS SECTION.A 1) ( + ) ) ( ) ) (b + 1) ) 6(n 1) ) (6w 7) 6) (6p + q) 7) (a + ) ) ( ) 9) 7( + 1) ) t(t + 7) 11) (9p p + 1) 1) (9b b + ) 1) ( + 1) 1) (w + w 1) 1) (w w 6) 16) 7(m m + ) 17) v(v v 7) 1) ( ) 19) m (m 9m + ) 0) n (n n ) 1) Not equivalent ) Equivalent ) = ( ), -intercept: (0, ) verte: (, 0) ) = ( 7 + ) -intercept: (0, 1) verte: (.,.7) ) = ( 6 + ) -intercept: (0, ) verte: (, 1) SECTION.B 6) = 6( 1) -int: (0, 6), verte: (, 0) 7) 1 ) 6pr 9) q 0) 9 z 1) (9b + 6b + 7) ) (b 7b + 7) ) 7(r 6 + r + 9) ) n(1n n 0) ) ( + 7 + ) 6) 6n (7n + n + ) 7) 7(a 7a 6a ) ) ( + 0 1) 9) d (1d d + d ) 0) 7b (b + b b + ) 1) +, + ), ) +, ), + ) (b )(b + ) 6) (m + 7)(m + ) 7) (n )(n 7)) ) (a 9)(a + 6) 9) (n + )(n ) ) (v 6)(v ) 11) = ( + 6)( + ) -int: (,0), ( 6,0) 1) = ( + )( 1) -int. (, 0), (1, 0) 1) = ( + )( + ) -int: (, 0), (, 0) 1) = ( 9)( ) -int: (9, 0), (, 0) 1) (k + )(k + ) 16) (n + )(n + 1) 17) ( 1)( ) 1) (a 6)(a + ) 19) (v )(v ) 0) ( 7)( + ) 1) ( + )( ) ) (a + 7)(a + ) ) ( + 1)( + ) ) ( + )( ) ) Prime (not factorable) 6) ( + 1)( + ) 7) a. = b. f() = ( + 9)( + 1) c. -int: ( 9, 0), ( 1, 0) d. = e. es f. ) (b 7) 9) ( + 9)( ) 0) f() = ( )( ) -int: (, 0), (, 0) f() 16 1 - - -int: (, 0), (, 0) f() 1 6 - - 6-6 -1-1 - -0 6 1) f ( ) = ( ) ( + ) 6 APPENDICES

Unit Practice Problem ANSWERS SECTION.B (continued) ) = ( )( + ) -int: (, 0), (, 0) ) = ( )( + ) -int: (, 0), (, 0) ) a. 7 b. ( )( ) c. (, 0), (, 0) d. 7 e. es ) f. ) a. (0, ), (1, 0), (, ) 0 seconds meters 1 second 0 meters seconds meters b. meters c. h(t)= (t 9)(t + 1), 9 sec d. 1 meters at seconds e. h(t) 0 0 60 0 0 0 6 t 1) +, + ), ), + ) +, ) (n )(n ) 6) (k )(k 7) 7) (7 + 9)( + 6) ) (b )(b + ) 9) ( + )( 7) ) (p + )(p 7) 11) (a )(a ) 1) ( + )( + 1) 1) not equivalent 1) Equivalent 1) = ( + 1)( ) verte:(1., 6.1) -int: ( 0., 0), (, 0) 16) = ( + )( ) verte: ", # $ -int: ",0$,(,0) 17) = ( )( + 1) verte: ( 0.1,.) -int:(0.7, 0),( 1, 0) 1) = ( + )( + ) verte: (.0,.0) -int: " %,0$,",0$ 19) ( 1)( ) 0) (7v + )(v + ) 1) (r + 7)(r 1) ) (n )(n + ) ) D ) a. feet b. (t + 1)(t ), The ball hits the ground at t = seconds; t = 0. does not make sense in this contet. SECTION.C ) c. 9 feet at 1. seconds d. h(t) 6 1-1 1-1 ) ( + )( + 1) 6) ( + 1)( + ) 7) (7 )( 1) ) ( 1)( ) t 9) ( + 1)( + ) 0) (b + 1)(b + 1) 1) (b 1)(b 1) ) (6b + 1)(b + ) ) (b )(b 1) ) (b + )(b + ) ) (k 1)(k + 1) 6) (k + 1)(k ) 7) (k + )(k 1) ) (7k + 1)(k ) 9) (k + 1)(k 1) 0) (v + 1)(v ) 1) (v )(v + ) ) (v )(v ) ) (v )(v + 6) ) 1(v + )(v ) ) ( + )( + ) or ( + ) Appendi B ~ Selected Answers ~ Unit 7

Unit Practice Problem ANSWERS SECTION.D 1) ( + )( ) 16) = ( +) 1) ( + )( ) ) ( + )( ) verte: (, 0) verte: (0, ) ) ( 11)( 11) ) ( + )( + ) ) ( )( ) or ( ) -intercept: (, 0) 1 -intercept: " &,0$,",0$ & 6) ( + )( ) 7) ( )( + ) ) ( + 7)( + 7) or ( + 7) 9) ( 1)( + 1) ) ( +)( + ) or ( + ) 11) ( )( + ) 1) ( 9)( + 9) 1) Equivalent 1) No 1) = ( + 1)( 1) verte: (0, 1) -int: (1,0), ( 1,0) -16 16-1 - 17) = ( + 6) verte and -int: ( 6,0) 1-0. -0. 0. 0. - 19) (7 + 1)(7 1) 0) ( + ) or ( + )( + ) 1) ( )( + ) ) ( )( + ) ) (a + b)(a + b) or (a + b) ) ( + ) or ( + )( + ) ) ( + ) or ( + )( + ) 6) (k + w ) or (k + w ) (k + w ) -0-1 -6 - -0-10 1) 16,,? in grid is = 16 ),? in grid is = ), APPENDICES SECTION.E ) 0, 0 ) 1, 1 6) 9, 7) 1, 9 ) 6, 9) 6, ( 6) ) 1, 1 11) 9, ( + 7) 1) 6, ( 6) 1) %# ', " ( $ 1) & ',"+ & $ 1) a. = b. = ( + ) 1 c. (, 1) d. es

Unit Practice Problem ANSWERS SECTION.E (continued) 1) e. 9 7 6 f () 1 1 17 0 1 1 0 1 f () 0 17 1 1 17) f () = ( 7) 6 1 16 - -16-19) f () = ( 1) 7 16 - -0 - - 16) f () = ( + ) 11 6. 0. f () 0 11 0 Tables will var, but the graphs should match. 1) f () = ( + ) 1-1 - - - -16 - - -0 0) a. b. f () = ( + ) 16 c. (, 16) d. es e. 6 0 f () 0 1 16 1 0 Tables will var, but the graphs should match. - - -7 - -16-1 SECTION.et 1) a. = b. f () = ( + ) c. (, ) d. es e.. 6 0. f () 0 1 0 0 Tables will var, but the graphs should match. 1) e. (con t) 1) f. = g. Domain: All real numbers Range: Appendi B ~ Selected Answers ~ Unit 9

Unit Practice Problem ANSWERS SECTION.et (continued) ) f () = 1( + 1) + Possible table points (one of which should be the verte): (, 1), ( 1, ), (1, 1) ) f () = ( + 1) + 7 Possible table points (one of which should be the verte): (, ), ( 1, 7), (1, ) ) a. f () = ( + )( ) b. (, 0) and (, 0) c. f () = ( + 1) d. Verte: ( 1, ) e. ) f () = ( + ) 0 Possible table points (one of which should be the verte): ( 6, ), (, 0), (0, ) ) f () = ( + ) + Possible table points (one of which should be the verte): (, ), (, ), (0, ) 11) a. (, 0), (, 0) b. f () = 6 c. f () = ( 1) 7 d. (1, 7) e. 6) )() = "+ & ' $ '# % 7) )() = "+ $ + ) )() = "+ & $ *' & 9) )() = "+ ( ' $ %# % 1) a. A() = 1( 1) + b. 1 feet b 1 feet c. sq. feet 1) Up or down: Up Vertical str/comp: stretch verte: (1, 0) Ais of Sm: = 1 -intercept: (0, ) UNIT REVIEW 1) f() 1) -intercept(s): (1,0) Domain: all real numbers Range: 0 - - 0 APPENDICES

Unit Practice Problem ANSWERS UNIT REVIEW (continued) ) Up or down: Up vertical str/comp: compression verte: (1, ) Ais of Smmetr: = 1 -intercept: (0, 1.) -intercept: (, 0) and ( 1, 0), Domain: all real numbers Range: ) Up or down: Down vertical str/comp: stretch verte: (1, ) Ais of Smmetr: = 1 -intercept: (0, ), -intercept: none Domain: all real numbers Range: ) = + 9 ) = + 0 6) = 9 + 7 + 1 7) = + + 1 ) = ( )( ) 9) = ( + )( ) ) = ( )( 1) 11) = ( + ) 1) = ( ) 1) = ( + )( ) 1) = ( + 1)( ) 1) = ( 7) 1 16) = " $ #* ' 17) = ( + ) + 1) = " # $ + # ' 19) 6 feet 0) seconds 1) seconds ) 0 t, the time has to be positive and stops in the contet of this problem when the ball comes back to the ground. ) 0 h(t) 6, the height cannot be negative because that would be below ground and the highest the ball gets it 6 feet. ) 1. feet ) 0.6 seconds 6) feet 7) 1. seconds, the positive -int. ) 6 feet, that is the -intercept 9) a. h+= 16+ +1++ b. b. t h (t) 0 0.97 00.97 00 9. 1. 1.7 00 19.7 0 c. (19.7, 0); the rocket returns to the ground at 19.7 seconds. d. (, 0); the rocket was feet above ground when launched. e. (9., 1.); at 9. seconds the rocket reached its maimum height of 1. feet. f. 0 feet g..97 1.7 h. Domain: 0 19.7; the rocket is in the air during these times. Range: 0 1.; this represents the possible heights of the rocket at an given time. Appendi B ~ Selected Answers ~ Unit 1

Unit Practice Problem ANSWERS 1) Find the zeros of the function means to find the -value(s) that make the function f() = 0. (answers will var) ) The -intercepts (answers will var) ) 1 solution real roots (rational) zeros: = ) solutions roots are comple-no real roots zeros: none ) solutions real roots (rational) zeros: = 0 or = 6) solutions: = or = check: f() = () + () 1 = 0 f( ) = ( ) + ( ) 1 = 0 both solutions create a zero 7) solutions: = 0 or = check: f(0) = (0) + (0) = 0 f() = () + () = 0 both solutions create a zero ) 9) SECTION.1A solutions: = or = check: f( ) = ( ) + ( ) = 0 f() = () + () = 0 ) solutions: 1. or. 11) solutions: 1.67or = 1.0 1) solutions: 0.9 or. 1) solutions:.1 or.1 0..1.1,0 0..1.1,0 1) solutions:.16 or 1.16.16.16,6 1.16 1.16,6 1) solutions: 1.16 or.16 1.16 1.16,0.16.16,0 16) 17) solutions: = or = 1 1) solutions: = or = 1 19) solutions: = or = 6 check: f( ) = ( ) + ( ) + 1 = 0 f(6) = (6) + (6) + 1 = 0 solutions: = or = 1 solutions: = 0 or = APPENDICES

Unit Practice Problem ANSWERS SECTION.1A (continued) 0) solutions: = 0 or = 1) ) a. (16) f() = + 6 + =, = 1 (17) f() = + 0 + =, = 1 (1) f() = + 1 + 1 =, = 1 b. (19) f() = + = 0, = (0) f() = (½) + = 0, = (1) f() = + 16 = 0, = c. (answers will var) If one function is a multiple of another function, then the will have the same solutions. ) a. b. feet; the point (, ) is the verte of the graph of the equation. c. (answers will var) feet; this is the -intercept of the graph of the equation (that is greater than zero). solutions: = 0 or = 1) a. Javier is correct. The -intercept gives the time (t) when = 0, which is where the ball has a height of 0 on the ground. (eplanations will var) b. Meg s ball hit the ground first. It took. seconds for it to hit the ground. c. Javier s ball hit the ground second. It took.1 seconds for it to hit the ground. ) a. h(t) = 16t + t +. b.. feet c. After 0.1 seconds and then again after.0 seconds. d. After 1.09 seconds. SECTION.1B ) e..6 feet is the maimum that the football ever gets into the air. f.. seconds is when the ball hits the ground (the -intercept). ) Verte: (1.6, 1.06). The verte reveals how long it takes (1.6 seconds) for the ball to reach the maimum height (1.06 ft). (eplanations will var) ) t = 11 das, which is the -coordinate of the verte. ) a. 1. sections is the maimum speed (the -coordinate of the verte of the graph) b. 9 km/h (the -coordinate of the verte of the graph) ) c. 6. seconds (the -coordinate of the -intercept of the graph. 6) a. 0 is half the perimeter, so one length plus one width equals 0 feet of fencing. (eplanations will var) b. A(L) = L(0 L) A(L) = L + 0L c. (L, A(L)) = (, 6) when the length is, the width is 0 L =. So, Area = L W ft ft 6 ft Area of the tomato patch is 6 square feet. d. The tomato patch is a square shape (ft ft). Appendi B ~ Selected Answers ~ Unit

Unit Practice Problem ANSWERS SECTION.1B (continued) 7) a. According to #6d, Sharon should make a square with her 0 feet of fencing. So, 0 = feet on each side. b. A(L) = L(0 L) A(L) = L + 0L calc the ma, shows a verte at (, 0) 7) c. (L, A(L)) = (, 0) when the length is, the width is 0 L =. So, Area = L W ft ft = 0 ft The area of the fenced pla area is 0 square feet. ) Yes, the maimum height of the orange is 19 feet (which is the -coordinate of the verte of the graph). Jim can grab the orange either on the wa up for the orange or on its wa down (eplanations will var). 9) At approimatel 1 miles per hour will the car be able to stop for a sign 0 feet awa. ) a. 11.6 feet b..79 seconds 1) ( + ) ) ( ) ) 7(a + ) ) 1(z 1) ) b(b + 1) 6) r( r) 7) t(9t + 1) ) n(n ) 9) h(h + ) ) 9(1 ) 11) a(a + ) 1) d(d 6) 1) ( + 7)( + 6) 1) ( + )( + ) or ( + ) 1) ( + )( + ) SECTION.A 16) ( + )( ) 17) ( )( ) or ( ) 1) ( )( + ) 19) ( + )( 1) 0) ( )( + ) 1) ( )( ) or ( ) ) (6 1)( 1) ) ( + 1)( ) ) ( + 1)( + 6) ) (a )(a + ) 6) ( )( + ) 7) ( 6)( + 6) ) = ( + 1)( + ) 9) = ( + 7)( 7) 0) The -intercepts have a -coordinate of zero. Setting = 0 and solving for gives the solution(s), which are the -intercept(s). (answers will var) 1) = ( )( 0) (, 0) and (0, 0) ) = ( + )( 1) (1, 0) and (, 0) ) a. An arrow should be drawn to the location that the graph of the height of the ball in relationship to time touches the -ais. b. = 0 means the place where the ball hits the ground (where the height is zero). 1) =, = 9 ) = (double root) ) =, = ) =, = ) =, = 6) = 0, = 7) = 0, = 1) The legs of the triangle have lengths of units and units. ) The legs of the triangle have lengths of units and 1 units. ) The value of is (side lengths are 9 units and 1 units). SECTION.B ) = 1, = 9 9) =, = ) = (double root) 11) = 1, = 1) = 1, = 1) =, = 1) =, = SECTION.C ) The value of is 11 (side lengths are units and 1 units). ) The two numbers that satisf the situation are and 1. 6) The dimensions are 1 feet b feet. 1 1) =, = 1 16) =, = 1 17) =, = 7) The two positive numbers are and. ) The two negative integers are and 6. 9) ( = ) You can frame a ft ft square picture. APPENDICES

Unit Practice Problem ANSWERS 1) a. b. c. d. 6 e. 6 f. ) a. 9 ) a. b. 6+ 11 c. 7 d. e. f. 11+ b. SECTION.D ) a. b. c. d. 1 1 e. 0 f. 6 ) a. 1 + 7 b. 6 + c. + d. + e. 6 f. + 6 6) a. 19 + b. 1 c. 7 6 d. 76 + e. 1 7 f. 7 11 7) a. es, this -value is a solution b. no ( = 6 + ) is not a solution to the equation. c. es, ( = ) is a solution to the equation. ) a. a + b. + 9) c. 7 6 d. + 6 + 7 + 7? 6 = 0 9 + 6 7 + 7 + 7? 6 = 0 16 66 + 6 7 9 7? + 6 = 0 16 + 7 9 7? + 6 = 0 + 7 9 7? + 6 = 0 9 + 7 7? 6 = 0? 6 = 0? 6 6 = 0 0 = 0 1) =, = ( = ± ) ) =, = ( = ± ) ) =, = ( = ± ) ) =, = 7 ) = 7, = 1 6) = 1, = SECTION.E 7) It takes seconds to reach the ground. ) A sign should be posted stating the maimum speed is km per hour. 9) The other leg of the triangle And created is 9 feet long. Appendi B ~ Selected Answers ~ Unit

Unit Practice Problem ANSWERS 1) The error occurs when Omar tried to add to both sides before he takes the square root of both sides. The correct solutions are = 6 or = 0 ) Yes, both of the values given are solutions. ) Yes, both of the values given are solutions. ) Yes, both of the values given are solutions. ) No, neither value given is a solution. SECTION.F 6) Both answers are correct. The balloon could have hit a 6 foot tall student on the wa up after 0.1 seconds, or it could have hit the student on its descending path after 1.1 seconds. 7) = ± ) = 1± 9) = ± ) =, = 11) = ± 1) = 1) =, = 1) = ± 1) t.1seconds 16) ± 6 miles, but onl 6,7.9 miles makes sense. 1) a. 1 b. 1 c. i d. 1 e. i f. i g. i h. 1i i. i 7 j. 1i k. i l. 0i ) a. 9 + i b. 1 6i c. + i d. 11+ i e. 9 i f. 1 19i ) a. 6 b. 1 c. i d. 60i e. i f. 1 ) (all answers in # ma var) a. In method 1, Kasem never used the definition on i = 1 to rewrite either radical. Method uses the definition correctl. b. (1) Method combines 9 = 6 instead of simplifing each separatel like in method. 6 APPENDICES (eplanations will var) SECTION.G ) c. If there is a negative number under a radical, ou must simplif it to include the i s before combining with other radicals. ) a. b. 1 c. 6 6) a. 1i b. 9 + i c. 6 60i d. 1 6i e. + 0i f. + i 7) a. 16 + 0i b. 6 i c. 7 0i ) a. 6 + i b. 6 i 7 c. 7i 11 9) a. Yes ( i ) + 6 = 6 6 = 6 b. Yes ( i ) ( i ) + + + 0 = = 9) c. Yes ( i ) ( i ) ( i ) ( i ) i 1 1 + = 1 1 + 6 + = ( ) ( ) ( ) 1 6i + 9 1 + 6i + = ) a. + i 11) b. + i c. 7i 6 d. 6 + i? + i + i 6 + = 9 + 6i + i? ( + i ) + = ( ) 9 + 6i + 1? 9 i + = 6i? + 1 i = i? + 1 i = + i 1 i =? = =

Unit Practice Problem ANSWERS 1) i ) i ) i ) 0i ) i 6) 11i 7) 1 ) 9 9) ) 1 11) 16 0i 1) 1 + i 1) 0 + i 1) Yes (/) + = 1 1=1 1) = 7, = ) = 0, = ) = 1, = ) From line 1 to line, the person forgot to add to both sides of the equation. ) 9; =, = 1) = 7 ± 1 ) = 9 ± ) = ± ) ± = ) 7 ± 7 = 6) = 1 ± 6 7) = ) 1± = 1) = ± 1i ) = 7 ± i 6 ) = 1 ± i ) = 6 ± i ) = 1 ± 6i 1) Yes SECTION.H 0+/ 1 1= 17 17= 17 16) Yes / +11=7 7=7 17) Yes 0 +/+1 = = 1) =± 19) =1±/ 6 0) =± 1) = 1±/ ) + 1.6 seconds. The balloon lands on the ground after approimatel 1.6 seconds. ( 0.96 does not work when verified). SECTION.I 6) 1; =, = 7) = 1, = 9 ) = 1, = 9) = 6 ) =7,= SECTION.J 1± 9) = ) When > 0, there will be real solutions. When = 0 there will be 1 real solution. 11) = 1. ; The width of the pool is 1 meters and the length is 19 meters. 1) t = 6 ; The arrow will strike the ground after 6 seconds. SECTION.K 6) 1± i 1 = 7) = ± i ) ± i 1 = (#-6, verification of solutions should be shown b student.) ) =± ) = 7 ) =±/ 6) =±/ 6 7) s 7.071 inches; The length of one side of the square is approimatel 7.071 inches. ) =±/. With these solutions being imaginar, it reinforces the reason wh the graph does not cross the -ais. There are no real solutions, there are two comple solutions. 11) = The two numbers that satisf the conditions are and. 1) = The length of each side of the original garden is ards and the area of the garden is square ards. 1) a. h ( ) = ; The bottle rocket is feet high after seconds. b. t 1.1 or t 1.7 ; The rocket will be at 00 feet in the air at two different times; once on the wa up and once on the wa down during its flight. ± i 1 9) = ) = 0 ; Emma hits the golf ball 0 ards. Appendi B ~ Selected Answers ~ Unit 7

Unit Practice Problem ANSWERS SECTION.K (continued) 1 1 11) = or = 1 ; At eactl ½ second and 1½ seconds, the tape measure is at eactl 17 feet above the ground. Gail can catch the tape measure anwhere between ½ second and 1½ seconds after Veronica tosses the tool. 1) When < 0 there will be imaginar solutions for a quadratic equation. 1) Create the quadratic formula: a b c + + = 0 a b c + + 0 = a a b c + + = 0 a a b c + = a a b b c b + + = + a a a a b b ac + = a a a 1) (continued) b b ac + = a a b b ac + = a a + = or + = + a a a a b b ac b b ac = or = + a a a a b b ac b b ac = ± a a b b ac ± = b b ac a 1) If the quadratic equation does not factor, the roots are either irrational or imaginar. (Answers will var) ) a: b: c: 1 = or = solve b factoring? Yes, b ac (9) is a perfect square number. 1) i ) 6 ) i ) = 1, = ) = 6) ± i = 7) 6 ± i 11 = ) = 1± i 9) =, = SECTION.L ) a: 1 b: 7 c: 9 7 ± 1 = solve b factoring? No, b ac (1) is not a perfect square number. ) a: b: c: 1 ± 9 = solve b factoring? No, b ac (9) is not a perfect square number. SECTION.M ) Error on line 6, need to divide both terms of the numerator b the denominator value of ; = ± i 11) = ± 7 1) = 1) 1 ± i 11 = ) a: 1 b: 1 c: 1 = 1± solve b factoring? No, b ac () is not a perfect square number 6) a: 9 b: 6 c: 1 1± = 7) a: b: c: 1 ± 17 = ) 1.6 seconds 9) a. seconds b. 11 seconds 1) a..6 seconds b. Since t = 0 means time when object is initiall thrown, positive time means time after the throw was started. Negative time implies time before the object was thrown, which doesn t make sense here. (Answers will var) c. The object was at a height of feet after 0.7 seconds and again at 1.7 seconds. APPENDICES

Unit Practice Problem ANSWERS 1)» Use a graphing utilit to find real zeros.» Factor and use the zero product propert.» Square root propert» Complete the square» Quadratic formula ) Graph: The solutions are the -intercepts of =, = 1 -----1-1 1 - - - - -6-7 Factor: ( )( + ) = 0 = or = Quadratic Formula: 1± = 1± = = or = Solutions: = or = Method:: The factoring method is the fastest and ields rational solutions with fewer possibilities of careless errors. (eplanations ma var) ) Graph: The solutions are the -intercepts of ±1.7 ) (continued) SECTION.N Square Root: = 0 = = = ± Quadratic Formula: 0 ± 1 = ± = = ± Solutions: = ± Method: The square root method. There is no linear term (the term). The b-value is equal to zero. (Answers will var) ) a. Graphing calculator (but it can t be used to find comple/imaginar solutions) b. Completing the Square and Quadratic Formula c. Factoring or Square Root propert ) Factoring it s a trinomial that factors nicel. (method and eplanation ma var) 1 =, = 6) Factoring it is a binomial with a common factor. (method and eplanation ma var) = 0, = 00 7) Factoring it is a trinomial with a leading coefficient of 1. (method and eplanation ma var) = 9, = ) There are two scenarios for this situation: (1) Numbers and 9 () Numbers and 9 9) width: 7 in. length: in. ) Suzie needs 0 feet of fencing. 11) Mike needs 0 feet of fencing. 1) Solutions: = ± Methods included here ma var Method 1: Square Root + = 9 = 6 = 1 = ± Method : Quadratic Formula 0 ± = ± 1 = = ± Wh? Using square roots has the fewest number of steps and is the easiest. The function is not also factorable, so the quadratic formula was the onl other method to use. 1) Solutions: = ; so the width is m and length is m Method 1:Complete the Square + = 7 ( ) + + = 7 + + + = 900 ( ) + = 900 continue solving from here Note that is etraneous. Method : Factoring + = 7 ( ) + 7 = 0 ( ) ( ) + = 0 continue solving from here Wh? Complete the Square ma be more efficient since " $ was a whole number, and the square root propert is well engrained. Factoring might have taken longer to find integers with a product of 7. (answers/methods ma var) Appendi B ~ Selected Answers ~ Unit 9

Unit Practice Problem ANSWERS SECTION.N (continued) ± i 1 1) = ; solutions should be verified 6 ± 9 1) = ; solutions should be verified 16) = ( ) ( ) + 6 + 9 = 0 0 = 0 i 17) = ± 9 i 1 + 1 = 0 9 0 = 0 and i 1 + 1 = 0 9 0 = 0 1) a. The ball was in the air sec. b. The ball reaches the ma height at. seconds. c. The ma height is 0 feet. d. Graph a second line at = 60 and use calculate intersect to get the - coordinate that creates a function value of 60. The ball is at 60 feet again at approimatel.0 seconds. (1) () (answers will var) Square Root Propert: Eq n #: ( ) = Solutions: = ± (( ) ) ( ) + = (( ) ) ( ) 0 APPENDICES = = and = = = Factoring: Eq n #: 1 = 0 Solutions: = or = ( ) ( ) 1 = 0 1 = 0 0 = 0 and ( ) ( ) 1 = 0 9 + 6 1 = 0 0 = 0 Factoring: Eq n #: + 1 = 0 Solutions: = 6 ± 1 ( ) ( ) 6 + 1 + 1 6 + 1 = 0 and ( ) ( ) 0 = 0 6 1 + 1 6 1 = 0 0 = 0 SECTION.O Quadratic Formula: Eq n #1: + 7 1 = 0 Solution(s): = or = + 7 1 = 0 0 = 0 and ( ) ( ) + 7 1 = 0 0 = 0 Graphing: Eq n #: h t = 16t + t + 7 ( ) h(t) 000 - -00 Solution: t 1.00 It took approimatel 1 seconds for the rock to hit the ground. ( ) ( ) 16 1.00 + 1.00 + 7 = 0 0 = 0 t 6) a. = 7 or = 1 b. graph: c. The intersection points of the graphs in part b) are the solutions for in part a). 9 = (7 ) 9=9 and 9=1 9=9 7) a. The. feet is how far the hose s nozzle is above the ground where the water begins to shoot out. b.. feet c. No. At 7.9 feet from the nozzle, the stream would hit the top of the 6 foot fence. However, at the foot distance, the water height is lower, reaching onl.96 feet high. ) (answers will var) Xmin: 0 Xma: 0 Ymin: 00 Yma: 00

Unit Practice Problem ANSWERS 1) In the quadratic formula, b ± b ac =, the a discriminant is the value of the epression b ac that is under the radical. This number is used to determine the number and tpe of solutions of a quadratic equation. ) discriminant: number of solutions: tpe: comple ) discriminant: 16 number of solutions: tpe: real, rational ) discriminant: 0 number of solutions: 1 tpe: real, rational 1) a. 6 b. c. real, rational d. =, = e. - - f. (1, 9) g. minimum h. (0, ) i. all real numbers j. 9 SECTION.A ) discriminant: 17 number of solutions: tpe: real, irrational 6) discriminant: 0 number of solutions: 1 tpe: real, rational 7) discriminant: number of solutions: tpe: comple ) discriminant: 1 number of solutions: tpe: real, rational 9) discriminant: 7 number of solutions: tpe: comple ) the second line of work, the negative 6 must be in parentheses ( 6) ( 1) ( ) discriminant: 16 # of solutions tpe: real, rational SECTION.B ) a. 16 b. c. real, rational d. = ± e. - - f. (0, ) g. maimum h. (0, ) i. all reals j. 11) a. 0000 = 0.00 + 1 + 7760 b. 0.00 + 1 = 0 c. 90. d. Yes, the discriminant > 0, which ields two real solutions. Onl the positive solution would pertain to this stor. helmets. 1) D = 977. Yes, the discriminant is greater than zero and ields two real solutions. 1) D = 0,000. No, since the discriminant is less than 0 (negative) there are no real solutions. ) D ) discriminant = solutions, real, irrational ) discriminant = 0 1 solution, real, rational 6) discriminant = 11 comple solutions 7) a. negative discriminant b. zero discriminant c. positive discriminant ) Discriminant is a positive perfect square number real, rational solutions Possible equation: ( ) ( 7) = = + 1 (Answers ma var) Appendi B ~ Selected Answers ~ Unit 1

Unit Practice Problem ANSWERS 1) a. Not a sol. b. Yes, a sol. ) a. Yes, a sol. b. Not a sol. ) a. Not a sol. b. Not a sol. ) (test points ma var) Use (0, 0) Is 0? No, so points outside the parabola are solutions. SECTION.A 7) < ) 9) > ) C 11) A 1) F 1) E 1) B 1) D 16) - - - 0) 17) - 1) ) (test points ma var) Use (0, 0) Is 0? Yes, so points outside the parabola are solutions. ) - - -6 - - - - -6 1) - 6 - - 6) (test points ma var) Use (0, 0) Is 0 > 6? No, so graph outside the parabola. 19) - 6 - - ) 6 6-6 - - - -6 - - - APPENDICES

Unit Practice Problem ANSWERS 1) a. When < or > b. < < SECTION.B ) < or > Function values are MORE than 0 above -ais 9) 0 > 0 < or > Function values are MORE than 0 above -ais - ) a. < < b. < or > 6) 0 or ) - 1 + 7 < 0 < < 9 Function values are MORE than 0 above -ais ) Answers will var, but -intercepts and concavit must match. Eample: 7) < < Function values are LESS than 0 below -ais 11) # or ) Answers will var, but -intercepts and concavit must match. Eample: - Function values are LESS than 0 below -ais - ) + + 0 1 or Function values are MORE than 0 above -ais 1) < < -6 6 Function values are LESS than 0 below -ais - Function values are LESS than 0 below -ais Appendi B ~ Selected Answers ~ Unit

Unit Practice Problem ANSWERS SECTION.B (continued) 1) $0 < < $0 The -intercepts of the graph are at (0, 0) and (0, 0) with a verte is at (60, 1600). 1) a. 0.0ft. and 1. ft. The -intercepts are at (.0, 0) and (1., 0) and the verte is at (1., 1.7) 16) d inches. Values in the domain d 0 are etraneous (a rope cannot have a negative diameter). The -intercepts of the graph are at (, 0) and (, 0) with the verte is at (0, 0). The graph of W d intercepts the horizontal line of = 90 at (, 90). The compan will make a profit if the sell the coats for a selling price between $0 and $0. 1) 1 t seconds The curve intercepts the horizontal line h(t) = are at t = 1 and t =, verte is at (, 6) b. No. From 1 feet awa, the height of the ball would be 9. feet, so the ball would go over the top of the net. Also, from the graph and algebraic solution inequalities above, 1 is not in the correct ranges (eplanations ma var). The diameter of the rope would need to be inches or more to support 90 pounds. 17) Drivers with ages between 6.6 ears and 70 ears (inclusive on 70) have a reaction time slower than milliseconds. R 0 0 The ball will be greater than or equal to feet above the ground from 1 second after hitting the ball to seconds after hitting the ball. 0 0 0 0 60 70 APPENDICES

Unit Practice Problem ANSWERS 1) Answers will var a. There is no -term, such as = or if it is in verte form, like ( ) + = 0. b. If it is not factorable and a = 1 and b is even c. a, b, and c are all integers and fairl small numbers d. If it is not factorable and a, b, and c are larger numbers and/or decimals. ) Answers ma var Suggest A, J and K for square root method Suggest C, D, and E for completing the square Suggest B, F and I for factoring Suggest G, H and L for quadratic formula ) Square Root Method [A] = ± [J] = 1± i [K] = 1 or = 1 ) Completing the Square [C] = ± i [D] = ± 6 [E] = or = 6 ) Factoring [B] = or = [F] = or = [I] = UNIT REVIEW 6) Quadratic Formula [G] ± = or 1± [H] ± i = 6 [L] = or = 7) a. = ± b. 1 = 1 or = c. = ± i 7 ± i d. = 1 ) t = seconds 9).6 seconds ) a. Discrim. is negative. There are two comple solutions. b. Discrim. is zero. There is one real, rational solution. c. Discrim. is positive. There are two real solutions. 11) a. Discrim. is 16. There are two comple solutions. b. Discrim. is 61. There are two real, irrational solutions. c. Discrim. is 0. There is one real, rational solution. 1) B 1) < < 1 1) Algebraicall find the -intercepts. Sketch the graph of a parabola that has these -intercepts and opens up if a > 0 or down if a < 0. Also determine dashed or solid line. Identif the -values (critical values) for which the graph lies below the - ais (#1a) or above (or on) the -ais (#1b). For or include the -intercepts in the solution. 1) a. < < 1 Function values less than 0 (below -ais) b. or Function values are 0 (on or above the -ais) 16) Graph B 17) 1 or 1) a. ( ) h t = 16t + t + b. 6 feet c. 6 feet d. seconds e..6 seconds f. No doesn t make sense to talk about negative time Appendi B ~ Selected Answers ~ Unit