Answers to Algebra 2 Unit 2 Practice

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1 LESSON a. A(l ) 0l l. C Answers to Algebra Unit Practice b. The Area of a Rectangle with Perimeter 0 Area (cm ) A(l) Length (cm) c. Yes; the length of a rectangle that has an area of cm is cm. The width is cm. Both and satisf the equation, 0l l. d. cm 0 cm or 0 cm cm e. domain: 0,, 0, (0, 0), {, 0,, 0}; range: 0, A # 00, (1, 00), {A A, 0, A # 00} f. 00 cm ; the graph shows the maimum to be at 00. This occurs when the length is 0 cm. The width will also be 0 cm since the perimeter will be 0 cm when both the length and width are 0 cm. This rectangle is a square.. horses; with 0 ft of fencing, the maimum area of the corral would be 00 sq ft. This is enough space for horses.. 00 sq ft; A(0) 0(0) The maimum value is represented b the -coordinate of the highest point on the graph. l 7. a. ( 7)( 1 ). D b. ( 9)( ) c. ( )( ) or ( ) d. ( 11)( 1 11) e. ( )( 1 ) f. ( )( 1 ) g. ( )( ) h. ( )( 1 ) i. ( 1 )( 1 ) j. ( 1 )( ) 9. Since c is negative, the constant terms in the factored form are different. Since b is positive, the constant term with the greater absolute value will be positive.. No; there are no factors of whose sum is. LESSON a. 1, b., c., d. 1, e., f., 1 g., 1 LESSON h., 1 i. 7, j., 7 01 College Board. All rights reserved. A1 SpringBoard Algebra, Unit Practice

2 1. a A b c. 1 0 d. 1 0 e. 1 0 f g. 1 0 h To solve a quadratic equation b factoring, ou use the Zero Product Propert. First, set the equation equal to 0. Net, factor the equation. Since the equation is equal to 0, one or both of the factors must be equal to 0. Set each factor equal to 0 and solve. 1. a. 1 ft Garden b c. ( 1 1)( ) 0 ft ft d. The solution shows that the width of the path is ft. The solution 1 should be discarded because a negative measure for the path does not make sense. ft 1. C 19. a.,, b., 1 or. c., 0. or.. d. # # e., 1 or. f.,, 1 g., or. h. # # 0. a. l (7 l ) $ 00 b. l 10l # 0 c. (l 0)(l 0) # 0 d. 0 # l # 0, # w # 0 LESSON a. i b. 11i c. i d. i e. i f. 7i g. i h. 0i. A. imaginar ais LESSON or, ; product is positive when the value of results in both factors having the same sign. 17. a. # or $ 0 b.,, 0 i 1i i i real ais 01 College Board. All rights reserved. A SpringBoard Algebra, Unit Practice

3 . a. 1 i b. i c. d. i e. 7i. a. 1 ; (1 ) 0 b. 1 i, i LESSON -. a. 1 1 i b. 1 1 i c. 1 i d. 1 i e. 1 i f. 11i g. i h. 7. a i b i c. 9 d. 1 i e. 0i f. 1 7i g. 7 17i h. 7i 1 1. a. i 1 b. 1 i c. 1 i 9 d. 1 i e. 1 i f. i 1 g. i h. 1 i 9. Answers will var; accept an comple number with an imaginar part of i. Sample answer: i; ( i) ( i) 1 0i. 0. C LESSON - 1. a. ( 1 i)( i) or ( 1 i)( i) b. ( 1 7i)( 7i) c. ( 1 i)( i) d. ( 1 i)( i). a. i, i b. 1 i, 1 i c. i, i d. i, i. D. a. 7i, 7i b. i, i c. i, i d. 9 i, 9 i. i, i ; 9 1 () 1 ( ) ( 1 (i )( (i ) 0 1 i 0, i 0 i, i i, i 01 College Board. All rights reserved. A SpringBoard Algebra, Unit Practice

4 LESSON ; sample eplanation: Step 1: Add 7 to both sides. ( ) 7 Step : Divide both sides b. ( ) 7 Step : Take the square root of 7 both sides. Step : Rationalize the denominator. 1 Step : Add to both sides a.. C b. c. d. 7i e., 1 i f. 9 0 g. h. i a ; ( 1 ) b ; ( 7) 0. a. 17 b. c. d. i LESSON 9-1. b b ac a. a. 1, 1. D b. 1 c. 1 d. 1, e., i 1 f. 0 g. h a. ; completing the square; the variable terms are alread isolated on one side, the coefficient of the -term is 1, and the coefficient of the -term is even. This makes completing the square eas. b., ; factoring; the left side of the equation can easil be factored as ( 1 )( ) c. ; Quadratic Formula; the coefficient of the -term is not 1. This makes the other methods of solving a quadratic equation difficult. d., 1; taking the square root of both sides. When ou add 1 to both sides of the equation, each side is a perfect square. 17. a. t b. about 0. s and 0.7 s after the ball is thrown 01 College Board. All rights reserved. A SpringBoard Algebra, Unit Practice

5 c. Sample answer: Substitute the times from part b into the original equation to check that the make the left side of the equation approimatel equal to. LESSON 9-1(0.) 1 17(0.) (0.7) 1 17(0.7) b ac; the value of the discriminant reveals the number and nature of the roots: one real rational root (discriminant is zero), two real rational roots (discriminant is positive and a perfect square), two real irrational roots (discriminant is positive but not a perfect square), two comple conjugate roots (discriminant is negative). 7. a. 7; since b ac is positive and not a perfect square, there are two real, irrational roots.. B 9. a. 0 b. 7; since b ac is negative, there are two comple conjugate roots. c. 1; since b ac is positive and a perfect square, there are two real, rational roots. d. 0; since b ac is zero, there is one real, rational root. b. sample answer: ( 1) 0. a. The radicand will be positive when b. ac. b. If the radicand is positive, there are two real solutions. c. When the radicand is positive and a perfect square, the solution will be rational. LESSON A. a. ; the directri is horizontal, so the ais of smmetr is a vertical line through the focus. The focus has an -coordinate of, so the ais of smmetr is the line. b. (, ); the verte is the midpoint of the segment that connects the focus and the directri. The endpoints of this segment have coordinates (, ) and (, 1), so the verte has coordinates (, ). c. Opens up; the ais of smmetr is vertical and the focus is above the directri, so the parabola opens up.. a. 1 1 ( 1 ) 1 1 b. ( 1) 1 c. 1 1 d. 1 ( ) 1. verte: (, 1); ais of smmetr: ; focus: (, 0); directri: 01 College Board. All rights reserved. A SpringBoard Algebra, Unit Practice. LESSON -. a. 1 7 b. 1 1 c. 1 d f () 1. B

6 a. (0, ); sample eplanation: For a quadratic function, the ais of smmetr is a vertical line that passes through the verte, so the ais of smmetr is 1. The point (, ) is 1 unit to the right of the ais of smmetr. The point one unit to the left of the ais of smmetr with the same -coordinate as (, ) is (0, ). b.. A c. Rocket B will hit the ground around seconds sooner than Rocket A. Sample eplanation: I set the height of each quadratic model equal to 0 and used the Quadratic Formula to solve for the time. The solutions show that Rocket B will hit the ground after about 1 seconds and Rocket A will hit the ground after about seconds, or about seconds later.. Three. Three noncollinear points determine a parabola, so ou can perform a quadratic regression if ou have at least data points.. a b. Yes; the monthl revenue increases and then decreases as the selling price increases, which indicates a quadratic model could be a good fit for the data. c. Answers will var but should be close to $0. LESSON a. translated units down f() LESSON - 1. a. A quadratic model is a better fit. Sample justification: A graph shows the data points are closer to the quadratic model. The values of first increase and then begin to decrease as increases, which indicates the shape of a quadratic model. b. A linear model is a better fit. Sample justification: The values of increase as increases without ever decreasing, which indicates the shape of a linear model.. a. Rocket A: Rocket B: b. Rocket B. Sample eplanation: Graph both quadratic models on the same coordinate grid. The graphs show that Rocket B reaches a greater height than Rocket A. 01 College Board. All rights reserved. A SpringBoard Algebra, Unit Practice

7 b. translated units to the right f() d. translated units right and units down f() c. translated units left and units up f() 7. a. translated units up; g() 1. B b. translated units left; h() ( 1 ) c. translated 1 unit right and units down; j() ( 1) d. translated units left and units down; k() ( 1 ) 9. (, ). Sample eplanation: The graph of g() is the graph of f() translated units to the left and units down. The verte of f() is (0, 0), so the verte of g() will be units to the left and units down from (0, 0) at (, ) Sample eplanation: The graph of h() is the graph of f() translated units to the right and 1 unit up. The ais of smmetr of f() is 0, so the ais of smmetr of h() will be units to the right of 0 at. 01 College Board. All rights reserved. A7 SpringBoard Algebra, Unit Practice

8 LESSON a. a translation of units up and a reflection over the -ais f() c. a horizontal stretch b a factor of and a translation down units f() b. a vertical stretch b a factor of and a translation down 1 unit f() d. a vertical shrink b a factor of 1, a translation up 1 unit, and a reflection over the -ais f() 7. a. translated units right, stretched verticall b a factor of, and reflected over the -ais, g() ( ) b. translated 1 unit left and verticall shrunk b a A factor of 1, h() 1 ( 1 1) c. horizontall stretched b a factor of and translated up 1 unit, j() ( 1 ) 1 1 d. translated down units and verticall stretched b a factor of, k() 01 College Board. All rights reserved. SpringBoard Algebra, Unit Practice

9 7. Sample answer: The graph of g() 1 is a 7. C vertical shrink of f () b a factor of 1. The graph of is a horizontal stretch of f () b a h() ( 1 ) factor of. 7. (1, ). Sample eplanation: The graph of h() is a translation 1 unit to the left of the graph of f(). This translation moves the verte from (0, 0) to (1, 0). The translation is followed b a vertical stretch b a factor of. This stretch does not change the position of the verte. The stretch is then followed b a translation units up. This moves the verte from (1, 0) to (1, ). LESSON a. g() ( ) 1 ; translated units right and units up g() b. h() ( 1 ) 1; translated units left and 1 unit down h() c. j() ( 1 1) ; verticall stretched b a factor of, and translated 1 unit left and units down j() 01 College Board. All rights reserved. A9 SpringBoard Algebra, Unit Practice

10 d. k() ( ) ; verticall stretched b a factor of, and translated units right and units down k() 77. (, ). Sample eplanation: In the verte form of the equation, the value of h is and the value of k is. 7. a. verte form: h() ( 1 ) 79. A verte: (, ) ais of smmetr: graph opens: up b. verte form: h() ( ) 1 verte: (, ) ais of smmetr: graph opens: up c. verte form: h() ( ) 1 1 verte: (, 1) ais of smmetr: graph opens: down d. verte form: h() ( 1 ) verte: (, ) ais of smmetr: graph opens: up 0. Sal should write 1 in the first bo because adding 1 completes the square for the quadratic epression inside the parentheses. Write in the second bo because subtracting outside the parentheses keeps the epression on the right side of the function balanced. LESSON The value of f () increases for, and decreases for.. Sample eplanation: If the graph opens downward, the verte is the highest point. So, the value of f () increases as the graph moves from left to right toward the verte. The -coordinate of the verte is, so the function increases for,. The value of f () decreases as the graph moves from left to right awa from the verte, so the function decreases for... Sample eplanation: The equation is written in standard form, a 1 b 1 c 0, so I would use the formula b. Substituting 1 for b and a for a, I find that. Substituting for in the original equation, I find that f () is 9 when is. The verte is (, 9).. a. (, ). B b. (7, 17) c. (, 1) d. (1, ). a. G() 1 0 b G() College Board. All rights reserved. A SpringBoard Algebra, Unit Practice 0 0

11 c. (, 0); the -coordinate of represents the width in feet that gives the greatest area. The -coordinate of 0 represents the greatest area in square feet. d. ft b ft; the -coordinate of the verte,, is the width that results in the greatest area. Substituting this value of into the epression for the length gives the length that results in greatest area: 0 () ft. LESSON 1-. a. -intercepts: and -intercept: b. -intercepts: and -intercept: 1 c. -intercepts: and -intercept: 1 d. -intercepts: and -intercept: 7. An -intercept is the -coordinate of a point where a graph intersects the -ais. Quadratic functions can have 0, 1, or -intercepts.. C 9. The graph of a quadratic function has onl one -intercept when the verte of the graph is on the -ais. 90. a. ; the -intercept represents the profit the tour compan would make for selling the tickets for 0 each. The -intercept is negative, which indicates a loss of mone. The tour compan would lose if it gave the tickets awa for free. b. and ; the -intercepts represent selling prices that would result in a profit of 0. The tour compan would break even but make no profit if it were to sell the tickets for or for. c. Reasonable domain: # # ; reasonable range: 0 # # 00. Sample eplanation: A graph of T() shows that the tour compan s profit is greater than or equal to 0 when the selling price is between and, so the reasonable domain is # #. The graph also shows that the maimum value of the tour compan s profit is 00. Because the profit must be greater than or equal to 0, the reasonable range of the function is 0 # # 00. d. ; the graph of T() opens downward and its verte is (, 00). The verte indicates that the touring compan will make a maimum profit of 00 b selling the tickets for. LESSON a. verte: ( 7, 9 ) -intercept: -intercept(s): 1 and ais of smmetr: 7 minimum: 9 f() 01 College Board. All rights reserved. A11 SpringBoard Algebra, Unit Practice

12 b. verte: (1, ) -intercept: -intercept(s): 1 and ais of smmetr: 1 maimum: 9. a. The ticket price of either 1 or would result in a profit of 00. Sample eplanation: Set T() equal to 00; Subtract 00 from both sides to get ( 1)( ). So, 1 or, which means the selling price of 1 and will result in a profit of 00. b. No. The graph shows that the verte of the profit function is (, 00), so the maimum profit the tour compan can earn is 00. To confirm, set f () equal to 00: Subtract 00 from both sides to write the equation in standard form: Use the Quadratic Formula to solve for, which shows that i. Because the equation has comple solutions, there is no real value of that results in a profit of 00. c. 17. Sample eplanation: Evaluate T() for : T() () 1 0() 17. The tour compan can epect to make a profit of 17 if it sells tickets for each. 9. The -intercept occurs when the graph intersects the -ais at 0. Evaluate the function for 0. f (0) (0) 0(0) 1. The -intercept is. 9. D 9. The ais of smmetr is the vertical line through the verte, so the -coordinate of the verte will give ou the ais of smmetr. LESSON 1-9. The value of the discriminant reveals the number of -intercepts of the graph of the function: when the discriminant is zero, there is one -intercept; when the discriminant is positive, there are two -intercepts; when the discriminant is negative, there are no -intercepts. 97. There are two real, irrational solutions and two -intercepts. 9. a. discriminant: 9 two real, irrational solutions -intercepts:. and College Board. All rights reserved. A1 SpringBoard Algebra, Unit Practice

13 b. discriminant: 0 one real, rational solution -intercept: d. discriminant: two real, rational solutions -intercepts: and 1 c. discriminant: 1 two comple conjugate solutions 99. C no real solutions 0. The discriminant is less than zero. no -intercepts LESSON 1-1. a. 01 College Board. All rights reserved. A1 SpringBoard Algebra, Unit Practice

14 b. d. c.. D. B. a. not a solution b. not a solution c. solution d. solution. a. f (DBH) r; sample eplanation: The equation is linear because the diameter is two times the radius, which is a linear measure. b. f (a) # r # 0; sample eplanation: The inequalit is quadratic because the area is a function of the square of the radius. 01 College Board. All rights reserved. A1 SpringBoard Algebra, Unit Practice

15 LESSON 1-1. a. c. b. solutions: (1., 1.) and (.,.) d. solutions: (.,.) and (.,.) solutions: (, 1) and (, ) solution: (, 1) 01 College Board. All rights reserved. A1 SpringBoard Algebra, Unit Practice

16 7. Mari is correct. If the line touches the curve at eactl one point, the sstem will have onl one solution. An eample of a sstem with onl one solution is. ( ). B 9. a. The demand might increase if he lowers wthe price. 1. b. He will onl realize a profit if the demand function is greater than the suppl function. c. The break-even point will be the point where the graphs of the demand and suppl functions intersect LESSON Check students work. Possible answers: The linear equation is solved for. Substitute for in the quadratic equation. Net write the resulting equation in standard form and solve for. Then substitute the value for into the original equation(s) to find the corresponding value of. Substitute the resulting values for both and into each of the original equations to check. 11. a. (, 1), (, ) 11. B b. (, 17) c. (, 9), (, ) d. no real solutions 11. The - and -coordinates of the two solutions are the same. The sstem of equations has one real solution. 11. The graphs of the equations intersect at two points, (, 9) and (, ). The sstem has two real solutions solutions: (7., 77) and (., 1) 01 College Board. All rights reserved. A1 SpringBoard Algebra, Unit Practice