Answers Investigation 2

Transcription

1 Applications 1. a. Square Side Area Rectangle Length Width Area b. (See Figure 1.) c. The graph and the table both show that the area of the rectangle increases as the area of the square increases. The area of the square is alwas 16 cm 2 greater than the area of the rectangle. This constant difference of the two can be seen on the graph, but the table shows the eact value of the difference. d. Area of the square is A = 2 where is the side length and the area of the new rectangle is A = ( + 4)( - 4) or A = Figure 1 Rectangles and Squares With Equal Perimeters Area Square Rectangle 1

2 e. (See Figure 2 and Figure 3.) c. = ( + ) 2. a. Area (cm 2 ) b. A = ( + ) and A = 2 + O Length (cm) Figure 2 Area of a Square 90 Area (cm 2 ) O Length (cm) Figure 3 Area of a Rectangle 90 Area (cm 2 ) O Length (cm) 2

3 ( - 1) 2 7 Note: In Eercises 4, 6, and 7, students ma reverse the shaded/unshaded portions of the square ( + 6) ( + 10) 9. ( - 6) 10. ( + 11) 11. ( - 2) ( + ) and and ( + )( + ) 1. ( - 4) and (2 + 3) and ( + )( + 6) and a ( - ) b. ( + )( + ) or , which is equivalent to

4 22. a. c. A = ; this equation is quadratic because 2 is the highest power of. = O 1 3 Note: To see the parabola shape we need a window which includes negative values which could not, in practical terms, represent lengths. c. This equation is quadratic because 2 is the highest power of O b. A = ( + )( + 4) and A = Note: To see the parabola shape we need a window that includes negative values which could not, in practical terms, represent lengths or or or , or , or or or or ( + 7)( + 7) = or (3 + 4)(3 + 4) = or

5 34. (3-4)(3-4) = or a b. ( + 3)( + 4) ( + 2)( + ) 36. a. ( + 12)( + 1); = = ( + 12) + 1( + 12) = ( + 12)( + 1) b. ( - 12)( - 1); = = ( - 12) + (-1)( - 12) = ( - 12)( - 1) c. ( + 6)( + 2); = = ( + 6) + 2( + 6) = ( + 6)( + 2) d. ( - 6)( - 2); = = ( - 6) + (-2)( - 6) = ( - 6)( - 2) e. ( + 3)( + 4); = = ( + 3) + 4( + 3) = ( + 3)( + 4) f. ( - 3)( - 4); = = ( - 3) + (-4)( - 3) = ( - 3)( - 4) g. ( + 12)( - 1); = = ( + 12) + (-1)( + 12) = ( + 12)( - 1) h. ( - 12)( + 1); = = ( - 12) + 1( - 12) = ( - 12)( + 1) i. ( + 6)( - 2); = = ( + 6) + (-2)( + 6) = ( + 6)( - 2) j. ( - 6)( + 2); = = ( - 6) + 2( - 6) = ( - 6)( + 2) k. ( + 4)( - 3); = = ( + 4) + (-3)( + 4) = ( + 4)( - 3) l. ( - 4)( + 3); = = ( - 4) + 3( - 4) = ( - 4)( + 3) 37. a. ( + 1)( + 1) = b. ( + )( + ) = c. ( - )( - ) = The pattern is squaring a binomial when the coefficient of is 1, so the pattern has the form, ( + a) 2. The square of a binomial is the square of plus 2(a)() plus the square of a. Smbolicall, this is represented b: ( + a) 2 = ( + a)( + a) = 2 + a + a + a 2 or 2 + 2a + a 2. A similar pattern holds when the coefficient of is not 1. (a + c) 2 = (a + c)(a + c) = (a) 2 + ac + ac + c 2 or (a) 2 + 2ac + c 2. Students eplored this pattern in Problem a. ( + 1)( - 1) = 2-1 b. ( + )( - ) = 2-2

6 c. ( + 1.)( - 1.) = The pattern is multipling the sum and difference of two numbers. The result is the difference of the squares of the two numbers. Smbolicall, this is represented b: ( + a)( - a) = 2 + a - a - a 2 or 2 - a 2. A similar pattern holds when the coefficient of is not 1: (a + c)(a - c) = (a) 2 - c 2. Students eplored this pattern in Problem a = ( + 3) 2 b = ( - 3) 2 c. 2-9 = ( + 3)( - 3) d = ( + 4)( - 4) 40. a = (2 + 3)( + 1) b = (2 + 3)(2-3) c = (2 + 3)(2 + 3) 41. a = ( - 7)( + 7) b = (2-7)(2 + 7) c = ( - 1.2)( + 1.2) 42. Quadratic; since it has an 2 term and this is the highest power of. 43. Not quadratic; it is linear. 44. Quadratic; because it is the product of two linear factors, neither of which is constant. 4. Quadratic; it is the product of two linear factors, neither of which is constant. 46. Not quadratic; it is eponential. 47. Quadratic; since it has an 2 term and this is the highest power of. 4. Quadratic; it is the product of two linear factors, neither of which is constant. 49. Not quadratic; it is linear. 0. Quadratic; since it has an 2 term and this is the highest power of. 1. a. = 2-9; -intercepts: 3 and -3; -intercept: -9; Minimum: (0, -9); Line of smmetr: = 0 b. = 2 + ; -intercepts: 0 and -; -intercept: 0; Minimum: ( - 2, -2 4 ); Line of smmetr: = - 2 c. = ; -intercepts: -3 and -; -intercept: 1; Minimum: ( -4, -1); Line of smmetr: = -4 d. = ; -intercepts: 3 and -; -intercept: -1; Minimum: ( -1, -16); Line of smmetr: = -1 e. = ; -intercepts: -3 and ; -intercept: -1; Minimum: (1, -16); Line of smmetr: = 1 f. = 2-3; -intercepts: 0 and 3; -intercept: 0; Minimum: (1., -2.2); Line of smmetr: = a. = ( + 3)( + 2) b. -intercept: 6; -intercepts: -3 and -2 c. Minimum: ( -2., -0.2) d. = -2. e. The factored form can be useful in predicting the -intercepts and the ais of smmetr. The epanded form can be useful in predicting the -intercept. Students ma have different preferences in equation forms; however, the should be able to justif their choices. 3. a. = ( + )( - ) b. -intercept: -2; -intercepts: - and c. Minimum: (0, -2) d. = 0 e. The factored form can be useful in predicting the -intercepts and the ais of smmetr. The epanded form can be useful in predicting the -intercept. Students ma have different preferences in equation forms; however, the should be able to justif their choices. 6

7 4. a. Students ma choose to draw a rectangle to help them answer this problem. The can represent the area as A = (2 + 3). c. The -intercepts are (0, 0) and ( - 3 2, 0 ). To find the -intercept on a graph ou find the point(s) where the parabola hits the -ais. To determine the -intercepts from the equation, find the values for that make the factors and equal to zero. 3 b. = O 2 2 Connections. a. P = 00 n b. This is an inverse relationship: as the number of friends increases, the amount of mone each person receives decreases, n 7 0. c. A graph would help ou answer questions about how the amount of mone each person receives changes with the number of people sharing the prize. A table would help answer questions about how much mone each person would receive given a specific number of friends. An equation would help answer specific questions about an value of n. d. This relationship is inverse, which can be seen from the graph or the equation. Students investigated inverse relationships in Thinking With Mathematical Models. 6. a. C is the cost for t minutes. Stellar Cellular: C = t, Call An Time: C = 0.9t Cost of Cell Phone Plans Cost $0 $40 $30 $20 $10 $ Minutes Stellar Cellular Call An Time Frogs, Fleas, and Painted Cubes 7 Investigation 2

8 (See Figure 4.) b. Neither of these plans is quadratic. Both are linear. This can be seen in the equations since t is not multiplied b another factor of t in either equation. Both equations are in the linear form = m + b. In the table, ou can see that both have a constant rate of change, which means the are linear. For the Stellar Cellular, the cost increases +1.9 for ever minutes. In the Call An Time plan, the increase is +4.7 ever minutes. Both graphs look like straight lines, so the are not quadratic. c. The plans are equal when the number of minutes is about 2 reading from the table. Solving the equation t = 0.9t for t gives an eact answer of about minutes. 7. a. A = 2(2) or 4 2 b. The area of the new square is 4 times the area of the original square. Students ma choose to make a drawing to help them see this relation between the areas. c. Yes; the angles are still 90, and the ratios of pairs of corresponding sides are 2 : 1. Figure 4 Calls per Minute in Cell Phone Plans Time in Minutes Cost in Dollars Stellar Cellular Call An Time $13.9 $1.90 $4.70 $17. $9.0 $19.0 $14.2 $21.7 $19.00 $23.70 $23.7 $2.6 $2.0 $27.60 $33.2 $29. $3.00 $31.0 $42.7 $33.4 $47.0

9 . a. A = 2( + 1)(2) or b. The area of the new rectangle is 4 times the area of the original rectangle. It can be seen on the drawing below. 1 1 c. Yes; the angles are still 90, and the ratios of pairs of corresponding sides are 2 : a. Recall C = pd, where d is the diameter. So, = pd. Or, we can sa that d = p. b. The radius is one half of the diameter, so radius = 1 2 * p or r = 2p. c. A = pr 2, where r is the radius; A = p ( 2p) 2. d. This is a quadratic relation since the value is squared. e. C = 10 ft, d = 10 p 3.1 feet, r = 10 2p = p 1.9 feet and A = p ( 2p) feet. 60. Rectangle: A = /(10 - /) = 10/ - / 2 and P = / + /(10 - /) + (10 - /) = 20 Parallelogram: Area cannot be determined since ou are not given the height. P = 20. Smmetric Kite: P = 20; area cannot be determined. We can make two triangles b drawing diagonals, but we don t know the bases or heights, so comparing area is not possible. Nonisosceles Trapezoid: Area and perimeter cannot be determined. Area cannot be determined because ou are not given the length of one of the bases or the height. The perimeter cannot be determined because ou are not given the length of the other two sides. Isosceles Right Triangle: Since the triangle is isosceles right, the base is 10 - / and the height is 10 - /. So, A = 1 2 (10 - /)(10 - /) = 0-10/ /2 and P = / + (10 - /) + (10 - /) = 20 - /. 61. a. = b. No; given two points, there is onl one line that ou can draw through them. 62. If = 2, then ( - ) = - 6. If = 3, then ( - ) = If = 1, then = 2. If = 1 3, then = If = 2, then = 1. If = - 4, then = If = - 2, then ( - 7)( + 2) = 0. If = 2, then ( - 7)( + 2) =

10 Etensions 66. C 67. (2 + 1)( + 1) 6. (2 + 3)(2 + 2) 69. = O = a. The graphs have the same size and shape. The are both parabolas, and the both open upward. b. The graphs have different locations on the coordinate plane. The also have different - and -intercepts and lines of smmetr. c. The -intercept for = is (0, 0). For = 2 + 2, it is (0, 2). d. The graph of = has -intercepts of 0 and - 2. It is not possible to find the -intercepts for the equation = because there is no value of that that ou could square and add 2 and get zero. e. Yes; a parabola will alwas cross the -ais. If ou etend the end of the parabola out to the right and out to the left, eventuall it is going to cross the -ais