GEORGIA MATHEMATICS 1 - Carnegie Learning

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End of Chapter Test PAGE 4Read the scenario below. Use the scenario to answer Questions 11 through 15.Tanya’s jump rope forms the shape of a parabola as she jumps. The shape can be modeled bythe equation y 2 (x 3)(x 3) , where x is the number of feet to the right of Tanya’s waist,3and y is the height of the jump rope above Tanya’s waist as she swings the rope over her head.11. Create a graph of the quadratic function on the grid below. First, choose your bounds andintervals. Be sure to label your graph clearly.Answers may vary. Sample answers are provided below.Variable quantity Lower bound Upper bound Interval3Horizontal distanceHeight–3.5 4.0 0.5–1.0 6.5 0.5Jumping Ropey5.55.0y = − 2 (x + 3)(x − 3)34.54.0Height (feet)3.53.02.52.01.51.00.5−3.0 −2.5 −2.0 −1.5 −1.0 −0.5 0.5 1.0 1.5 2.0 2.5 3.0 3.5−0.5Horizontal distance (feet)x© 2008 Carnegie Learning, Inc.12. What is the y-intercept of the graph? What does it represent in the problem situation?Use complete sentences in your answer.Sample Answer: The y-intercept is 6. It is the height of the jump rope above Tanya’s waist,as she swings the rope over her head.240 Chapter 3 ■ Assessments Georgia Mathematics 1
End of Chapter Test PAGE 5Name ___________________________________________________Date _____________________13. What are the x-intercepts of the graph? What do they represent in the problem situation?Use complete sentences in your answer.The x-intercepts are –3 and 3. They represent the horizontal position of the ends of thejump rope which are level with Tanya’s waist.14. What is the horizontal distance between the ends of the jump rope? Use a complete sentencein your answer.The horizontal distance between the ends of the jump rope is 6 feet.15. What are the zeroes of this function? Use complete sentences to explain your reasoning.The solutions to the equation are –3 and 3, because when x = –3, then (x + 3) = 0 and zeromultiplied by any number is zero. When x = 3, then (x – 3) = 0 and zero multiplied by anynumber is zero.316. What are the solutions to the equation x 2 95? Give your answers in radical form, anduse a complete sentence in your answer.The solutions are 95 and 95.Use the following scenario for Questions 17 through 20.© 2008 Carnegie Learning, Inc.Jude builds a water balloon launcher that launches the balloon straight up. The motion of aballoon that is released by the launcher can be modeled by using the vertical motion modely 16t 2 vt h, where t is the time that the object has been moving in seconds, v isthe initial velocity of the object in feet per second, h is the initial height of the object in feet,and y is the height of the object in feet at time t seconds.17. The launcher is designed to launch the water balloon from a height of 2.75 feet at aninitial velocity of 20 feet per second. Write a quadratic function that models the heightof the water balloon in terms of time.y 16t 2 20t 2.75Georgia Mathematics 1 Chapter 3 ■ Assessments 241
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Page 54: 5. y x 2 3x 5x 32(1) 1.5;y ( 3

GEORGIA MATHEMATICS 1 - Carnegie Learning

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