Problem Solving with i Quadratic and Variation Function Models ^OBJECTIVES ACTIVITY 4.1 The Amazing Property of Gravity ':\. Evaluate functions of the form y = ax 2. f'2. Graph functions of the form y = ax. 3. Interpret the coordinates of points on the graph of y = ax 2 in context. 4. Solve an equation of the form ax 2 = c graphically. 5. Solve an equation of the form ax 2 = c algebraically by taking square roots. Note: a + 0 in objectives 1-5. In the sixteenth century, scientists such as Galileo were experimenting with the physical laws of gravity. In a remarkable discovery, they learned that if the effects of air resistance are neglected, any two objects dropped from a height above earth will fall at exactly the same speed. That is, if you drop a feather and a brick down a tube whose air has been removed, the feather and brick will fall at the same speed. Surprisingly, the function that models the distance fallen by such an object in terms of elapsed time is a very simple one: 5 = 16/ 2, where t represents the number of seconds elapsed and s represents distance (in feet) the object has fallen., The function defined by s = I6t 2 operations: indicates a sequence of two mathematical Start with a value for t» square the value» multiply by 16 > to obtain values for s. 1. Use the given equation to complete the table. \ 0 1 2 3 2. a. How many feet does the object fall one second after being dropped? b. How many feet does the object fall two'seconds after being dropped? 3. a. Determine the average rate of change of distance fallen from time t = 0 tor = 1. 399
400 CHAPTER 4 PROBLEM SOLVING WITH QUADRATIC AND VARIATION FUNCTION MODELS b. What are the units of measurement of the average rate of change? / c. Explain what the average rate of change indicates about the falling object. 4. Determine and interpret the average rate of change of distance fallen from time t = 1 to t = 2. 5. Is-the function s = 16/ a linear function? Explain your answer. 6. a. If the object hits the ground after 5 seconds, determine the practical domain \ of the function. b. On the following grid, plot the points given in the table in Problem 1 andj sketch a curve representing the distance function through the points. 400 f 350 «22 300 â 250 8 200 B 2 150 S 10 50 0»-r 1 2 3 4 5 Time (in seconds) c. Use your graphing calculator to verify the graph in part b. Your graph should resemble the one below. WINDOW Xnin=0 Xnax=5 Xscl=.5 Vnin=0 Yriax=400 Vscl=50 Xres-ll
ACTIVITY 4.1 THE AMAZING PROPERTY OF GRAVITY 401 7. a. Confirm that the point (2.5, 100) lies on the graph. What do the coordinates of this point indicate about the falling object? b. Confirm that the point (4.5, 324) lies on the graph. What do the coordinates of this point indicate about the falling object? 8. Use the graph to estimate the amount of time it takes the object to fall 256 feet. 9. Use s = I6t 2 to write an equation to determine the amount of time it takes the object to fall 256 feet. To solve the equation in Problem 9, you need to reverse the order of operations indicated by the function rule, replacing each operation by its inverse. Here, one of the operations is "square a number." The inverse of squaring is to take a square root. Start with a value for Í > divide by 16 > take its square root > to obtain t. In particular, if s = 256 feet: Start with 256 > divide by 16 r take its square root > to obtain /. 256-16 = 16 VÎ6 = 4 t = 4 Therefore, yoii can conclude that it takes four seconds for the object to fall 256 feet. 10. Reverse the sequence of operations indicated by s = I6t to determine the amount of time it takes an object to fall 1296 feet, approximately the height of a 100-story building. /
402 CHAPTER 4 PROBLEM SOLVING WITH QUADRATIC AND VARIATION FUNCTION MODELS Graph of a Parabola Some interesting properties of the function defined by s = 16í 2 arise when yonl ignore the falling object context and consider just the algebraic rule itself. Replace t with x and s with y and consider the general equation y = \6x 2. First, byf ignoring the context, you can allow x to take on a negative, positive, or zero value 1 For example, suppose x = 5. Then y = l6{-5) 2 = 16 25 = 400. 11. a. Use y = 16.x to complete the table. -3-2.5-2.r-1.5 -j -0,5 0 0 0.5 1 16-1.5 2 64 2.5 3 144 b. What pattern (symmetry) do you notice from the table? 12. a. Sketch the graph of y = 16.x by using the table in Problem 11. Plot the] points and then draw a curve through them. Scale the axes appropriately. b. Use a graphing calculator to produce a graph of this function in the window -5 < x < 5, -100 <y< 400. Your graph should resemble the one belowi WINDOW Xnin=-5 Xmax=5 Xscl=.5 Y"min=-100 Vnax=480 Vscl=50 Xres=l V, / X - i.u.2 The U-shaped graph of y = 16x is called a parabola.
ACTIVITY 4.1 THE AMAZING PROPERTY OF GRAVITY 403 13. a. The graph of the squaring function defined by y = x is a parabola. How is the graph of y = 16x related to the graph of y = x 2? b. The graphs of y = 16x and y = x 2 are both U-shaped. Is the graph of y = 16x" wider or narrower than the graph of y = x 2? Explain. c. Is the graph of y = 0.5* wider or narrower than the graph of y = x 2? Explain. Verify using a graphing calculator. d. If 0 < a < 1, then is the graph of y = ax~ wider or narrower than the graph of y = x? e. If a > 1, then is the graph of y = ax wider or narrower than the graph of y -x 2? Solving Equations of the Form ax 2 = c, a = 0 14. a. In the table in Problem 11a, how many points on the graph of y = 16.x 2 lie 256 units above the x-axis? b. Identify the points. What are their coordinates? The x-values of the points on the graph of y = 16x 2 that lie 256 units above the "i x-axis can be determined algebraically by solving the equation 256 = 16x. Example 1 Solve the equation 16x 2 = 256 algebraically. SOLUTION Step 1. Divide both sides by 16: 16x 2 256 16 16 to obtain x2 = 16 f
ACTIVITY 4.1 THE AMAZING PROPERTY OF GRAVITY 405 17. a. Refer to the graph in Problem 12 to determine how many points on the graph of y = 16x 2 lie 16 units below the x-axis. b. Set up an equation that corresponds to the question in part a. c. How many solutions does this equation have? Explain. 3 18. What does the graph of y = 16x 2 (Problem 12) indicate about the number of solutions to the following equations? (You do not need to solve these equations.) a. 16x 2-100 b. 16x 2 = 0 c. 16x 2 = -96 19. Solve the following equations: a. 5x 2 = 20 b. 4x 2 = 0 c. 3x 2 = -12 SUMMARY Activity 4.1 1. The graph of a function of the form y = ax 2, a = 0, is a U-shaped curve and is called a parabola. 2. If a > 0, then the larger the value of a, the narrower the graph of y = ax 2. 3. An equation of the form ax 2 = c, a + 0, is solved algebraically by dividing both sides of the equation by a and then taking the positive and negative square roots of both sides. EXERCISES Activity 4.1 1. On the Earth's Moon, gravity is only one-sixth as strong as it is on Earth, so an object on the Moon will fall one-sixth the distance it would fall on Earth in the same time. This means that the gravity distance function for a falling object on the Moon is s = 16\ 2 8 2 where r/represents time since the object is released, in seconds, and 5 is the distance fallen, in ft.