2. Parabolic Motion Math 112

3. Instructional Objectives…….…..…..3 Driving Question…………………………..4 Maximum Height and Time……….….5 Completing the Square………………..6 Coordinates of Vertex…………….…….9 Graphing Technology:Table……..…10 Graph……………………………....11 Methods to Solve for Height……….12 Transformational Form…………….13 General Form………………………….…15 Quadratic Formula………….…16 Graphing Technology……………..…17

4.  C9 – translate between different forms of quadratic functions  C23 – solve problems involving quadratic equations in a variety of ways  C31 – analyze and describe the characteristics of quadratic functions

5. A baseball is batted into the air and follows the parabolic path defined by the equation h = - 2t 2 +16t+1, where t is time in seconds and h is height in feet. A)What is the maximum height reached by the ball, and at what time does it reach maximum height? B)When is the ball at a height of 25 feet?

6. We will answer question A by finding the vertex. We will focus on two methods to determine the coordinates of the vertex: 1) Write the equation in transformational form by completing the square. 2) Use graphing technology.

7.  Begin with the equation in general form as it was given. h = - 2t 2 + 16t + 1 Step 1: Move all constants to the left side. h – 1 = - 2t 2 + 16t Step 2 : Factor out the coefficient of t 2 from both terms involving t variable h – 1 = - 2(t 2 - 8t)

8. Step 3: Determine what value must be added in to create a perfect square on the right side. - 2(t 2 - 8t +16) Step 4: Balance the equation by adding an equivalent amount to the left side. h – 1- 32 = - 2(t 2 - 8t + 16) Divide the factored “b” term by 2 and square it

9. Step 5: Factor the perfect square. h - 33 = -2(t - 4) 2 Step 6: Rearrange the formula into transformational form. -1/2(h - 33) = (t - 4) 2

10.  The vertex which occurs at a maximum here has coordinates (4, 33)  4 is in brackets with t, therefore represents time when the ball is at maximum height.  33 is in brackets with h, therefore represents the maximum height. Therefore the ball reaches a maximum height of 33 feet at 4 seconds.

11. Because we know the equation is quadratic and we see symmetry in the table, then we can be sure that the vertex has coordinates (4, 33). 4 is an x-value which represents time, and 33 is a y-value which represents height. Therefore the ball reaches a maximum height of 33 feet at 4 seconds. 1. Insert the equation into [Y=] 2. Generate a table of values to determine the greatest y-value, and its corresponding x-value. XY 115 225 331 433 531

12. Therefore the ball reaches a maximum height of 33 feet at 4 seconds. 1. Insert the equation y = - 2x 2 + 16x + 1 into [Y=] 2. Adjust the window settings so the graph can be viewed on the screen. Then press GRAPH.Let’s do it together! 3. Use the CALCULATE feature to set the left and right boundaries for the maximum. The calculator will generate the coordinates for the maximum. Let’s try.

13. Transformational Form General Form Graphing Calculator

14. -1/2(h - 33) = (t - 4) 2 Step 1: Substitute h as 25 and simplify the left side of the equation. -1/2(25 - 33) = (t - 4) 2 -1/2(-8) = (t - 4) 2 4 = (t - 4) 2 Step 2: Square root both sides. +/- 2 = t – 4 Step 3: Add 4 to both sides. +/- 2 + 4 = t

15. +2 + 4 = t 6 = t The ball is at a height of 25 feet at 6 seconds. -2 + 4 = t 2 = t The ball is at a height of 25 feet at 2 seconds Does it make sense that there are two answers to this question? Explain.

16. h = - 2t 2 +16t+1 Because we want to know the time the ball is at a height of 25 feet, we must:  substitute 25 as h,  set the left side equal to zero, and  solve using the quadratic formula. 25 = - 2t 2 +16t+1 0 = - 2t 2 + 16t - 24 For a closer look how to apply the quadratic formula, let’s view this short explanation.

17. Substitute the values of a, b, and c into the formula above. Again there are two solutions: x = (-16+8)/(-4)ORx = (-16-8)/(-4) x = (-8)/(-4)x = (-24)/(-4) x = 2x = 6 Therefore the ball reaches a height of 25 feet at 2 seconds and at 6 seconds. a = -2 b = 16 c = -24

18. We can use the table feature in much the same way we did to determine the coordinates of the vertex. 1.Write the equation in [Y =] 2.Generate the table. 3.Locate all y-values of 25 in the table. 4.Read the corresponding x-values when y=25. In the table we see the ordered pair (2, 25) and (6, 25), therefore the ball reaches a height of 25 feet at 2 seconds and 6 seconds. XY 225 331 433 531 625