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3. When the trace cursor is moving upward the particle is moving to the right, and when the cursor is moving downward the particle is moving to the left. Again we find the same values of t for when the particle reverses direction. 4. We can represent the velocity by graphing the parametric equations x 4 ( t ) 5 x 1 9 ( t ) 5 12 t 2 2 32 t 1 15, y 4 ( t ) 5 2 (part 1), x 5 ( t ) 5 x 1 9 ( t ) 5 12 t 2 2 32 t 1 15, y 5 ( t ) 5 t (part 2), x 6 ( t ) 5 t , y 6 ( t ) 5 x 1 9 ( t ) 5 12 t 2 2 32 t 1 15 (part 3) [ 2 8, 20] by [ 2 3, 5] ( x 4 , y 4 ) [ 2 8, 20] by [ 2 3, 5] ( x 5 , y 5 ) [ 2 2, 5] by [ 2 10, 20] ( x 6 , y 6 ) For ( x 4 , y 4 ) and ( x 5 , y 5 ), the particle is moving to the right when the x -coordinate of the graph (velocity) is positive, moving to the left when the x -coordinate of the graph (velocity) is negative, and is stopped when the x -coordinate of the graph (velocity) is 0. For ( x 6 , y 6 ), the particle is moving to the right when the y -coordinate of the graph (velocity) is positive, moving to the left when the y -coordinate of the graph (velocity) is negative, and is stopped when the y -coordinate of the graph (velocity) is 0. Exploration 3 Seeing Motion on a Graphing Calculator 1. Let t Min 5 0 and t Max 5 10. 2. Since the rock achieves a maximum height of 400 feet, set y Max to be slightly greater than 400, for example y Max 5 420. 4. The grapher proceeds with constant increments of t (time), so pixels appear on the screen at regular time intervals. When the rock is moving more slowly, the pixels appear closer together. When the rock is moving faster, the pixels appear farther apart. We observe faster motion when the pixels are farther apart. Quick Review 3.4 1. The coefficient of x 2 is negative, so the parabola opens downward. Graphical support: [ 2 1, 9] by [ 2 300, 200] 2. The y -intercept is f (0) 52 256. See the solution to Exercise 1 for graphical support. 3. The x -intercepts occur when f ( x ) 5 0. 2 16 x 2 1 160 x 2 256 5 0 2 16( x 2 2 10 x 1 16) 5 0 2 16( x 2 2)( x 2 8) 5 0 x 5 2 or x 5 8 The x -intercepts are 2 and 8. See the solution to Exercise 1 for graphical support. 4. Since f ( x ) 16( x 2 2 10 x 1 16) 16( x 2 2 10 x 1 25 2 9) 16( x 2 5) 2 1 144, the range is ( 2‘ , 144]. See the solution to Exercise 1 for graphical support. 5. Since f ( x ) 16( x 2 2 10 x 1 16) 16( x 2 2 10 x 1 25 2 9) 16( x 2 5) 2 1 144, the vertex is at (5, 144). See the solution to Exercise 1 for graphical support. 6. f ( x ) 5 80 2 16 x 2 1 160 x 2 256 5 80 2 16 x 2 1 160 x 2 336 5 0 2 16( x 2 2 10 x 1 21) 5 0 2 16( x 2 3)( x 2 7) 5 0 x 5 3 or x 5 7 f ( x ) 5 80 at x 5 3 and at x 5 7. See the solution to Exercise 1 for graphical support. 7. } d d y x } 5 100 2 32 x 1 160 5 100 60 5 32 x x 5 } 1 8 5 } } d d y x } 5 100 at x 5 } 1 8 5 } Graphical support: the graph of NDER f ( x ) is shown. [ 2 1, 9] by [ 2 200, 200] 8. } d d y x } . 0 2 32 x 1 160 . 0 2 32 x . 2 160 x , 5 } d d y x } . 0 when x , 5. See the solution to Exercise 7 for graphical support. Section 3.4 81

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9. Note that f 9 ( x ) 52 32 x 1 160. lim h 0 } f (3 1 h h ) 2 f (3) }5 f 9 (3) 32(3) 1 160 5 64 For graphical support, use the graph shown in the solution to Exercise 7 and observe that NDER ( f ( x ), 3) < 64.
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