Example

2. Example • Suppose a firework is launched with an upward velocity of about 80 ft/sec from a height of 224 feet. Its height in feet, h(t), after t seconds is given by • h(t) = 16t2 + 80t + 224 • After peaking how long will it be before the shell is 224 feet above ground again? • 1. Introduction. We can make a drawing and label it using the information • provided. Note that the time cannot be negative. 2. Body. The relevant function has been provided. Since we are asked to determine how long it will be before the firework is 224 feet above ground, we are interested in the value of t for which h(t) = 224. 16t2 + 80t + 224 = 224 Solve. Solve algebraically and graphically. 16t2 + 80t + 224 = 224 16t2 + 80t = 0 16t(t  5) = 0 t = 0 or t = 5 The solutions are t = 0 or t = 5.

3. Graphical - Solution The solutions are t = 0 and t = 5. 3. Conclusion. The firework will reach 224 feet 5 seconds after it is launched.

6. Solution 1. Introduction. 2. Body.

7. 3. Conclusion.

8. 3. Conclusion.

9. Example The length of a rectangular piece of sheet metal is 6 inches more than four times the width. A 3 inch square is cut from each corner of the metal, and the sides are bent up so that the metal forms an open box. If the volume of the box is 389 cubic inches, what are the original dimensions of the metal? (The volume of a box is given by the formula V = lwh.) 4x x - 6 Solution 1. Introduction. x = width sheet metal (inches) 4x + 6 = length of sheet metal (inches) Height of box is 3 inches Volume of box is 389 cubic inches x - 6 = width of box (inches) 4x = length of box (inches)

10. 2. Body. Using the formula V = lwh we get the equation Negative answer does not make sense in the context of the problem. Solve graphically. We get The solution is x = 9.436, the length is 4x + 6, so Here is a check: 3. Conclusion. The dimensions of the sheet metal are 9.436 inches by 43.742 inches.