330 CHAPTER 3 Polynomial and Rational Functions Quadratic Models Because the vertex of a vertical parabola is the highest or lowest point on the graph, equations of the form y = ax2 + bx + c are important in certain problems where we must find the maximum or minimum value of some quantity. • When a 60, the y-coordinate of the vertex gives the maximum value of y. • When a 70, the y-coordinate of the vertex gives the minimum value of y. The x-coordinate of the vertex tells where the maximum or minimum value occurs. If air resistance is neglected, the height s (in feet) of an object projected directly upward from an initial height s0 feet with initial velocity v0 feet per second is given by the quadratic function s1t2 = −16 t2 +v 0 t +s0 , where t is the number of seconds after the object is projected. The coefficient of t2 1that is, -162 is a constant based on the gravitational force of Earth. This constant is different on other surfaces, such as the moon and the other planets. EXAMPLE 5 Solving a Problem Involving Projectile Motion A ball is projected directly upward from an initial height of 100 ft with an initial velocity of 80 ft per sec. (a) Give the function that describes the height of the ball in terms of time t. (b) After how many seconds does the ball reach its maximum height? What is this maximum height? (c) For what interval of time is the height of the ball greater than 160 ft? (d) After how many seconds will the ball hit the ground? ALGEBRAIC SOLUTION (a) Use the projectile height function. s1t2 = -16t2 + v 0 t + s0 Let v 0 = 80 and s0 = 100. s1t2 = -16t2 + 80t + 100 (b) The coefficient of t2 is -16, so the graph of the projectile function is a parabola that opens down. Find the coordinates of the vertex to determine the maximum height and when it occurs. Let a = -16 and b = 80 in the vertex formula. t = - b 2a = - 80 21-162 = 2.5 s1t2 = -16 t2 + 80 t + 100 s12.52 = -1612.522 + 8012.52 + 100 s12.52 = 200 Therefore, after 2.5 sec the ball reaches its maximum height of 200 ft. GRAPHING CALCULATOR SOLUTION (a) Use the projectile height function as in the algebraic solution, with v0 = 80 and s0 = 100. s1t2 = -16t2 + 80 t + 100 (b) Using the capabilities of a calculator, we see in Figure 11 that the vertex coordinates are indeed 12.5, 2002. Be careful not to misinterpret the graph in Figure 11. It does not show the path followed by the ball. It defines height as a function of time. Here x = t and y1 = s1t2. y1 = −16x 2 + 80x + 100 0 300 −70 9.4 Figure 11
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