331 3.1 Quadratic Functions and Models (c) If we graph y1 = -16x 2 + 80x + 100 and y 2 = 160, as shown in Figures 12 and 13, and locate the two points of intersection, we find that the x-coordinates for these points are approximately 0.92 and 4.08. Therefore, between 0.92 sec and 4.08 sec, y1 is greater than y2, and the ball is greater than 160 ft above the ground. (c) We must solve the related quadratic inequality. -16t2 + 80t + 100 7160 -16t2 + 80t - 60 70 Subtract 160. 4t2 - 20t + 15 60 Divide by -4. Reverse the inequality symbol. Use the quadratic formula to find the solutions of 4t2 - 20t + 15 = 0. t = -1-202 {21-2022 - 41421152 2142 t = 5 - 210 2 ≈0.92 or t = 5 + 210 2 ≈4.08 These numbers divide the number line into three intervals: 1-∞, 0.922, 10.92, 4.082, and 14.08, ∞2. Using a test value from each interval shows that values in the interval 10.92, 4.082 satisfy the inequality. The ball is greater than 160 ft above the ground between 0.92 sec and 4.08 sec. (d) The height is 0 when the ball hits the ground. We use the quadratic formula to find the positive solution of the equation -16t2 + 80t + 100 = 0. Here, a = -16, b = 80, and c = 100. t = -80 {2802 - 41-16211002 21-162 t ≈ -1.04 or t ≈6.04 Reject The ball hits the ground after about 6.04 sec. (d) Figure 14 shows that the x-intercept of the graph of y = -16x2 + 80x + 100 in the given window is approximately 16.04, 02, which means that the ball hits the ground after about 6.04 sec. S Now Try Exercise 61. Here a = 4, b = -20, and c = 15. −6 0 −70 300 9.4 y2 = 160 y1 =−16x 2 + 80x + 100 Figure 12 −6 0 −70 300 9.4 y2 = 160 y1 =−16x 2 + 80x + 100 Figure 13 0 −70 300 9.4 y1 =−16x 2 + 80x + 100 Figure 14 EXAMPLE 6 Modeling the Number of Hospital Outpatient Visits The number of hospital outpatient visits (in millions) for selected years is shown in the table. Year Visits Year Visits 3 563.2 10 651.4 4 571.6 11 656.1 5 584.4 12 675.0 6 599.6 13 678.0 7 603.3 14 693.1 8 624.1 15 722.1 9 642.0 16 747.1 Data from American Hospital Association.
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