University:
Montgomery College
Course:
MATH 092: Foundations of Mathematical Reasoning
Academic year:

2022
Views:

184

Pages:

19
Author:

gartelpabv

Read the power point presentation 25-APP-Quadratic Functions Example A fireworks shell is launched into the air. The shell's height (in feet (7) at time (seconds) is modeled well by the function f(t) = -16t² + 200t + 1. When should the shell explode so that it goes off at the maximum height? What is that height? a = -16 b = 200 c = 1 a) Show here the work to answer the questions on this problem. i. When should the shell explode so that it goes off at the maximum height? t = x = -b / 2a = -200 / 2(-16) = -200 / -32 = 6.25 seconds ii. What is that maximum height? h = f(6.25) = -16(6.25)² + 200(6.25) + 1 = -625 + 1250 + 1 = 626 ft b) Now find h(4) and interpret in context. h(4) = -16(4)² + 200(4) + 1 = 545 (4, 545) After 4 seconds the height of the shell is 545 ft. c) Find the h-intercept and interpret in context. h(0) = -16(0)² + 200(0) + 1 = 1 (0, 1) The initial height of the shell is 1 ft. Math 093 and 117 A - Quadratic Functions - Applications 1) A batter hits a baseball into the air. The height h in feet of the baseball after t seconds is given by the quadratic function: h = -16t² + 64t + 3. a) Complete the table of values, plot the ordered pairs and sketch the graph. t h 0 3 1 51 2 67 3 51 4 3 5 -77 [Graph with t on the x-axis and h on the y-axis. A parabola opens downwards, with the vertex at (2, 67). Points from the table are plotted.] b) What is the initial height of the ball? → at t=0 h=3 ft c) How high is the ball after 1.5 seconds? → at t=1.5 h=63 ft d) What is the maximum height of the ball? → y-coord of vertex → 67 ft e) When did it reach the maximum height? → x-coord of vertex → 2 seconds f) When is the ball at a height of 63 feet? I. Solve graphically II. Solve algebraically g) When will the ball hit the ground? h) When is the ball at a height of 80 feet? (c) h = -16(1.5)² + 64(1.5) + 3 = 63 ft (d) x = -b / 2a = -64 / 2(-16) = -64 / -32 = 2 h = -16(2)² + 64(2) + 3 = 67 ft (e) x = -b / 2a = 2 sec (f) ii [rest of solution not shown] (a) x = # of years since 1973 y = med. # of leisure hrs per week (b) y = 0.040x² - 1.215x + 26.026 (c) x-coord of vertex x = -(-1.215) / 2(0.040) = 15.1875 ≈ 15 years after 1973 In 1988 (d) y = 0.040(15.1875)² - 1.215(15.1875) + 26.026 y = 16.8 hours per week [Graph with x-axis labeled "yrs since 1973" and y-axis labeled "Med # of Leisure (hrs per week)". A parabola opens upwards with vertex at approximately (15.2, 16.8)] Interpret Vertex (15.2, 16.8) / week yrs since 1973 The lowest amount of leisure time was 16.8 hrs per week & it happened in 1988.