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messersmith_power_intermediate_algebra_1e_ch4_7_10

Solution a) Because a 1, the graph of f (x) will open downward. Therefore, the vertex will be the highest point on the parabola. The function will attain its maximum value at the vertex. b) Use x b 2a to fi nd the x-coordinate of the vertex. For f (x) x2 4x 2, x b 2a (4) 2(1) 2 The y-coordinate of the vertex is f (2). f (2) (2)2 4(2) 2 4 8 2 6 The vertex is (2, 6). c) f (x) has no minimum value. The maximum value of the function is 6, the y-coordinate of the vertex. (The largest y-value of the function is 6.) We say that the maximum value of the function is 6 and that it occurs at x 2 (the x-coordinate of the vertex). d) From the graph of f (x), we can see that our conclusions in parts a)c) make sense. x 10 10 10 10 (2, 6) YOU TRY 1 Let f(x) x2 6x 7. Repeat parts a)d) from Example 1. 2 Given a Quadratic Function, Solve an Applied Problem Involving a Maximum or Minimum Value EXAMPLE 2 A ball is thrown upward from a height of 24 ft. The height h of the ball (in feet) t sec after the ball is released is given by h(t) 16t2 16t 24. a) How long does it take the ball to reach its maximum height? b) What is the maximum height attained by the ball? Solution a) Begin by understanding what the function h(t) tells us: a 16, so the graph of h would open downward. Therefore, the vertex is the highest point on the parabola. The maximum value of the function occurs at the vertex. The ordered pairs that satisfy h(t) are of the form (t, h(t)). To determine how long it takes the ball to reach its maximum height, we must fi nd the t-coordinate of the vertex. t b 2a 16 2(16) 1 2 The ball will reach its maximum height after 1 2 sec. SECTION 10.6 Applications of Qu www.mhhe.com/messersmith adratic Functions and Graphing Other Parabolas 669


messersmith_power_intermediate_algebra_1e_ch4_7_10
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