Explanation Given Information: The equation of parabola is f ( x ) = − 1 2 ( x − 1 ) 2 + 2 . Graph: Consider the parabola, f ( x ) = − 1 2 ( x − 1 ) 2 + 2 The standard equation of shifted parabola is f ( x ) = a ( x − h ) 2 + k where the vertex of the parabola is ( h , k ) and a tells whether the graph opens up or down and whether it is wider or narrower than the graph of f ( x ) = x 2 . The graph opens up if a > 0 and down if a < 0 . In case 0 < | a | < 1 , the graph of the function will be wider than that of the function f ( x ) = x 2 while for | a | > 1 the graph will be narrower than that of the function f ( x ) = x 2 . Now, rewrite the given function as below, f ( x ) = − 1 2 ( x − 1 ) 2 + 2 Compare the above function with the standard equation of shifted parabola. Therefore, a = − 1 2 h = 1 k = 2 As a < 0 , thus the parabola opens down. Now, the absolute value of a is, | − 1 2 | = 1 2 As 0 < | a | < 1 , thus the graph is wider than that of f ( x ) = x 2 . Now, to graph the function, construct a table with the values of the ordered pairs, by substituting different value of x in the function. For x = 0 , substitute 0 for x, f ( 0 ) = − 1 2 ( 0 − 1 ) 2 + 2 = − 1 2 ( − 1 ) 2 + 2 = − 1 + 4 2 = 3 2 For x = 1 , substitute 1 for x , f ( 1 ) = − 1 2 ( 1 − 1 ) 2 + 2 = − 1 2 ( 0 ) 2 + 2 = 0 + 2 = 2 For x = − 1 , substitute − 1 f

Beginning and Intermediate Algebra (6th Edition)

6th Edition

ISBN: 9780321969163

Author: Margaret L. Lial, John Hornsby, Terry McGinnis

Publisher: PEARSON

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Have you ever seen a bridge in the shape of a parabola? Where is its maximum on the bridge? The maximum of that bridge is the highest point on that bridge. The process of finding the maximum of the bridge here is called the maximization.

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Chapter 11.6, Problem 35E

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To graph: The parabola f(x)=−12(x−1)2+2 and find its vertex, axis of symmetry, domain and range.

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Chapter 11.6, Problem 34E

Chapter 11.6, Problem 36E

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Beginning and Intermediate Algebra (6th Edition)

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To graph: The parabola f ( x ) = − 1 2 ( x − 1 ) 2 + 2 and find its vertex, axis of symmetry, domain and range.

Solution Summary: The author compares the given function with the standard equation of shifted parabola.

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To graph: The parabola f(x)=−12(x−1)2+2 and find its vertex, axis of symmetry, domain and range.

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Chapter 11 Solutions