Explanation Given: The equation of the parabola x = 1 2 ( y − 2 ) 2 + 4 . Graph: Consider the equation of the parabola x = 1 2 ( y − 2 ) 2 + 4 which is of the form x = y 2 . Then, the provided equation gives a horizontal parabola symmetric about the x − axis. Next, plot the points. For y = 0 , x = 1 2 ( y − 2 ) 2 + 4 = 1 2 ⋅ ( − 2 ) 2 + 4 = 2 + 4 = 6 For y = 2 , x = 1 2 ( y − 2 ) 2 + 4 = 1 2 ⋅ ( 2 − 2 ) 2 + 4 = 0 + 4 = 4 For y = 4 , x = 1 2 ( y − 2 ) 2 + 4 = 1

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ISBN: 9780134188515

Author: Crawford

Publisher: YUZU

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Maximization

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 9, Problem 8RP

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To graph: The parabola of the equation x=12(y−2)2+4 and label its vertex and plot at least one intercept.

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Chapter 9, Problem 7RP

Chapter 9, Problem 9RP

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Chapter 9 Solutions

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To graph: The parabola of the equation x = 1 2 ( y − 2 ) 2 + 4 and label its vertex and plot at least one intercept.

Solution Summary: The author graphs the parabola of the equation x=12(y-2) 2+4 and labels its vertex and plots its intercept.

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To graph: The parabola of the equation x=12(y−2)2+4 and label its vertex and plot at least one intercept.

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Chapter 9 Solutions