Explanation Calculation: Consider the equation of the parabola, − 12 − x 2 = 12 y + 4 x Separate x terms on one side of the equation. − 12 − 12 y = x 2 + 4 x x 2 + 4 x = − 12 − 12 y Complete the square, x 2 + 4 x + 4 = − 12 − 12 y + 4 x + 2 2 = − 12 y − 8 Write the above equation in the standard form of the parabola, x − h 2 = 4 p y − k . x + 2 2 = − 12 y − 8 x + 2 2 = − 12 y + 8 12 x + 2 2 = 4 − 3 y + 2 3 Compare the above equation with the standard form of the parabola, h = − 2 , k = − 2 3 and p = − 3 The axis of the parabola is vertical and since, p is negative, the parabola opens downwards

College Algebra Real Mathematics Real People Edition 7

18th Edition

ISBN: 9781305752368

Author: Ron Larson

Publisher: CENGAGE C

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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 4.4, Problem 34E

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To calculate: The vertex, focus and directrix of the parabola, −12−x2=12y+4x and then, plot the parabola.

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Chapter 4.4, Problem 33E

Chapter 4.4, Problem 35E

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Chapter 4 Solutions

College Algebra Real Mathematics Real People Edition 7

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The vertex, focus and directrix of the parabola, − 12 − x 2 = 12 y + 4 x and then, plot the parabola.

Solution Summary: The author calculates the vertex, focus, and directrix of the parabola and plots it.

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To calculate: The vertex, focus and directrix of the parabola, −12−x2=12y+4x and then, plot the parabola.

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Chapter 4 Solutions