Explanation Given Information: The equation of the parabola is y = − 2 ( x + 1 ) 2 − 4 . Consider the equation, y = − 2 ( x + 1 ) 2 − 4 The parabola having equation y = a ( x + h ) 2 + k will have vertical axis of symmetry. So that, the provided equation has vertical axis of symmetry. Now the parabola having equation of the form y = a ( x − h ) 2 + k , then if a > 0 the parabola will open upward and if a < 0 parabola opens downward. So, rewrite the provided equation as y = − 2 ( x + 1 ) 2 − 4 and compare to find a = − 2 . Therefore, as a < 0 the parabola will open downward. Now consider the provided equation y = − 2 ( x + 1 ) 2 − 4 to make a required table to draw a graph, For x = − 2 , y = − 2 ( − 2 + 1 ) 2 − 4 = − 2 − 4 = − 6 Therefore, the coordinate is ( − 2 , − 6 ) . For x = − 1 , y = − 2 ( − 1 + 1 ) 2 − 4 = 0 − 4 = − 4 Therefore, the coordinate is ( − 1 , − 4 ) . For x = 0 . y = − 2 ( 0 + 1 ) 2 − 4 = − 2 − 4 = − 6 Therefore, the coordinate is ( 0 , − 6 )

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ISBN: 9781264367702

Author: Miller

Publisher: MCGRAW-HILL HIGHER EDUCATION

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Chapter 13.2, Problem 40PE

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The parabola y=−2(x+1)2−4 has axis vertical or horizontal axis of symmetry and then whether it opens upward, downward, left or right.

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Chapter 13.2, Problem 39PE

Chapter 13.2, Problem 41PE

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Chapter 13 Solutions

BEGINNING AND INTERMEDIATE AGEBRA

Ch. 13.1 - Find the distance between the points ( − 4 , − 2 )...Ch. 13.1 - Prob. 2SP Ch. 13.1 - Prob. 3SP Ch. 13.1 - Prob. 4SP Ch. 13.1 - Prob. 5SP Ch. 13.1 - Prob. 6SP Ch. 13.1 - Prob. 7SP Ch. 13.1 - Prob. 8SP Ch. 13.1 - Prob. 1PE Ch. 13.1 - Prob. 2PE

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The parabola y = − 2 ( x + 1 ) 2 − 4 has axis vertical or horizontal axis of symmetry and then whether it opens upward, downward, left or right.

Solution Summary: The author explains that the parabola y=-2(x+1)2-4 has axis vertical or horizontal symmetry and will open upward, downward, left or right.

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The parabola y=−2(x+1)2−4 has axis vertical or horizontal axis of symmetry and then whether it opens upward, downward, left or right.

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Chapter 13 Solutions