What type of conic section is the following equation?
5x2 - y = 12

The type of conic section for the equation 5x^2 - y = 12 is Parabolas. The "vertex" form of a parabola with its vertex at (h, k) is: y = a(x – h)^2 + k

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Question

Asked 3/25/2014 6:03:42 AM

Updated 3/25/2014 7:47:43 PM

1 Answer/Comment

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The type of conic section for the equation 5x^2 - y = 12 is Parabolas.

The "vertex" form of a parabola with its vertex at (h, k) is:

y = a(x – h)^2 + k

The "vertex" form of a parabola with its vertex at (h, k) is:

y = a(x – h)^2 + k

Added 3/25/2014 7:47:43 PM

This answer has been confirmed as correct and helpful.

Confirmed by jeifunk [3/25/2014 7:51:08 PM], Unconfirmed by jeifunk [3/25/2014 7:51:08 PM], Confirmed by jeifunk [3/25/2014 7:51:10 PM], Unconfirmed by jeifunk [3/25/2014 7:51:10 PM], Confirmed by jeifunk [3/25/2014 7:51:11 PM], Unconfirmed by jeifunk [3/25/2014 7:51:11 PM], Confirmed by jeifunk [3/25/2014 7:51:12 PM]

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1 Answer/Comment

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Updated 4/21/2014 9:22:43 PM

2 Answers/Comments

x^2 + 2x - 1 = 0

a = 1, b = 2, c = -1

b^2 - 4ac = 2^2 - 4(1)(-1) = 8

x = [-2 ± sqrt(8)]/2

= -1 ± sqrt(2)

The solution for the equation x^2 + 2x - 1 = 0 is x = -1 + sqrt(2) or x = -1 - sqrt(2)

a = 1, b = 2, c = -1

b^2 - 4ac = 2^2 - 4(1)(-1) = 8

x = [-2 ± sqrt(8)]/2

= -1 ± sqrt(2)

The solution for the equation x^2 + 2x - 1 = 0 is x = -1 + sqrt(2) or x = -1 - sqrt(2)

Added 4/21/2014 9:19:47 PM

This answer has been confirmed as correct and helpful.

Confirmed by jeifunk [4/21/2014 9:21:29 PM]

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