Asked by 95
Parabola:
Rewrite 3y^2 + 6y + 108x - 969=0 in standard form, show work.
Rewrite 3y^2 + 6y + 108x - 969=0 in standard form, show work.
Answers
Answered by
Reiny
3(y^2 + 2y + ...) - 969 = -108x
3(y^2 + 2y + 1) - 969 - 3 = -108x
3(y+1)^2 - 972 = -108x
(-1/36) (y+1)^2 + 9 = x
I will assume you realize that this is a "horizontal" parabola and that you know how to read the vertex from that form.
3(y^2 + 2y + 1) - 969 - 3 = -108x
3(y+1)^2 - 972 = -108x
(-1/36) (y+1)^2 + 9 = x
I will assume you realize that this is a "horizontal" parabola and that you know how to read the vertex from that form.
Answered by
Reiny
Just realized that Steve had already answered this question before you reposted it
http://www.jiskha.com/display.cgi?id=1324054249
Have patience and always check if your question has been answered before re-posting it.
It saves unnecessary work on our part.
http://www.jiskha.com/display.cgi?id=1324054249
Have patience and always check if your question has been answered before re-posting it.
It saves unnecessary work on our part.
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