SOLUTION: 3x^2+10x+4y+11=0 It's a parabola; and we have to find the range

SOLUTION: 3x^2+10x+4y+11=0 It's a parabola; and we have to find the range

Algebra.Com

Question 861474: 3x^2+10x+4y+11=0
It's a parabola; and we have to find the range
Answer by Fombitz(32388) (Show Source): You can put this solution on YOUR website!

So the vertex occurs at (,) and the parabola opens downwards.
The maximum is
Range : (,]

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