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

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: 3x^2+10x+4y+11=0 It's a parabola; and we have to find the range      Log On


   

Question 861474: 3x^2+10x+4y+11=0
It's a parabola; and we have to find the range
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
3x%5E2%2B10x%2B4y%2B11=0+
4y=-3x%5E2-10x-11
4y=-3%28x%5E2%2B%2810%2F3%29x%29-11
4y=-3%28x%5E2%2B%2810%2F3%29x%2B25%2F9%29%29-11%2B3%2825%2F9%29
4y=-3%28x%2B%285%2F3%29%29%5E2-33%2F3%2B25%2F3
4y=-3%28x%2B%285%2F3%29%29%5E2-8%2F3
y=-%283%2F4%29%28x%2B%285%2F3%29%29%5E2-2%2F3
So the vertex occurs at (-5%2F3,-2%2F3) and the parabola opens downwards.
The maximum is y=-2%2F3
Range : (-infinity,-2%2F3]