How do you solve $y^3-2y^2-9y+18>=0$ using a sign chart?

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Answer 1 · 2018-08-04T14:42:17

Solution: $-3 <= y <= 2 and y>=3 or [-3,2] uu [3,oo)$

$f(y)= y^3-2 y^2-9 y+18>=0$

$y^3-2 y^2-9 y+18$

$= y^2(y-2) -9( y-2)$

$= (y-2)(y^2 -9) =(y-2)(y+3)(y-3) :.$ zeros are

$y=-3, y=2 , y=3:. f(-3)=0 , f(2)=0 , f(3)=0$

$f(y)=(y-2)(y+3)(y-3)$

Sign chart:

When, $y< -3$ , sign of $f(y)$ is $(-) * (-)* (-) = (-) ; < 0$

When, $-3 < y < 2$ , sign of $f(y)$ is $(-) * (+) *(-) = (+) ; > 0$

When, $2 < y < 3$ , sign of $f(y)$ is $(+) * (+)* (-) = (-) ; < 0$

When $y> 3$ , sign of $f(y)$ is $(+) * (+)* (+) = (+) ; > 0$

Solution: $-3 <= y <= 2 and y>=3 or [-3,2] uu [3,oo)$

graph{x^3-2 x^2-9 x+18 [-80, 80, -40, 40]} [Ans]

How do you solve y^3-2y^2-9y+18>=0y^3-2y^2-9y+18>=0 using a sign chart?