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

Question

1632 views around the world

You can reuse this answer
Creative Commons License

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]