How do you solve #x^2+6x>=0# using a sign chart?

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Answer 1 · 2018-02-18T12:30:00

Solution: # x <= -6 and x >=0 # . In interval notation:

#x| (-oo,-6]uu [0,oo)#

#x^2+6x>=0 or x(x+6) >=0#

Critical points are #x=0 and x=-6 #

Sign chart:

When #x <=-6 # sign of #f(x)# is #(-)(-)=(+) ; >=0#

When #-6 < x <0 # sign of #f(x)# is #(-)(+)=(-) ; <0#

When # x >=0 # sign of #f(x)# is #(+)(+)=(+) ; >=0#

Solution: # x <= -6 and x >=0 # . In interval notation:

#x| (-oo,-6]uu [0,oo)#

graph{x^2+6x [-20, 20, -10, 10]} [Ans]