How do you solve #-s^2+4s-6<0#?

Question

1154 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-30T12:45:39

Solution: # x in RR # or #(-oo,oo) #

# -s^2 +4s -6 < 0 a= -1 ,b = 4 , c= -6 #

Vertex #(x) = -b/2a= -4/-2= 2# .

Vertex #(y) = - 2^2+4*-2 -6= -4+8-6 =-2#

Vertex is # 2,-2 # . The parabola opens down wards since #a <0#

The range is #< -2 # i.e #<0# . The domain is any real value

i.e # x in RR # or #(-oo,oo)# . The graph also shows

#- s^2 +4s -6 # is # < -2 # i.e # < 0# .

Solution: # x in RR # or #(-oo,oo) #

graph{-x^2+4x-6 [-12.66, 12.65, -6.33, 6.33]}