Solve for x^2+3x-18=0 and the resulting values for x form the reference points for 2 sets of values (domains -> input)
Set x^2+3x-18=0
Note that 3xx6=18" and that "6-3=3 giving:
(x+6)(x-3)=x^2+3x-18=0
So for this condition x=-6" and "x=+3
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As the x^2 term is positive the graph is of form uu. So we are looking for all those values of x that relate to the plot where it is cuts and is above the x-axis. In other words they will be 'traveling' away from the origin in both the positive and negative direction.
x<=-6" and "x>=3
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The set x such that it is all values from and including -6 to approaching -oo
The set x such that it is all values from and including +3 to approaching +oo
{x:x in [3" to "+oo)}
{x:x in [-6" to "-oo)}
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