Question 988697
I assume the first one is 3r^2-6R+12
at r=0 y=12
The vertex is -b/2a; r=1, from -(-6/6) That is the smallest value y will reach.
at r=1; y=3-6+12=9.
r has no bounds.  Domain is all reals.  Range is r> or equal to 9
{{{graph(300,300,-10,10,-10,20,3x^2-6x+12)}}}
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second can be done by graphing.
{{{graph(300,200,-10,10,-10,10,(1/x)-4)}}}
Domain is all x except x=0. Range is all y except y=4.
Note, if the problem is y=1/(x-4), I will just show the graph.  It shifts it, but you get the same type of answer, only shifted.
{{{graph(300,200,-10,10,-10,10,1/(x-4))}}}
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7/-x^2
x can be anything but zero
x^2 can never be negative, and -x^2 therefore can never be positive. The range is y (-oo,0).  Y can never equal zero but approaches it asymptotically,
{{{graph(300,300,-10,10,-10,10,-7/x^2)}}}