Question 892421
{{{g(f)=2f+1}}}
{{{g(f)=2(x^2+3x)+1}}}
{{{g(f)=2x^2+6x+1}}}
Complete the square to convert to vertex form,
{{{g(f)=2(x^2+3x)+1}}}
{{{g(f)=2(x^2+3x+9/4)+1-2(9/4)}}}
{{{g(f)=2(x+3/2)^2+2/2-9/2}}}
{{{g(f)=2(x+3/2)^2-7/2}}}
There are no constraints on the domain : ({{{-infinity}}},{{{infinity}}}).
The function has a minimum value of {{{-7/2}}} so the range is 
[{{{-7/2}}},{{{infinity}}})
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.
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{{{graph(300,300,-10,10,-10,10,2x^2+6x+1)}}}