Question 354090
how to find range of f(x)=x(square)-3x+2/(x+1) algebraically?
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f(x) = x^2-[(3x+2)/(x+1)]
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Range???
x^2 is always >=0
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(3x+2)/(x+1) has a vertical asymptote at x=-1 so all Real Numbers
are subtracted from x^2.
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Graph:
{{{graph(400,300,-10,10,-10,10,x^2-((3x+2)/(x+1)))}}}
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Range is "All Real Numbers"
Cheers,
Stan H.