Question 1052526
{{{(3x^2+4x-3)/(x^2+3)=(3+4/x-3/x^2)/(1+3/x^2)}}}
In the limit as x goes to {{{infinity}}} and {{{-infinity}}},
{{{g(x)=3}}}
So the horizontal asymptote is {{{y=3}}}.
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Crossing the horizontal asymptote,
{{{(3x^2+4x-3)/(x^2+3)=3}}}
{{{3x^2+4x-3=3x^2+9}}}
{{{4x=12}}}
{{{x=3}}}
Then,
{{{y=(3(3)^2+4(3)-3)/(3^2+3)=(27+12-3)/(9+3)=36/12=3}}}
(3,3)
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*[illustration dfc18.JPG].