Question 363165
THe only restrictions are when the denominator equals zero.
{{{x^2+x-12=0}}}
{{{(x+4)(x-3)=0}}}
Two solutions:
{{{x+4=0}}}
{{{highlight(x=-4)}}}
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.
.
{{{x-3=0}}}
{{{highlight(x=3)}}}
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.
Domain:({{{infinity}}},{{{-4}}})U({{{-4}}},{{{3}}})U({{{3}}},{{{infinity}}})