Question 1151452
If {{{y=x^2-x-12}}}, find the range of the value of {{{x }}}for which {{{y>=0}}}

{{{x^2-x-12>=0}}}..........solve for {{{x}}}

{{{x^2-x-12>=0}}}.......factor

{{{x^2+3x-4x-12>=0}}}

{{{(x^2+3x)-(4x-12)>=0}}}

{{{x(x+3)-4(x-3)>=0}}}

{{{(x - 4)(x + 3)>=0}}}

solutions:

{{{(x - 4)>=0}}} =>{{{x>=4}}}
{{{(x + 3)>=0}}} =>{{{x<=-3}}}

interval:

({{{-infinity}}},{{{-3}}}] U [{{{4}}},{{{infinity}}})


{{{ graph( 600, 600, -10, 10, -10, 10,x^2-x-12, (x - 4)(x + 3)>=0) }}}


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