Question 411584

{{{y=-(2/3)x-2}}}..........here, you have a slope {{{m=-(2/3)}}} and the graph 

of this line crosses {{{y-axis}}} at {{{-2}}}


to graph it you need two points


let points be : 


({{{0}}},{{{-2}}}).....your {{{y-intercept}}}...point where {{{x=0}}}


now find your {{{x-intercept}}}...point where {{{y=0}}}


{{{0=-(2/3)x-2}}}


{{{(2/3)x=-2}}}


{{{x=-2/(2/3)}}}


{{{x=-(2/1)/(2/3)}}}


{{{x=-(2*3)/(2*1)}}}


{{{x=-(cross(2)*3)/(cross(2))}}}


{{{x=-3}}}


so, {{{x-intercept}}} is at


({{{-3}}},{{{0}}})..


{{{ graph( 500, 500, -6, 5, -10, 10, -(2/3)x-2) }}}