Question 1205693


If a line has an {{{x}}}-intercept at {{{x=-3}}} we have a point

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

 and if a {{{y}}}-intercept at {{{y=2}}} => {{{b=2}}} and we have a point

({{{0}}},{{{2}}})


now we can find a slope

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


 equation in the form 

{{{y=mx+b}}}...substitute {{{m}}} and  {{{b}}}

{{{y=(2/3)x+2}}}


{{{ graph( 600, 600, -10, 10, -10, 10, (2/3)x+2) }}}