<PRE><B>Question 662763</B>
the {{{slope-intercept}}} form is: {{{y=mx+b}}}  where {{{m}}} is a slope and {{{b}}} is a {{{y-intercept}}}


the slope {{{m}}} and {{{y-intercept}}} {{{b}}} of {{{y-1=(-2/3)x}}}} will be:


{{{y-1=(-2/3)x}}}}..solve for {{{y}}}

{{{y=(-2/3)x+1}}}}.....the {{{slope-intercept}}} form 

so, {{{m=(-2/3)}}} and {{{b=1}}}

negative slope shows us that a graph goes through II and IV quadrant, and {{{b=1}}} shows us that a graph intercepts {{{y-axis}}} at {{{y=1}}}

{{{ graph( 600, 600, -10, 10, -10, 10, (-2/3)x+1) }}}

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