Question 23102
If the slope is {{{-2/3}}}, and it passes through the x-intercept at (3,0), then 
y=mx+b
{{{0=-2/3(3) + b}}}
{{{0 = -2 + b}}}
(((2=b}}}, which is the y intercept.


The equation of the line is therefore {{{y= (-2/3)x+2}}} and the graph looks like this:

{{{graph (300, 300, -10, 10, -10, 10, (-2/3)x+2) }}}

R^2 at SCC