Looking at the equation, it is apparent that it would follow a linear model as there are no exponents or powers (eg #x^2#)
Then what we can do if find to points on the graph and draw a line through these points. The easiest points to find are usually the #y"-intercept"# and #x"-intercept"#
The #y"-intercept"# is when #x=0#, so to find that point, make #x=0# in the equation, and solve for #y#
#x=0#
#=>-2(0)+y=2#
#=>y=2#
Hence, when #x=0#, #y=2#. This represents the point #(0,2)#. So we should plot this point:
Then we calculate the other point, the #x"-intercept"#
The #x"-intercept"# is when #y=0#, so to find that point, make #y=0# in the equation, and solve for #x#
#y=0#
#=>-2x+(0)=2#
#=>-2x=2#
#=>x=2/-2#
#=>x=-1#
Hence, when #y=0#, #x=-1#. This represents the point #(-1,0)#. So we should plot this point with the other point:
Now join the two points with a straight line:
And you have the graph of your equation
You can use any two poinst on the line though. The only difference is that instead of finding when #x=0# or #y=0#, you would find when #x="another number like "3# or when #y="another number like "4#
It is easier to find the #y"-intercept"# and #x"-intercept"# but if you want to challenge yourself, find two points other than the #y"-intercept"# and #x"-intercept"#
