Question 153957
I'm confused. 
You say there are no parentheses around -1/3 but you show them.
Is it 
{{{y=-(1/3)x+2}}}
or
is it
{{{y=-1/(3x)+2}}}
I'll graph them both and you can choose the one you need. 
Let's start with
{{{y=-(1/3)x+2}}}
To graph, pick a range of x that you want.
I'll let x vary from -3 to 3 in steps of 1 and calculate y values.
{{{y(-3)=-(1/3)(-3)+2=3}}}
{{{y(-2)=-(1/3)(-2)+2=8/3}}}
{{{y(-1)=-(1/3)(-1)+2=7/3}}}
{{{y(0)=-(1/3)(0)+2=2}}}
{{{y(1)=-(1/3)(1)+2=5/3}}}
{{{y(2)=-(1/3)(2)+2=4/3}}}
{{{y(3)=-(1/3)(3)+2=1}}}
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Now plot the x,y pairs.

{{{drawing( 300, 300, -4, 4, -4, 4,grid( 1 ),
circle( -3, 3, .1 ),
circle( -2, 8/3, .1 ),
circle( -1, 7/3, .1 ),
circle( 0, 2, .1 ),
circle( 1, 5/3, .1 ),
circle( 2, 4/3, .1 ),
circle( 3, 1, .1 ))}}}
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Then connect the points with a straight line. 
{{{drawing( 300, 300, -4, 4, -4, 4,grid( 1 ),circle( -3, 3, .1 ),
green(line(-6,4,6,0)),
circle( -2, 8/3, .1 ),
circle( -1, 7/3, .1 ),
circle( 0, 2, .1 ),
circle( 1, 5/3, .1 ),
circle( 2, 4/3, .1 ),
circle( 3, 1, .1 ))}}}
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Next, it's the same procedure, just a different function.
{{{y(-3)=-1/(3(-3))+2=19/9}}}
{{{y(-2)=-1/(3(-2))+2=13/6}}}
{{{y(-1)=-1/(3(-1))+2=7/3}}}
{{{y(0)=-1/(0)+2}}} This value is undefined
{{{y(1)=-1/(3(1))+2=5/3}}}
{{{y(2)=-1/(3(2))+2=11/6}}}
{{{y(3)=-1/(3(3))+2=17/9}}}

{{{drawing( 300, 300, -4, 4, -4, 4,grid( 1 ),
circle( -3, 19/9, .1 ),
circle( -2, 13/6, .1 ),
circle( -1, 7/3, .1 ),

circle( 1, 5/3, .1 ),
circle( 2, 11/6, .1 ),
circle( 3, 17/9, .1 ))}}}
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The function is not linear and goes off to {{{infinity}}} as x approaches 0 from the left and goes off to {{{-infinity}}} as x approaches 0 from the right. 
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{{{drawing( 300, 300, -4, 4, -4, 4,grid( 1 ),
circle( -3, 19/9, .1 ),
circle( -2, 13/6, .1 ),
circle( -1, 7/3, .1 ),

circle( 1, 5/3, .1 ),
circle( 2, 11/6, .1 ),
circle( 3, 17/9, .1 ), graph( 300, 300, -4, 4, -4, 4, -1/(3x)+2)) }}}
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To get better definition of the function, you would use more points. I shortcutted(cheated) and used the graphing software.