How do you graph #y=-3x+3# using a table?

1 Answer
Jan 14, 2017

Answer:

See explanation.

Explanation:

Create a blank table with two columns. Label the first column #x#, and the second one #"-"3x+3#, like this:

#[(ul("  "x"  "),ul("-"3x+3)),(,),(,),(,),(,),(,)]#

Next, choose some values of #x# and fill these in the #x# column. (Typically, we choose some values near 0.) Your table should now look something like this:

#[(ul("  "x"  "),ul("-"3x+3)),("-"2,),("-"1,),(0,),(1,),(2,)]#

Then, find the matching values of #"-"3x+3# for each of these #x#-values, and place them in the other column. For example, when #color(red)(x="-"2)#, we have

#color(white)="-"3color(red)x+3#
#="-"3(color(red)("-"2))+3#
#=6+3#
#=9#

Once the table is full, it should look something like this:

#[(ul("  "x"  "),ul("-"3x+3)),("-"2,9),("-"1,6),(0,3),(1,0),(2,"-"3)]#

These are some #(x,y)# pairs that you can plot to help you draw the line represented by the equation #y="-"3x+3#. The plot of these points is below:

graph{((x+2)^2+(y-9)^2-0.025)((x+1)^2+(y-6)^2-0.025)((x)^2+(y-3)^2-0.025)((x-1)^2+(y)^2-0.025)((x-2)^2+(y+3)^2-0.025)=0 [-15.55, 15.63, -4.79, 10.8]}

From here, the line that passes through these points is easy to see. We simply connect the dots with a straight line to finish the job:

graph{(-3x+3-y)((x+2)^2+(y-9)^2-0.025)((x+1)^2+(y-6)^2-0.025)((x)^2+(y-3)^2-0.025)((x-1)^2+(y)^2-0.025)((x-2)^2+(y+3)^2-0.025)=0 [-15.55, 15.63, -4.79, 10.8]}

And we're done!