Question

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

Answer 1

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!