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

Jan 14, 2017

See explanation.

#### Explanation:

Create a blank table with two columns. Label the first column $x$, and the second one $\text{-} 3 x + 3$, like this:

$\left[\left(\underline{\text{ "x" "),ul("-} 3 x + 3}\right) , \left(,\right) , \left(,\right) , \left(,\right) , \left(,\right) , \left(,\right)\right]$

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 $\text{-} 3 x + 3$ for each of these $x$-values, and place them in the other column. For example, when $\textcolor{red}{x = \text{-} 2}$, we have

$\textcolor{w h i t e}{=} \text{-} 3 \textcolor{red}{x} + 3$
="-"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 $\left(x , y\right)$ pairs that you can plot to help you draw the line represented by the equation $y = \text{-} 3 x + 3$. 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!