- #1
Saladsamurai
- 3,020
- 7
Homework Statement
I am the worst human being ever right now. I cannot believe how much trouble I am having trying to sketch this by hand. I am used to just setting the coordinates equal to zero for 2 of them and finding the coordinate intercepts of the third; but that fails in particular case.
I tried to set up a table like so:
[tex]\begin{center} \begin{tabular}
{| 1 | c | }
\hline z & y = z - 2x \\
\hline 0 & y = -2x \\
\hline 1 & y = 1 - 2x \\
\hline 2 & y = 2 - 2x \\
\hline \end{tabular} \end{center}
[/tex]
So I thought that "at" each value of z I could plot the line y = z - 2x and get something that works...but I cannot seem to make it happen.
For example: at z = 0, I would plot y = -2x, which is a line whose slope is -2 and whose y-intercept is 0.
at z = 2, I would plot y = 2 - 2x, which is the line whose slope is -2 and whose y-intercept is at 2.
It is just freaking me out that I have 2 lines with the same slope, yet they intercept y at different points.
Maybe someone can unconfuse me; but then again, maybe not.