Sketching this Plane f(x,y) = y + 2x

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SUMMARY

The discussion centers on the challenges of sketching the plane defined by the equation f(x,y) = y + 2x. The user attempts to find intercepts by setting z values and plotting corresponding lines, specifically y = z - 2x for z values of 0 and 2. The confusion arises from the realization that lines with the same slope can have different y-intercepts, leading to difficulty in visualizing the three-dimensional plane. The correct approach involves understanding that the plane extends vertically along the z-axis, representing all values of z for the given lines.

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Saladsamurai
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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}<br /> {| 1 | c | } <br /> \hline z & y = z - 2x \\ <br /> \hline 0 & y = -2x \\ <br /> \hline 1 & y = 1 - 2x \\<br /> \hline 2 & y = 2 - 2x \\ <br /> \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.
 
Physics news on Phys.org
plot the line y = 2x in the xy plane (z=0), then z can take any value so the plane will bethe vertical extension of that line along the z axis.
 

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