How Do You Plot Level Curves for the Function z=(x²-2y+6)/(3x²+y)?

[ronho1234]

sketch the level curve z=(x^2-2y+6)/(3x^2+y) at heights z=0 and z=1

i have already compute the 2 equations for the 2 z values and drawn it in 2d but when it comes to plotting it with the extra z axis i don't know what to do. please help...

 

[jacobrhcp]

As your title suggests, what you need is a 'level curve', not a level surface.

Maybe I'm wrong, but if I had to do this exercise, I'd find y(x) for z=1 and z=0 and draw them, as you say have done already. I think you're done already.

 

[HallsofIvy]

There is no "extra z-axis". There is no z-axis at all! The level curves you are asked to draw are in the xy-plane. Graph (x^2-2y+6)/(3x^2+y)= 0 and (x^2-2y+6)/(3x^2+y)= 1.

The first is easy- it is the same as the graph of x^2- 2y+ 6= 0 which is just the parabola y= (1/2)x^2+ 3. The second is not much harder: x^2- 2y+ 6= 4x^2+ y or y= 6- 3x^2, also a parabola.