(a) For z=1 you have gone wrong somewhere. You should have and with y=0 .
For your sketch, maybe you could draw the sphere inside a cube like this (x-1)^2+(y+2)^2+z^2=2 - Wolfram|Alpha.
Here is the given question:
(a) Sketch the level curves of z = (x^2 - 2y +6)/(3x^2 + y) at heights z = 0 and z =1.
(b) Sketch the surface (x−1)^2 + (y+2)^2 + z^2 = 2 in R^3. Write down a point which is on the surface.
(a) I set z = 0 and then sketched the level curve. I got a positive parabola, intersecting the y-axis at 3 (no roots). Is that correct?
For z = 1, I got a negative parabola, intersecting the y-axis at 2, with roots -2 and 2. Is that correct? Also, would I draw each level curve on separate axes or on the same one?
(b) I set two variables, x = 2, y = -2 and therefore, z = 1. I then substituted them into the equation and with it equal to 2, so (2, -2, 1) as a point on the surface of this sphere. Moreover, center is (1, -2, 0) and the radius is √2. However, I am confused with sketching the curve. How would I arrange the axes to sketch the sphere? Since y = -2, it exists on the negative y axis. & I assume you would arrange the axes (appropriately and correctly) so that it forms a cube for the sphere to be within, right? That, I'm unsure of how to do as would all the axes still meet at the origin?
Sorry, I do not understand. I checked x^2 - 2y + 6 = 3x^2 + y with wolfram alpha and I get the same parabola: negative, intersects at y = 2 with roots x = -2, 2. So, what is the error then? Moreover, what do you mean with y = 0 2x^2 = 6?
& With the sketch of the surface, would it be correct to sketch the sphere like that, showing the center (1, -2, 0) and a point on the surface (2, -2, 1)?
Look more closely at the graph.
x^2-2y+6=3x^2+y, x=3^.5, y=0 - Wolfram|Alpha
Okay, so does the parabola still intersect at y = 2 but the roots are not x = 2, -2 but x = √3, -√3 instead? Is the level curve for z = 0 correct though?