1. The problem statement, all variables and given/known data Sketch the level curve of the surface [tex]z = \frac{x^2 - 2y + 6}{3x^2 + y}[/tex] belonging to height z = 1 indicating the points at which the curves cut the y−axis. 2. Relevant equations 3. The attempt at a solution I put [tex]1 = \frac{x^2 - 2y + 6}{3x^2 + y}[/tex] but then don't know how to proceed. The answer shows an inverted parabola at y = 2, but I don't know how to get that.
From [tex] 1 = \frac{x^2 - 2y + 6}{3x^2 + y} [/tex] it follows that [tex] 3x^2 + y = x^2 - 2y + 6 [/tex] Can you cast this in the form y = f(x) ?
Do you mean like this... [tex]3x^2 + y = x^2 - 2y + 6[/tex] [tex]3y = -2x^2 + 6[/tex] [tex]y = \frac{-2x^2 + 6}{3}[/tex] [tex]y = \frac{-2x^2}{3} + 2[/tex] ???
Yes. Now, that's a parabola, isn't it? It's of the general form y = ax^2 + bx + c with a = -2/3, b = 0, c = 2; you can see that it is inverted (like a mountain top) because a < 0.