# Sketching level curves of f(x,y)

 P: 48 1. The problem statement, all variables and given/known data Sketch the level curve of the surface $$z = \frac{x^2 - 2y + 6}{3x^2 + y}$$ 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 $$1 = \frac{x^2 - 2y + 6}{3x^2 + y}$$ 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.
 Sci Advisor HW Helper P: 4,300 From $$1 = \frac{x^2 - 2y + 6}{3x^2 + y}$$ it follows that $$3x^2 + y = x^2 - 2y + 6$$ Can you cast this in the form y = f(x) ?
 P: 48 Do you mean like this... $$3x^2 + y = x^2 - 2y + 6$$ $$3y = -2x^2 + 6$$ $$y = \frac{-2x^2 + 6}{3}$$ $$y = \frac{-2x^2}{3} + 2$$ ???