Question

Can someone help me ??? 3x^2 -y^2+6x+z=0
I already know that it is a hyperbolic paraboloid but I do not know how to draw it

Answer 1

best thing to do is to start with a center axis, and draw what it would look like if it were projected onto the 3 common planes

Answer 2

(3x^2+6x) -y^2 +z = 0

3(x^2+2x) -y^2 +z = 0

3(x^2+2x+1-1) -y^2 +z = 0

3(x+1)^2 -y^2 +z = 3

Answer 3

when z=0, the xy plane satisfies: 3(x+1)^2 -y^2 = 3 which is a hyperbola shape

when y=0, the zy plane satisfies: 3(x+1)^2 +z = 3 which is a parabola shape

when x=0, the zy plane satisfies: 3 -y^2+z = 3 which is a parabola shape

Answer 4

looking like saddle back to me, but lets see if the wolf can help visualize it better