# Thread: graph an equation with x, y, z variables

1. ## graph an equation with x, y, z variables

-x^2 - y^2 - 16z^2= 0

how do i go about graphing this? all i know is it has to do with looking at "sections"

2. Err... unless $x,y,z \in \mathbb{Z}$, there is obviously only one solution to this: $x=y=z=0$ seeing as $x^2, y^2, 16z^2$ are all non-negative numbers...

3. i know thats what i thought that's why i'm so confused i dont know how to graph it

4. Well, I suppose we would be looking at a single point in $(0,0,0)$...

5. its just weird because they are supposed to be like spheres and paraboloids

6. Originally Posted by holly123
what is the general equation for an ellipsoid? i think -x^2-y^2-16z^6=0 might be an ellipsoid. trying to figure out how to graph it
The general equation for an ellipsoid is $\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}+\frac{(z-l)^2}{c^2}=1$, where $(h,k,l)$ is the center of the ellipsoid, and $a>0,\,b>0,\,c>0$.

Unfortunately, the equation $-x^2-y^2-16z^6=0$ isn't an ellipsoid. I'm not even sure what to call it since the degree of z is 6... Are you sure you copied it down right?

7. oh i'm sorry it's z^2!!

8. Originally Posted by Chris L T521
The general equation for an ellipsoid is $\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}+\frac{(z-l)^2}{c^2}=1$, where $(h,k,l)$ is the center of the ellipsoid, and $a>0,\,b>0,\,c>0$.

Unfortunately, the equation $-x^2-y^2-16z^6=0$ isn't an ellipsoid. I'm not even sure what to call it since the degree of z is 6... Are you sure you copied it down right?
Originally Posted by holly123
oh i'm sorry it's z^2!!
Ok, that's a little better. Are you sure its not $-x^2-y^2{\color{red}+}16z^2=0$? Because that would give us an elliptic [circular] cone!

Note that $-x^2-y^2+16z^2=0\implies z^2=\frac{x^2}{16}+\frac{y^2}{16}$, which is an elliptic cone centered at (0,0,0).

The graph would be as follows:

9. i double checked and it's a subtraction sign

10. Originally Posted by holly123
i double checked and it's a subtraction sign
That would cause some problems since $-x^2-y^2-16z^2=0\implies z^2=-\frac{x^2}{16}-\frac{y^2}{16}\implies z=\pm\sqrt{-\frac{x^2}{16}-\frac{y^2}{16}}$, which is imaginary since $x^2>0$ and $y^2>0$.

11. i know i'm so stuck..any suggestions?

12. Originally Posted by holly123
i know i'm so stuck..any suggestions?