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so you want the gradiennt to be <3,2,2>

F(x,y,z)=3x^2 -4y^2 -z = 0

dF/dx = 6x
dF/dy = -8y
dF/dz = -1

that is as far as i have gotten

6x = -3/2 ; x = -1/4
-8y = -1 ; y = -1/8
-1 = -1 ; z=-1

ok cool. thanks alot! i had no idea where to start

lets work this back in again to see if it gets us something good :)

-1=3((-1/4)^2) -4((-1/8)^2).. is it 3x^2 or 3(x^2)?

3(x^2)

been loooking at these to long...that the same thing lol

three x squared

3/16 -4/64 = -1?

what is that?

dunno, just a thought really

i think the point is (2,1,8)
8=3(2^2) -4(1^2)
8 = 12-4
8=8 seems better

how did you get that?

cant tell......

ok, so you are sure we want the gradient to be the normal vector?

well, the gradient and the normal are the same thing...

they would have to equal each other.....

suppose we scale the gradient fo match the normal; does that sound like a good thing to do?

sure, lets try it

if we scale it by -2; that will make the -1 = 2

yeah

i got the same thing

but that gets me the same results for x and y; <-1/4,1/8>

lets plug those in then and see what we get for a z i guess...

got a little hyper on that one lol

haha, so we leave z as a variable?

z is dependant on x and y right? and it will be the same ragardless if the combo we use for that

ok that makes sense

what does z look like...it seems hyperbolic to me....

its a saddle shape graph

z=1/8

the graph looks like we should only have 1 answer

...... you sure?

the vector can be positioned anywhere...its not anchored into anthing

im not sure, i just think it looks like that

checking it out on wolframs graphing thing

i have a different graphing software im using

well lets see yours :)

ok, let me see if i can send it in a word document

ok, well i need to go. Thanks alot for working with me on this problem!

the equation of the tangent plane is:
3(x+1/4) +2(y-1/8)+2(z-1/8)=0