0, the slope and y-intercept is 0
Hopes this helps:
Answer: x = 6
This type of sampling would be Random Sampling.
A three-dimensional vector field is conservative if it is also irrotational, i.e. its curl is
![\mathbf 0](https://tex.z-dn.net/?f=%5Cmathbf%200)
. We have
![\nabla\times\mathbf f(x,y,z)=-2e^{-x}\,\mathbf k](https://tex.z-dn.net/?f=%5Cnabla%5Ctimes%5Cmathbf%20f%28x%2Cy%2Cz%29%3D-2e%5E%7B-x%7D%5C%2C%5Cmathbf%20k)
so this vector field is not conservative.
- - -
Another way of determining the same result: We want to find a scalar function
![f(x,y,z)](https://tex.z-dn.net/?f=f%28x%2Cy%2Cz%29)
such that its gradient is equal to the given vector field,
![\mathbf f(x,y,z)](https://tex.z-dn.net/?f=%5Cmathbf%20f%28x%2Cy%2Cz%29)
:
![\nabla f(x,y,z)=\mathbf f(x,y,z)](https://tex.z-dn.net/?f=%5Cnabla%20f%28x%2Cy%2Cz%29%3D%5Cmathbf%20f%28x%2Cy%2Cz%29)
For this to happen, we need to satisfy
![\begin{cases}f_x=ye^{-x}\\f_y=e^{-x}\\f_z=2z\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7Df_x%3Dye%5E%7B-x%7D%5C%5Cf_y%3De%5E%7B-x%7D%5C%5Cf_z%3D2z%5Cend%7Bcases%7D)
From the first equation, integrating with respect to
![x](https://tex.z-dn.net/?f=x)
yields
![f_x=ye^{-x}\implies f(x,y,z)=-ye^{-x}+g(y,z)](https://tex.z-dn.net/?f=f_x%3Dye%5E%7B-x%7D%5Cimplies%20f%28x%2Cy%2Cz%29%3D-ye%5E%7B-x%7D%2Bg%28y%2Cz%29)
Note that
![g](https://tex.z-dn.net/?f=g)
*must* be a function of
![y,z](https://tex.z-dn.net/?f=y%2Cz)
only.
Now differentiate with respect to
![y](https://tex.z-dn.net/?f=y)
and we have
![f_y=-e^{-x}+g_y=e^{-x}\implies g_y=2e^{-x}\implies g(y,z)=2ye^{-x}+\cdots](https://tex.z-dn.net/?f=f_y%3D-e%5E%7B-x%7D%2Bg_y%3De%5E%7B-x%7D%5Cimplies%20g_y%3D2e%5E%7B-x%7D%5Cimplies%20g%28y%2Cz%29%3D2ye%5E%7B-x%7D%2B%5Ccdots)
but this contradicts the assumption that
![g(y,z)](https://tex.z-dn.net/?f=g%28y%2Cz%29)
is independent of
![x](https://tex.z-dn.net/?f=x)
. So, the scalar potential function does not exist, and therefore the vector field is not conservative.