Y intercept when x = 0
f(x) = (1/2)^0 = 1
so
y intercept (0,1)
answer
C. (0,1)
The tangent to 
through (1, 1, 1) must be perpendicular to the normal vectors to the surfaces 
and 
at that point.
Let 
. Then is the level curve 
. Recall that the gradient vector is perpendicular to level curves; we have
so that the gradient of 
at (1, 1, 1) is
For the surface , we have
so that 
. We can obtain a vector normal to by taking the cross product of the partial derivatives of 
, and evaluating that product for 
:
Now take the cross product of the two normal vectors to and :
The direction of vector (24, 8, -8) is the direction of the tangent line to at (1, 1, 1). We can capture all points on the line containing this vector by scaling it by 
. Then adding (1, 1, 1) shifts this line to the point of tangency on . So the tangent line has equation
Answer - 1/6
As there is one number out of 6 of them
